NCERT Solutions Class 12 Physics Chapter 9

Q.1 A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

Ans.

Given, size of candle, h₁ = 2.5 cm
Object distance, u = −27 cm
Radius of curvature of the concave mirror, R = −36 cm

$\begin{array}{l}\text{Focal length of the concave mirror\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}as:}\\ \text{f =}\frac{\text{R}}{\text{2}}\text{=}\frac{\text{-36}}{\text{2}}\text{\hspace{0.17em}cm=-18\hspace{0.33em}cm\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\end{array}$

Let size of image = h₂
Let image distance = v
From the mirror formula, the image distance can be obtained as:

$\begin{array}{l}\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\\ \therefore \frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}}\\ =\frac{1}{-18\mathrm{cm}}+\frac{1}{27\mathrm{cm}}=\frac{-3+2}{54\mathrm{cm}}=-\frac{1}{54\mathrm{cm}}\\ \therefore \text{Image\hspace{0.33em}distance},\mathrm{v}=-54\text{\hspace{0.33em}cm}\end{array}$

Therefore, in order to obtain sharp image of the candle on the screen, the screen should be placed 54 cm away from the mirror.
Magnification of the image is given as:

$\begin{array}{l}\text{m=}\frac{{\text{-h}}_{\text{2}}}{{\text{h}}_{\text{1}}}\\ \text{=}\frac{\text{-v}}{\text{-u}}\text{=}\frac{\text{v}}{\text{u}}\\ \therefore \frac{{\text{-h}}_{\text{2}}}{\text{2.5 cm}}\text{=}\frac{\left(\text{-54 cm}\right)}{\left(\text{-27 cm}\right)}\text{= 2}\\ \therefore {\text{Size\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}image,\hspace{0.33em}h}}_{\text{2}}\text{=-5\hspace{0.33em}cm}\\ \therefore \text{The\hspace{0.33em}image\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}candle\hspace{0.33em}is\hspace{0.33em}real,\hspace{0.33em}inverted\hspace{0.33em}and\hspace{0.33em}magnified.}\end{array}$

If the candle is moved closer to the mirror, then the screen should be moved away from the mirror in order to obtain the image on the screen.

Q.2 A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Ans.

$\begin{array}{l}\begin{array}{l}{\text{Height of the needle, h}}_{\text{1}}\text{= 4.5 cm}\\ \text{Object distance, u = -12 cm}\\ \text{Focal length of the convex mirror, f = 15 cm}\\ \text{Let image distance = v}\\ \text{From the mirror formula, the image distance can be obtained as:}\end{array}\\ \frac{\text{1}}{\text{v}}\text{+}\frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{f}}\\ \therefore \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{f}}\text{–}\frac{\text{1}}{\text{u}}\\ \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{15 cm}}\text{+}\frac{\text{1}}{\text{12 cm}}\\ \frac{\text{1}}{\text{v}}\text{=}\frac{\text{4 + 5}}{\text{60 cm}}\text{=}\frac{\text{9}}{\text{60 cm}}\\ \therefore \text{Image\hspace{0.33em}distance,\hspace{0.33em}v =}\frac{\text{60 cm}}{\text{9}}\text{= 6.7\hspace{0.33em}cm}\\ \text{Therefore, the virtual image is formed at 6.7 cm behind the mirror.}\\ \text{The magnification of the image is given as:}\\ \text{m =}\frac{{\text{h}}_{\text{2}}}{{\text{h}}_{\text{1}}}\text{= –}\frac{\text{v}}{\text{u}}\\ \therefore \frac{{\text{h}}_{\text{2}}}{\text{4.5 cm}}\text{= –}\frac{\text{6.7 cm}}{\text{12 cm}}\\ \therefore {\text{h}}_{\text{2}}\text{= –}\frac{\text{6.7}}{\text{12}}\text{× 4.5 cm = 2.5\hspace{0.33em}cm}\\ \therefore {\text{Size\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}image, h}}_{\text{2}}\text{= 2.5\hspace{0.33em}cm}\\ \text{m\hspace{0.17em} = \hspace{0.17em}}\frac{{\text{h}}_{\text{2}}}{{\text{h}}_{\text{1}}}\text{=}\frac{\text{\hspace{0.17em}2.5 cm}}{\text{4.5 cm}}\text{= 0.56}\\ \therefore \text{The\hspace{0.33em}image\hspace{0.33em}formed\hspace{0.33em}is\hspace{0.33em}virtual\hspace{0.33em}and\hspace{0.33em}erect.}\\ \text{If the needle is moved farther from the mirror, then the image moves away from the mirror up}\\ \text{to its principal focus and its magnification goes on decreasing.}\end{array}$

Q.3 A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

Ans.

Given, actual depth of the needle in water,

h₁ = 12.5 cm

Apparent depth of the needle in water,

h₂ = 9.4 cm

Let refractive index of water = μ

The refractive index of water is given as:

\begin{array}{l}\mu =\frac{\text{Real}\text{depth}}{\text{Apparent}\text{depth}}=\frac{12.5}{9.4}\\ =1.33\\ \text{For}\text{the}\text{second}\text{case:}\\ \text{Refractive}\text{index}\text{of}\text{the}\text{liquid},\mu =1.63\\ \text{Real}\text{depth}\text{of}\text{needle}=12.5cm\\ \text{Let}\text{apparent}\text{depth}=y\\ \text{The}\text{relation}\text{for}\text{refractive}\text{index}\text{is}\text{given}\text{as:}\\ \mu =\frac{\text{Real}\text{depth}}{\text{Apparent}\text{depth}}\\ \therefore 1.63=\frac{12.5}{y}\\ \therefore y=\frac{12.5}{1.63}\\ =7.67\text{cm}\end{array}

Therefore, new apparent depth of the needle =7.67 cm

Therefore, the distance by which the microscope has to be moved upwards = 9.4 – 7.67 = 1.73 cm

Q.4 Figures (a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45^o with the normal to a water-glass interface [Fig. (c)].

Ans.

From the given figure, for the glass − air interface:
Angle of incidence, i = 60°
Angle of refraction, r = 35°
According to Snell’s law, the relative refractive index of glass with respect to air is given as:

\begin{array}{l}\text{μ}_{\text{air}}{}_{\text{glass}}\text{=}\frac{{\text{sin60}}^{\text{o}}}{{\text{sin35}}^{\text{o}}}\\ \text{=}\frac{\text{0}\text{.8660}}{\text{0}\text{.5736}}\\ \text{= 1}\text{.51}\to \text{(i)}\end{array}

From the given figure, for the air − water interface:
Angle of incidence, i = 60°
Angle of refraction, r = 47°
According to Snell’s law, the relative refractive index of water with respect to air is given as:

\begin{array}{l}\text{μ}_{\text{air}}{}_{\text{water}}\text{=}\frac{{\text{sin60}}^{\text{o}}}{{\text{sin47}}^{\text{o}}}\\ \text{=}\frac{\text{0}\text{.8660}}{\text{0}\text{.7314}}\\ \text{= 1}\text{.184}\to \text{(ii)}\end{array}

From equation (i) and (ii), the relative refractive index of glass with respect to water is given as:

$\begin{array}{l}\mathrm{\mu }_{\text{glass}}^{\text{water}}=\frac{\mathrm{\mu }_{\text{glass}}^{\text{air}}}{\mathrm{\mu }_{\text{water}}^{\text{air}}}\\ =\frac{1.51}{1.184}\\ =1.275\to \left(\mathrm{iii}\right)\end{array}$

$\begin{array}{l}\text{From\hspace{0.17em}the\hspace{0.17em}given\hspace{0.17em}figure,\hspace{0.17em}for\hspace{0.17em}the\hspace{0.17em}water}-\text{glass\hspace{0.17em}interface:}\\ \text{Given,\hspace{0.17em}angle\hspace{0.17em}of\hspace{0.17em}incidence},\mathrm{i}={45}^{\circ }\\ \text{Let\hspace{0.17em}angle\hspace{0.17em}of\hspace{0.17em}refraction\hspace{0.17em}}=\mathrm{r}\end{array}$

$\begin{array}{l}\text{According\hspace{0.17em}to\hspace{0.17em}Snell’s\hspace{0.17em}law,\hspace{0.17em}the\hspace{0.17em}refractive\hspace{0.17em}index\hspace{0.17em}of\hspace{0.17em}glass\hspace{0.17em}}\\ \text{with\hspace{0.17em}respect\hspace{0.17em}to\hspace{0.17em}water\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as}:\\ \mathrm{\mu }_{\text{glass}}^{\text{water}}=\frac{\mathrm{sini}}{\mathrm{sinr}}\\ =\frac{\sin {45}^{\circ }}{\mathrm{sinr}}\to \left(\mathrm{iv}\right)\\ \text{From\hspace{0.17em}equation\hspace{0.17em}(iii)\hspace{0.17em}and\hspace{0.17em}(iv),\hspace{0.17em}we\hspace{0.17em}have:}\\ 1.275=\frac{\sin {45}^{\circ }}{\mathrm{sinr}}\\ \therefore \mathrm{sinr}=\frac{0.7071}{1.275}\\ =0.5546\\ \therefore \mathrm{r}={\sin }^{-1}(0.5546)\\ =33{.68}^{\circ }\end{array}$

Q.5 A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

Ans.

The given situation is shown in the following figure.

Here, actual depth of bulb, MO = 80 cm = 0.8 m

Refractive index of water, μ =1.33

Assuming the bulb to a point source,

\begin{array}{l}\text{Area of the surface through which the light from the bulb}\\ \text{can emerge out =Area of circle of radius, r =}\frac{\text{NP}}{\text{2}}\text{= ON = OP}\\ \text{For}\text{angle}\text{of}\text{incidence,}\text{i = Critical}\text{angle = C}\\ \text{Angle}\text{of}\text{refraction,}{\text{r = 90}}^{\text{o}}\\ \text{According}\text{to}\text{Snell’s}\text{law,}\text{the}\text{relation}\text{for}\text{refractive}\text{index}\text{of}\\ \text{water}\text{is}\text{given}\text{as:}\\ \mu =\frac{\sin r}{\sin i}\\ \therefore \text{1}\text{.33 =}\frac{{\text{sin90}}^{\text{o}}}{\text{sinC}}\text{=}\frac{\text{1}}{\text{sinC}}\\ \therefore \text{sin C =}\frac{\text{1}}{\text{1}\text{.33}}\text{= 0}\text{.75}\\ \therefore {\text{C = sin}}^{\text{-1}}\left(\text{0}\text{.75}\right)\text{= 48}{\text{.6}}^{\text{o}}\\ \text{From}\text{the}\text{given}\text{figure,}\text{in}\Delta OPM,\text{we}\text{have}:\\ \text{tani =}\frac{\text{OP}}{\text{OM}}\\ \therefore \text{OP = OM tan48}{\text{.6}}^{\text{o}}\\ \text{= 0}\text{.8 × 1}\text{.1345}\\ \text{= 0}\text{.907}\text{m}\\ \text{Area of the surface through which the light from the bulb}\\ \text{can emerge out}\text{=}\pi {\text{r}}^{\text{2}}\\ \text{= 3}\text{.14}{\left(\text{OP}\right)}^{\text{2}}\\ \text{= 3}\text{.14}{\left(\text{0}\text{.907}\right)}^{\text{2}}\\ \text{= 2}\text{.59}{\text{m}}^{\text{2}}\end{array}

Q.6 A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

Ans.

$\begin{array}{l}{\text{Angle of minimum deviation,δ}}_{\text{m}}\text{= 40°}\\ \text{Angle of the prism,A = 60°}\\ \text{Refractive index of water,μ = 1.33}\\ \text{Let refractive index of the material of the prism =The angle of deviation is related to refractive index as:}\\ \text{Relation\hspace{0.33em}connecting\hspace{0.33em}refractive\hspace{0.33em}index\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}material\hspace{0.33em}}\\ \text{of\hspace{0.33em}the\hspace{0.33em}prism\hspace{0.33em}and\hspace{0.33em}angle\hspace{0.33em}of\hspace{0.33em}deviation\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}as:}\\ {\text{μ}}_{\text{P}}\text{=}\frac{\text{sin}\left(\frac{{\text{A + δ}}_{\text{m}}}{\text{2}}\right)}{\text{sin}\frac{\text{A}}{\text{2}}}\\ \text{=}\frac{\text{sin}\left(\frac{{\text{60}}^{\text{o}}{\text{+ 40}}^{\text{o}}}{\text{2}}\right)}{\text{sin}\frac{{\text{60}}^{\text{o}}}{\text{2}}}\text{=}\frac{{\text{sin50}}^{\text{o}}}{{\text{sin30}}^{\text{o}}}\text{= 1.532}\\ \therefore \text{Refractive\hspace{0.33em}index\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}material\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}prism=1.532}\\ \text{When\hspace{0.33em}the\hspace{0.33em}placed\hspace{0.33em}in\hspace{0.33em}water:}\\ {\text{Let\hspace{0.33em}new\hspace{0.33em}minimum\hspace{0.33em}angle\hspace{0.33em}of\hspace{0.33em}deviation=δ}}_{\text{m}}\text{‘}\\ \text{Refractive\hspace{0.33em}index\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}material\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}prism\hspace{0.33em}with\hspace{0.33em}respect\hspace{0.33em}}\\ \text{to\hspace{0.33em}water\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}as:}\\ \text{μ}_{\text{P}}^{\text{w}}\text{=}\frac{{\text{μ}}_{\text{P}}}{\text{μ}}\\ \text{=}\frac{\text{sin}\left(\frac{{\text{A + δ}}_{\text{m}}\text{‘}}{\text{2}}\right)}{\text{sin}\left(\frac{\text{A}}{\text{2}}\right)}\\ \therefore \text{sin}\left(\frac{{\text{A+δ}}_{\text{m}}\text{‘}}{\text{2}}\right)\text{= sin}\frac{\text{A}}{\text{2}}\text{×}\frac{{\text{μ}}_{\text{P}}}{\text{μ}}\\ \therefore \text{sin}\left(\frac{{\text{A+δ}}_{\text{m}}\text{‘}}{\text{2}}\right)\text{= sin}\left(\frac{{\text{60}}^{\text{o}}}{\text{2}}\right)\text{×}\frac{\text{1.532}}{\text{1.33}}\\ \text{=}\frac{\text{1}}{\text{2}}\text{×}\frac{\text{1.532}}{\text{1.33}}\text{= 0.5759}\\ {\text{= sin35.16}}^{\text{o}}\\ \therefore \text{sin}\left(\frac{{\text{A+δ}}_{\text{m}}\text{‘}}{\text{2}}\right){\text{=sin35.16}}^{\text{o}}\\ \therefore \frac{{\text{A+δ}}_{\text{m}}\text{‘}}{\text{2}}{\text{=35.16}}^{\text{o}}\\ \therefore {\text{A+δ}}_{\text{m}}{\text{‘=70.32}}^{\text{o}}\\ \therefore {\text{δ}}_{\text{m}}{\text{‘=70.32}}^{\text{o}}\text{– A}\\ {\text{=70.32}}^{\text{o}}{\text{– 60}}^{\text{o}}{\text{=10.32}}^{\text{o}}\\ \text{Therefore, the new minimum angle of deviation of the parallel beam of light= 10.20°}\end{array}$

Q.7 Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?

Ans.

Here, refractive index of glass, μ =1.55
Focal length of the double-convex lens, f = 20 cm
Since radius of curvature of both the faces of the given double convex lens are the same,
Let radius of curvature of one face of the lens, R₁=R
Let radius of curvature of the other face of the lens, R₂ = -R
From lens maker’s formula, we have:

\begin{array}{l}\frac{\text{1}}{\text{f}}\text{=}\left(\text{μ – 1}\right)\left[\frac{\text{1}}{{\text{R}}_{\text{1}}}\text{–}\frac{\text{1}}{{\text{R}}_{\text{2}}}\right]\\ \frac{\text{1}}{\text{20 cm}}\text{=}\left(\text{1}\text{.55 – 1}\right)\left[\frac{\text{1}}{\text{R}}\text{+}\frac{\text{1}}{\text{R}}\right]\\ \frac{\text{1}}{\text{20 cm}}\text{= 0}\text{.55 ×}\frac{\text{2}}{\text{R}}\\ \therefore \text{R = 0}\text{.55 × 2 × 20 cm}\\ \text{= 22}\text{cm}\end{array}

Therefore, radius of curvature of the given double-convex lens = 22 cm

Q.8 A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm?

Ans.

In this situation, the point P on the right side of the lens acts as a virtual object.
Object distance, u = 12 cm
(a) Here, focal length of the convex lens, f = 20 cm
Let image distance = v
From the lens formula, we have:

$\begin{array}{l}\frac{\text{1}}{\text{v}}\text{–}\frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{f}}\\ \Rightarrow \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{f}}\text{+}\frac{\text{1}}{\text{u}}\\ \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{20 cm}}\text{+}\frac{\text{1}}{\text{12 cm}}\text{=}\frac{\text{3 + 5}}{\text{60 cm}}\\ \therefore \text{v =}\frac{\text{60 cm}}{\text{8}}\text{= 7.5 cm}\\ \therefore \text{The image is formed at 7.5 cm to the right of the lens.}\end{array}$

(b) Here, focal length of the concave lens,f = −16 cm
Let image distance = v
From the lens formula,

\begin{array}{l}\frac{\text{1}}{\text{v}}\text{–}\frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{f}}\\ \therefore \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{f}}\text{+}\frac{\text{1}}{\text{u}}\\ \frac{\text{1}}{\text{v}}\text{= –}\frac{\text{1}}{\text{16 cm}}\text{+}\frac{\text{1}}{\text{12 cm}}\\ \frac{\text{1}}{\text{v}}\text{=}\frac{\text{-3 + 4}}{\text{48 cm}}\text{=}\frac{\text{1}}{\text{48}}\\ \therefore \text{v =}\text{48}\text{cm}\\ \therefore \text{The}\text{image}\text{is}\text{formed}\text{at}\text{48}\text{cm}\text{to}\text{the}\text{right}\text{of}\text{the}\text{lens}\text{.}\end{array}

Q.9 An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

Ans.

$\begin{array}{l}{\text{Here, size of the object, h}}_{\text{1}}\text{= 3 cm}\\ \text{Object distance, u = -14 cm}\\ \text{Focal length of the concave lens, f = -21 cm}\\ \text{Let image distance = v}\\ {\text{Let size of the object=h}}_{\text{2}}\\ \text{From the lens formula, we have:}\end{array}$

$\begin{array}{l}\frac{\text{1}}{\text{v}}\text{–}\frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{f}}\\ \therefore \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{f}}\text{+}\frac{\text{1}}{\text{u}}\\ \text{= –}\frac{\text{1}}{\text{21 cm}}\text{–}\frac{\text{1}}{\text{14 cm}}\\ \text{=}\frac{\text{-2 – 3}}{\text{42 cm}}\text{=}\frac{\text{-5}}{\text{42 cm}}\\ \therefore \text{v = –}\frac{\text{42}}{\text{5}}\text{\hspace{0.33em}cm = -8.4\hspace{0.33em}cm}\\ \therefore \text{The\hspace{0.33em}image\hspace{0.33em}formed\hspace{0.33em}is\hspace{0.33em}vertual,\hspace{0.33em}erect\hspace{0.33em}and\hspace{0.33em}8.4\hspace{0.33em}cm\hspace{0.33em}away\hspace{0.33em}from\hspace{0.33em}}\\ \text{the\hspace{0.33em}lens\hspace{0.33em}on\hspace{0.33em}the\hspace{0.33em}same\hspace{0.33em}side\hspace{0.33em}as\hspace{0.33em}the\hspace{0.33em}object.}\\ \text{Magnification\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}lens\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}by\hspace{0.33em}the\hspace{0.33em}relation:}\\ \text{m =}\frac{{\text{h}}_{\text{2}}}{{\text{h}}_{\text{1}}}\text{=}\frac{\text{v}}{\text{u}}\\ \therefore {\text{h}}_{\text{2}}\text{=}\frac{\text{-8.4 cm}}{\text{-14 cm}}\text{× 3}\\ \text{= 0.6 cm × 3 = 1.8\hspace{0.33em}cm}\end{array}$

Q.10 What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

Ans.

Here, focal length of the convex lens,
f₁ = 30 cm
Focal length of the concave lens,
f₂ = −20 cm
Let focal length of the system of lenses = f
The equivalent focal length of the combination of two lenses in contact is given as:

$\begin{array}{l}\frac{\text{1}}{\text{f}}\text{=}\frac{\text{1}}{{\text{f}}_{\text{1}}}\text{+}\frac{\text{1}}{{\text{f}}_{\text{2}}}\\ \text{=}\frac{\text{1}}{\text{30 cm}}\text{–}\frac{\text{1}}{\text{20 cm}}\\ \text{=}\frac{\text{2 – 3}}{\text{60 cm}}\\ \text{=-}\frac{\text{1}}{\text{60 cm}}\\ \therefore \text{f = -60\hspace{0.33em}cm}\\ \begin{array}{l}\text{The negative sign indicates it is a diverging lens with}\\ \text{the focal length of the combination of lenses=60 cm}\end{array}\end{array}$

Q.11 A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Ans.

\begin{array}{l}{\text{Here, focal length of the objective lens, f}}_{\text{o}}\text{= 2}\text{.0 cm}\\ {\text{Focal length of the eyepiece, f}}_{\text{e}}\text{= 6}\text{.25 cm}\\ \text{Distance between the objective lens and the eyepiece, d = 15 cm}\\ \left(\text{a}\right)\text{Least distance of distinct vision, d’ = 25 cm}\\ {\text{Image distance for the eyepiece, v}}_{\text{e}}\text{= -25 cm}\\ {\text{Let object distance for the eyepiece = u}}_{\text{e}}\\ \text{From the lens formula, we have}\\ \frac{\text{1}}{{\text{v}}_{\text{e}}}\text{–}\frac{\text{1}}{{\text{u}}_{\text{e}}}\text{=}\frac{\text{1}}{{\text{f}}_{\text{e}}}\\ \therefore \frac{\text{1}}{{\text{u}}_{\text{e}}}\text{=}\frac{\text{1}}{{\text{v}}_{\text{e}}}\text{–}\frac{\text{1}}{{\text{f}}_{\text{e}}}\\ \frac{\text{1}}{{\text{u}}_{\text{e}}}\text{=}\frac{\text{1}}{\text{-25 cm}}\text{–}\frac{\text{1}}{\text{6}\text{.25 cm}}\text{=}\frac{\text{-1}}{\text{5 cm}}\\ \therefore {\text{u}}_{\text{e}}\text{= -5}\text{cm}\\ \text{Image distance for the objective lens is given as}\\ {\text{v}}_{\text{o}}{\text{= d + u}}_{\text{e}}\text{= 15 cm – 5 cm = 10 cm}\\ {\text{Object distance for the objective lens = u}}_{\text{o}}\\ \text{From the lens formula, we have}\\ \frac{\text{1}}{{\text{v}}_{\text{o}}}\text{–}\frac{\text{1}}{{\text{u}}_{\text{o}}}\text{=}\frac{\text{1}}{{\text{f}}_{\text{o}}}\\ \therefore \frac{\text{1}}{{\text{u}}_{\text{o}}}\text{=}\frac{\text{1}}{{\text{v}}_{\text{o}}}\text{–}\frac{\text{1}}{{\text{f}}_{\text{o}}}\\ \frac{\text{1}}{{\text{u}}_{\text{o}}}\text{=}\frac{\text{1}}{\text{10 cm}}\text{–}\frac{\text{1}}{\text{2 cm}}\text{=}\frac{\text{-2}}{\text{5 cm}}\\ \therefore {\text{u}}_{\text{o}}\text{= -2}\text{.5}\text{cm}\\ \text{Magnifying}\text{power}\text{of}\text{the}\text{compound}\text{microscope}\text{is}\text{given}\text{as:}\\ \text{m =}\frac{{\text{v}}_{\text{o}}}{\left|{\text{u}}_{\text{o}}\right|}\left(\text{1+}\frac{\text{d’}}{{\text{f}}_{\text{e}}}\right)\text{=}\frac{\text{10 cm}}{\text{2}\text{.5 cm}}\left(\text{1+}\frac{\text{25 cm}}{\text{6}\text{.25 cm}}\right)\text{= 20}\\ \therefore \text{Magnifying}\text{power}\text{of}\text{the}\text{compound}\text{microscope = 20}\end{array}

\begin{array}{l}\text{(b)}\text{Since}\text{the}\text{image}\text{is}\text{formed}\text{at}\text{infinity,}\\ \therefore \text{Image}\text{distance}\text{for}\text{the}\text{eyepiece,}{\text{v}}_{\text{e}}\text{=}\infty \\ \text{Let}\text{object}\text{distance}\text{for}\text{the}{\text{eyepiece be u}}_{\text{e}}\\ \text{From}\text{lens}\text{formula,}\text{we}\text{have}\\ \frac{\text{1}}{{\text{v}}_{\text{e}}}\text{–}\frac{\text{1}}{{\text{u}}_{\text{e}}}\text{=}\frac{\text{1}}{{\text{f}}_{\text{e}}}\\ \frac{\text{1}}{\infty }\text{–}\frac{\text{1}}{{\text{u}}_{\text{e}}}\text{=}\frac{\text{1}}{\text{6}\text{.25}}\\ \therefore {\text{u}}_{\text{e}}\text{=-6}\text{.25}\text{cm}\\ \text{Image}\text{distance}\text{for}\text{the}\text{objective}\text{lens}\text{is}\text{given}\text{as}\\ {\text{v}}_{\text{o}}{\text{= d + u}}_{\text{e}}\\ {\text{v}}_{\text{o}}\text{= 15 cm – 6}\text{.25 cm = 8}\text{.75}\text{cm}\\ \text{Let}\text{object}\text{distance}\text{for}\text{the}\text{objective}{\text{lens be u}}_{\text{o}}\\ \text{From}\text{lens}\text{formula,}\text{we}\text{have}\\ \frac{\text{1}}{{\text{v}}_{\text{o}}}\text{–}\frac{\text{1}}{{\text{u}}_{\text{o}}}\text{=}\frac{\text{1}}{{\text{f}}_{\text{o}}}\\ \therefore \frac{\text{1}}{{\text{u}}_{\text{o}}}\text{=}\frac{\text{1}}{{\text{v}}_{\text{o}}}\text{–}\frac{\text{1}}{{\text{f}}_{\text{o}}}\\ \frac{\text{1}}{{\text{u}}_{\text{o}}}\text{=}\frac{\text{1}}{\text{8}\text{.75 cm}}\text{–}\frac{\text{1}}{\text{2 cm}}\\ \therefore {\text{u}}_{\text{o}}\text{=-}\frac{\text{17}\text{.5}}{\text{6}\text{.75}}\text{cm = -2}\text{.59}\text{cm}\\ \text{Magnifying power of the microscope is given as}\\ \text{m =}\frac{{\text{v}}_{\text{o}}}{\left|{\text{u}}_{\text{o}}\right|}\text{×}\frac{{\text{v}}_{\text{e}}}{\left|{\text{u}}_{\text{e}}\right|}\\ \text{m =}\frac{\text{8}\text{.75 cm}}{\text{2}\text{.59 cm}}\text{×}\frac{\text{25 cm}}{\text{6}\text{.25 cm}}\text{= 13}\text{.5}\\ \text{Therefore, magnifying power of the microscope is 13}\text{.51}\text{.}\end{array}

Q.12 A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.

Ans.

\begin{array}{l}\text{Here, least distance of distinct vision, d = 25 cm}\\ \text{Focal length of the objective of the compound microscope, fo = 8 mm = 0}\text{.8 cm}\\ \text{Focal length of the eyepiece of the compound microscope, fe = 2}\text{.5 cm}\\ {\text{Object distance,u}}_{\text{o}}\text{= -9}\text{.0 mm = -0}\text{.9 cm}\\ \text{From the lens formula, we have:}\\ \frac{\text{1}}{{\text{v}}_{\text{e}}}\text{–}\frac{\text{1}}{{\text{u}}_{\text{e}}}\text{=}\frac{\text{1}}{{\text{f}}_{\text{e}}}\\ \therefore \frac{\text{1}}{{\text{u}}_{\text{e}}}\text{=}\frac{\text{1}}{{\text{v}}_{\text{e}}}\text{–}\frac{\text{1}}{{\text{f}}_{\text{e}}}\\ \text{Here,}\text{image}\text{distance,}\\ {\text{v}}_{\text{e}}\text{= -d = -25}\text{cm}\\ \therefore \frac{\text{1}}{{\text{u}}_{\text{e}}}\text{=}\frac{\text{1}}{\text{-25 cm}}\text{–}\frac{\text{1}}{\text{2}\text{.5 cm}}\\ \text{=}\frac{\text{-1 – 10}}{\text{25 cm}}\text{=}\frac{\text{-11}}{\text{25 cm}}\\ {\text{u}}_{\text{e}}\text{=}\frac{\text{-25}}{\text{11}}\text{= -2}\text{.27}\text{cm}\\ \text{In}\text{case}\text{of}\text{objective}\text{lens:}\\ \text{From}\text{lens}\text{formula,}\text{we}\text{have:}\\ \frac{\text{1}}{{\text{v}}_{\text{o}}}\text{–}\frac{\text{1}}{{\text{u}}_{\text{o}}}\text{=}\frac{\text{1}}{{\text{f}}_{\text{o}}}\\ \therefore \frac{\text{1}}{{\text{v}}_{\text{o}}}\text{=}\frac{\text{1}}{{\text{f}}_{\text{o}}}\text{+}\frac{\text{1}}{{\text{u}}_{\text{o}}}\text{=}\frac{\text{1}}{\text{0}\text{.8 cm}}\text{–}\frac{\text{1}}{\text{0}\text{.9 cm}}\\ \frac{\text{1}}{{\text{v}}_{\text{o}}}\text{=}\frac{\text{0}\text{.9 – 0}\text{.8}}{\text{0}\text{.72 cm}}\text{=}\frac{\text{0}\text{.1}}{\text{0}\text{.72}}\\ \therefore {\text{v}}_{\text{o}}\text{=}\frac{\text{0}\text{.72}}{\text{0}\text{.1}}\text{= 7}\text{.2}\text{cm}\\ \text{Thus, image}\text{distance}\text{for}\text{objective,}{\text{v}}_{\text{o}}\text{=7}\text{.2}\text{cm}\\ \therefore \text{Distance}\text{between}\text{the}\text{two}\text{lenses =}\left|{\text{u}}_{\text{e}}\right|{\text{+v}}_{\text{o}}\\ \text{2}\text{.27 cm + 7}\text{.2 cm = 9}\text{.47}\text{cm}\\ \text{Magnifying}\text{power}\text{of}\text{the}\text{microscope}\text{is}\text{given}\text{as:}\\ \text{m =}\frac{{\text{v}}_{\text{o}}}{\left|{\text{u}}_{\text{o}}\right|}\left(\text{1+}\frac{\text{d}}{{\text{f}}_{\text{e}}}\right)\\ \text{m =}\frac{\text{7}\text{.2 cm}}{\text{0}\text{.9 cm}}\left(\text{1+}\frac{\text{25 cm}}{\text{2}\text{.5 cm}}\right)\text{=88}\\ \therefore \text{Magnifying}\text{power}\text{of}\text{the}\text{microscope = 88}\end{array}

Q.13 A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

Ans.

Here, focal length of objective, f_o = 144 cm

Focal length of eyepiece, f_e = 6.0 cm

$\begin{array}{l}\text{m =}\frac{{\text{f}}_{\text{o}}}{{\text{f}}_{\text{e}}}\text{=}\frac{\text{144 cm}}{\text{6 cm}}=24\hfill \\ \begin{array}{l}\text{Therefore, for normal adjustment, the magnifying power of the telescope = 24}\\ \text{Separation between the objective and the eyepiece is given as: L = fo + fe}\\ \text{L = 144 cm + 6}\text{.0 cm = 150 cm}\end{array}\hfill \\ \begin{array}{l}\\ \end{array}\hfill \end{array}$

Q.14 (a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 10⁶ m, and the radius of lunar orbit is 3.8 × 10⁸ m.

Ans.

$\begin{array}{l}{\text{Here, focal length of the objective lens, f}}_{\text{o}}{\text{= 15 m = 15 × 10}}^{\text{2}}\text{cm}\\ {\text{Focal length of the eyepiece, f}}_{\text{e}}\text{= 1.0 cm}\\ \left(\text{a}\right)\text{Angular magnification of the telescope is given as:}\\ \text{m =}\frac{{\text{f}}_{\text{o}}}{{\text{f}}_{\text{e}}}\\ \text{=}\frac{{\text{15×10}}^{\text{2}}\text{cm}}{\text{10 cm}}\text{= 1500\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\\ \text{Therefore, angular magnification of the given refracting telescope = 1500}\\ {\text{b) Here, diameter of moon,D = 3.48 × 10}}^{\text{6}}\text{m}\\ {\text{Radius of lunar orbit,r}}_{\text{a}}{\text{= 3.8 × 10}}^{\text{8}}\text{m}\\ \text{Let diameter of the image of moon = d}\\ \text{Angle subtended by the diameter of the moon is given as:}\\ {\text{θ}}_{\text{m}}\text{=}\frac{\text{Diameter\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}moon}}{\text{Radius\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}lunar\hspace{0.33em}obbit}}\\ \text{=}\frac{\text{D}}{{\text{r}}_{\text{a}}}\text{=}\frac{{\text{3.48×10}}^{\text{6}}}{{\text{3.8×10}}^{\text{8}}}\\ \text{Angle\hspace{0.33em}subtended\hspace{0.33em}by\hspace{0.33em}the\hspace{0.33em}image\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}as:}\\ {\text{θ}}_{\text{I}}\text{=}\frac{\text{Diameter\hspace{0.33em}of\hspace{0.33em}image}}{\text{Focal\hspace{0.33em}length\hspace{0.33em}of\hspace{0.33em}objective\hspace{0.33em}lens}}\\ \text{=}\frac{\text{d}}{{\text{f}}_{\text{o}}}\text{=}\frac{\text{d}}{\text{15}}\\ \text{Since,\hspace{0.33em}angle\hspace{0.33em}subtended\hspace{0.33em}by\hspace{0.33em}the\hspace{0.33em}diameter\hspace{0.33em}of\hspace{0.33em}moon=angle}\\ \text{subtended\hspace{0.33em}by\hspace{0.33em}image}\\ \therefore {\text{θ}}_{\text{m}}{\text{= θ}}_{\text{I}}\\ \therefore \frac{{\text{3.48×10}}^{\text{6}}}{{\text{3.8×10}}^{\text{8}}}\text{=}\frac{\text{d}}{\text{15}}\\ \therefore \text{d = 15 ×}\frac{{\text{3.48×10}}^{\text{6}}}{{\text{3.8×10}}^{\text{8}}}\\ {\text{= 13.73×10}}^{\text{-2}}\text{\hspace{0.33em}m = 13.73\hspace{0.33em}cm}\\ \begin{array}{l}\text{Since, angle subtended by the diameter of the moon = angle subtended by the image}\\ \text{Therefore, the diameter of the image of the moon formed by the objective lens = 13.74 cm}\end{array}\end{array}$

Q.15 Use the mirror equation to deduce that:
(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
[Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]

Ans.

(a) The mirror equation is given as:

\begin{array}{l}\frac{\text{1}}{\text{v}}\text{+}\frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{f}}\\ \therefore \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{f}}\text{–}\frac{\text{1}}{\text{u}}\\ \text{For}\text{a}\text{concave}\text{mirror,}\text{the}\text{focal}\text{length}\text{is}\text{negative}\text{.}\\ \therefore \text{f < 0}\\ \text{Since}\text{object}\text{lies}\text{on}\text{the}\text{left,}\\ \therefore \text{Object}\text{distance,}\text{u < 0}\\ \text{Since}\text{the}\text{object}\text{is}\text{placed}\text{between}\text{f}\text{and}\text{2f,}\\ \therefore \text{2f < u < f}\\ \therefore \frac{\text{1}}{\text{2f}}\text{>}\frac{\text{1}}{\text{u}}\text{>}\frac{\text{1}}{\text{f}}\\ \therefore \text{–}\frac{\text{1}}{\text{2f}}\text{< –}\frac{\text{1}}{\text{u}}\text{< –}\frac{\text{1}}{\text{f}}\\ \therefore \frac{\text{1}}{\text{f}}\text{–}\frac{\text{1}}{\text{2f}}\text{<}\frac{\text{1}}{\text{f}}\text{–}\frac{\text{1}}{\text{u}}\text{< 0}\\ \therefore \frac{\text{1}}{\text{2f}}\text{<}\frac{\text{1}}{\text{v}}\text{< 0}\to \text{(i)}\left(\frac{\text{1}}{\text{f}}\text{–}\frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{v}}\right)\\ \text{Equation}\text{(i)}\text{shows}\text{that}\frac{\text{1}}{\text{v}}\text{is}\text{negative}\text{.}\\ \therefore \text{The}\text{value}\text{of}\text{v}\text{is}\text{negative}\text{.}\\ \text{It}\text{implies}\text{that}\text{the}\text{real}\text{image}\text{is}\text{formed}\text{.}\\ \text{Since,}\frac{\text{1}}{\text{2f}}\text{<}\frac{\text{1}}{\text{v}}\\ \therefore \text{2f > v}\\ \therefore \text{-v > -2f}\\ \therefore \text{Real}\text{image}\text{is}\text{formed}\text{beyond}\text{2f}\text{.}\\ \text{(b)}\text{From}\text{the}\text{mirror}\text{formula}\text{for}\text{a}\text{convex}\text{mirror,}\text{we}\text{have:}\\ \frac{\text{1}}{\text{f}}\text{=}\frac{\text{1}}{\text{v}}\text{+}\frac{\text{1}}{\text{u}}\\ \therefore \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{f}}\text{–}\frac{\text{1}}{\text{u}}\to \text{(a)}\\ \text{Since}\text{for}\text{a}\text{convex}\text{mirror,}\text{the}\text{value}\text{of}\text{focal}\text{length}\text{is}\\ \text{always}\text{positive}\text{and}\text{object}\text{distance}\text{(u)}\text{is}\text{always}\text{negative}\text{.}\\ \therefore \text{f > 0}\text{and}\text{u < 0}\to \text{(b)}\\ \text{From}\text{equation}\text{(a)}\text{and}\text{(b),}\text{we}\text{obtain:}\\ \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{f}}\text{–}\frac{\text{1}}{\left(\text{-u}\right)}\\ \text{=}\frac{\text{1}}{\text{f}}\text{+}\frac{\text{1}}{\text{u}}\text{>0}\\ \therefore \frac{\text{1}}{\text{v}}\text{> 0}\\ \therefore \text{The}\text{value}\text{of}\text{v}\text{is}\text{positive}\text{.}\\ \therefore \text{For}\text{a}\text{convex}\text{mirror,}\text{whatever}\text{may}\text{be}\text{the}\text{value}\text{of}\text{u,}\\ \text{the}\text{value}\text{of}\text{image}\text{distance}\text{(v)}\text{always}\text{remains}\text{positive}\text{.}\\ \therefore \text{Independent}\text{of}\text{the}\text{object}\text{distance,}\text{a}\text{convex}\text{mirror}\\ \text{always}\text{forms}\text{virtual}\text{images}\text{.}\\ \text{(c)}\text{Since}\text{focal}\text{length}\text{of}\text{a}\text{convex}\text{lens}\text{is}\text{positive}\text{.}\\ \therefore \text{f > 0}\\ \text{Since}\text{the}\text{object}\text{lies}\text{on}\text{the}\text{left,}\\ \therefore \text{Object}\text{distance,}\text{u < 0}\\ \text{From}\text{the}\text{mirror}\text{formula}\text{for}\text{convex}\text{mirror,}\text{we}\text{obtain:}\\ \frac{\text{1}}{\text{f}}\text{=}\frac{\text{1}}{\text{v}}\text{+}\frac{\text{1}}{\text{u}}\\ \therefore \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{f}}\text{–}\frac{\text{1}}{\text{u}}\\ \text{Since}\text{f > 0}\text{and}\text{u < 0}\\ \therefore \frac{\text{1}}{\text{v}}\text{>}\frac{\text{1}}{\text{f}}\\ \therefore \text{v < f}\\ \therefore \text{The}\text{image}\text{formed}\text{by}\text{the}\text{convex}\text{mirror}\text{is}\text{diminished}\text{and}\\ \text{located}\text{between}\text{the}\text{focus}\text{and}\text{the}\text{pole}\text{.}\\ \text{(d)}\text{Since}\text{the}\text{focal}\text{length}\text{of}\text{a}\text{concave}\text{mirror}\text{is}\text{negative,}\\ \therefore \text{f < 0}\\ \text{Since}\text{the}\text{object}\text{is}\text{placed}\text{on}\text{the}\text{left}\text{of}\text{the}\text{concave}\\ \text{mirror,}\\ \therefore \text{u < 0}\\ \text{From}\text{the}\text{mirror}\text{formula}\text{for}\text{concave}\text{mirror,}\text{we}\text{obtain:}\\ \frac{\text{1}}{\text{f}}\text{=}\frac{\text{1}}{\text{v}}\text{+}\frac{\text{1}}{\text{u}}\\ \therefore \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{f}}\text{–}\frac{\text{1}}{\text{u}}\\ \text{Since}\text{f < 0}\text{and}\text{u < 0}\\ \therefore \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{-f}}\text{–}\frac{\text{1}}{\text{-u}}\\ \therefore \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{u}}\text{–}\frac{\text{1}}{\text{f}}\to \text{(A)}\\ \text{Since}\left|\text{u}\right|\text{<}\left|\text{f}\right|\\ \therefore \frac{\text{1}}{\text{u}}\text{>}\frac{\text{1}}{\text{f}}\to \text{(B)}\\ \text{From}\text{equation}\text{(A)}\text{and}\text{(B),}\text{we}\text{obtain:}\\ \frac{\text{1}}{\text{v}}\text{> 0}\\ \therefore \text{v}\text{is}\text{positive}\text{.}\\ \text{It}\text{shows}\text{that}\text{the}\text{image}\text{is}\text{formed}\text{on}\text{the}\text{right}\text{side}\text{of}\\ \text{the}\text{mirror}\text{.}\\ \therefore \text{The}\text{virtual}\text{image}\text{is}\text{formed}\text{.}\\ \text{For}\text{u < 0}\text{and}\text{v > 0}\text{.}\text{we}\text{have:}\\ \frac{\text{1}}{\text{u}}\text{>}\frac{\text{1}}{\text{v}}\\ \therefore \text{v > u}\\ \text{Magnification}\text{is}\text{given}\text{as:}\\ \text{m =}\frac{\text{v}}{\text{u}}\\ \text{Since}\text{v > u}\\ \therefore \text{Magnification}\text{is}\text{positive}\text{and}\text{the}\text{enlarged}\text{image}\text{is}\\ \text{formed}\text{.}\\ \therefore \text{When}\text{an}\text{object}\text{is}\text{placed}\text{between}\text{the}\text{pole}\text{and}\text{focus}\\ \text{of}\text{the}\text{concave}\text{mirror,}\text{the}\text{virtual}\text{and}\text{enlarged}\text{image}\\ \text{is}\text{formed}\text{.}\end{array}

Q.16 A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?

Ans.

Here, actual depth of pin,

d = 15 cm

Refractive index of glass,

μ =1.5

Lat apparent depth of pin=d’

The refractive index of glass is given as the ratio of actual depth to the apparent depth.

\begin{array}{l}\text{Refractive}\text{index}\text{of}\text{glass,}\text{μ =}\frac{\text{Real}\text{depth}}{\text{Apparent}\text{depth}}\\ \therefore \text{μ =}\frac{\text{d}}{\text{d’}}\\ \therefore \text{d’ =}\frac{\text{d}}{\text{μ}}\\ =\frac{15\text{cm}}{1.5}=10cm\\ \therefore \text{Distance through which the pin appears to be raised}\\ \text{= Real}\text{depth – Apparent}\text{depth = 15 cm – 10 cm}\\ \text{= 5}\text{cm}\end{array}

Q.17 (a) Figure 9.32 shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure

(b) What is the answer if there is no outer covering of the pipe?

Ans.

(a) Refractive index of the glass fibre, μ₁ =1.68
Refractive index of the material of outer covering, μ₂ = 1.44
Refractive index of the interface of the inner and outer core of the pipe is given as:

$\begin{array}{l}\text{μ =}\frac{{\text{μ}}_{\text{2}}}{{\text{μ}}_{\text{1}}}\text{=}\frac{\text{1}}{{\text{sini}}_{\text{c}}}\\ {\text{Here,\hspace{0.33em}i}}_{\text{c}}\text{= Critical\hspace{0.33em}angle}\\ \therefore {\text{sin i}}_{\text{c}}\text{=}\frac{{\text{μ}}_{\text{1}}}{{\text{μ}}_{\text{2}}}\\ \text{=}\frac{\text{1.44}}{\text{1.68}}\text{= 0.8571}\\ \therefore {\text{i}}_{\text{c}}{\text{= sin}}^{\text{-1}}\left(\text{0.8571}\right){\text{= 59}}^{\text{o}}\\ \text{Total\hspace{0.33em}internal\hspace{0.33em}reflection\hspace{0.33em}takes\hspace{0.33em}place\hspace{0.33em}only\hspace{0.33em}when,}\\ \text{Angle\hspace{0.33em}of\hspace{0.33em}incidence},\mathrm{i}>{\mathrm{i}}_{\mathrm{c}}\\ {\text{i.e.,\hspace{0.33em}i>59}}^{\text{o}}\\ \text{The\hspace{0.33em}maximum\hspace{0.33em}value\hspace{0.33em}of\hspace{0.33em}angle\hspace{0.33em}of\hspace{0.33em}reflection\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}as:}\\ {\text{\hspace{0.33em}r}}_{\text{max}}{\text{=90}}^{\text{o}}{\text{-59}}^{\text{o}}{\text{=31}}^{\text{o}}\\ \text{From\hspace{0.33em}Snell’s\hspace{0.33em}law,\hspace{0.33em}we\hspace{0.33em}obtain:}\\ {\text{μ}}_{\text{1}}\text{=}\frac{{\text{sin\hspace{0.33em}i}}_{\text{max}}}{{\text{sinr}}_{\text{max}}}\\ \therefore {\text{sin\hspace{0.33em}i}}_{\text{max}}{\text{= μ}}_{\text{1}}{\text{sin r}}_{\text{max}}\\ {\text{=1.68 × sin31}}^{\text{o}}\\ \text{= 0.8662}\\ \therefore {\text{\hspace{0.33em}i}}_{\text{max}}{\text{= sin}}^{\text{-1}}\left(\text{0.8662}\right){\text{= 60}}^{\text{o}}\\ \therefore {\text{Range\hspace{0.33em}of\hspace{0.33em}angles\hspace{0.33em}of\hspace{0.33em}incident\hspace{0.33em}rays\hspace{0.33em}with\hspace{0.33em}the\hspace{0.33em}axis\hspace{0.33em}of\hspace{0.33em}pipe£60}}^{\text{o}}\\ \text{The\hspace{0.33em}lower\hspace{0.33em}limit\hspace{0.33em}of\hspace{0.33em}angle\hspace{0.33em}of\hspace{0.33em}incidence\hspace{0.33em}is\hspace{0.33em}decided\hspace{0.33em}by\hspace{0.33em}the\hspace{0.33em}ratio}\\ \text{of\hspace{0.33em}diameter\hspace{0.33em}to\hspace{0.33em}length\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}pipe.}\\ \text{(b)\hspace{0.33em}If\hspace{0.33em}there\hspace{0.33em}is\hspace{0.33em}no\hspace{0.33em}outer\hspace{0.33em}covering\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}pipe:}\\ {\text{Refractive\hspace{0.33em}index\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}outer\hspace{0.33em}core\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}pipe,\hspace{0.33em}μ}}_{\text{1}}\text{= 1}\\ {\text{Refractive\hspace{0.33em}index\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}inner\hspace{0.33em}core\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}pipe,\hspace{0.33em}μ}}_{\text{2}}\text{= 1.68}\\ \text{From\hspace{0.33em}Snell’s\hspace{0.33em}law,\hspace{0.33em}we\hspace{0.33em}obtain:}\\ \text{μ =}\frac{{\text{μ}}_{\text{2}}}{{\text{μ}}_{\text{1}}}\\ \text{=}\frac{\text{1}}{{\text{sini}}_{\mathrm{c}}\text{‘}}\\ \therefore {\text{sini}}_{\mathrm{c}}\text{‘=}\frac{{\text{μ}}_{\text{1}}}{{\text{μ}}_{\text{2}}}\\ \text{=}\frac{\text{1}}{\text{1.68}}\\ \text{=0.5952}\\ \therefore {\mathrm{i}}_{\mathrm{c}}‘={\sin }^{-1}\left(0.5952\right)\\ =36{.5}^{\mathrm{o}}\\ \text{Here,\hspace{0.33em}for\hspace{0.33em}angle\hspace{0.33em}of\hspace{0.33em}incidence},\mathrm{i}={90}^{\mathrm{o}}\\ \text{We\hspace{0.33em}obtain,\hspace{0.33em}angle\hspace{0.33em}of\hspace{0.33em}reflection,\hspace{0.33em}}\mathrm{r}=36{.5}^{\mathrm{o}}\\ \therefore \mathrm{i}‘={90}^{\mathrm{o}}-\mathrm{r}\\ ={90}^{\mathrm{o}}-36{.5}^{\mathrm{o}}\\ =53{.5}^{\mathrm{o}}\\ \text{Since\hspace{0.33em}the\hspace{0.33em}value\hspace{0.33em}of\hspace{0.33em}}\mathrm{i}‘>{\mathrm{i}}_{\mathrm{c}}‘\\ \therefore \text{All\hspace{0.33em}the\hspace{0.33em}rays\hspace{0.33em}of\hspace{0.33em}light\hspace{0.33em}incident\hspace{0.33em}in\hspace{0.33em}the\hspace{0.33em}range\hspace{0.33em}}\le {90}^{\mathrm{o}}\text{\hspace{0.33em}will\hspace{0.33em}}\\ \text{undergo\hspace{0.33em}total\hspace{0.33em}internal\hspace{0.33em}reflection.}\end{array}$

Q.18 Answer the following questions:

(a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.
(b) A virtual image, we always say, cannot be caught on a screen. Yet when we ‘see’ a virtual image, we are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a contradiction?
(c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is?
(d) Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease?
(e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter?

Ans.

(a) Yes, plane and convex mirrors can produce real images if the object is virtual. For example: If the rays of light converging at a point behind the mirror are reflected to a point on the screen in front of the mirror. In this way, a real image is formed on the screen.

(b) No, there is not any contradiction. Since our eye lens is a double convex lens, it produces real and inverted images of the objects on the retina of the eye. In human eye, the virtual image acts as the object for the eye lens to produce the real image on the retina.

(c) Since the diver is under water and the fisherman is in the air, the light travels from the denser medium to the rarer medium. Therefore, it bends away from the normal and the fisherman looks taller to the diver.

(d) Yes, the apparent depth of tank of water will decrease further, when viewed obliquely as compared to the depth when seen normally. This is because, the direction of light changes when it travels from one medium to another.

(e) Since the refractive index of diamond is much greater as compared to that of an ordinary glass, the critical angle for diamond is much smaller as compared to that of an ordinary glass. Therefore, a skilled diamond cutter will put to use a large range of angles of incidence (less than critical angle) of light to ensure that the light entering the diamond undergoes multiple total internal reflections inside the diamond.

Q.19 The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?

Ans.

Here, distance between the object and the image, s = 3 m

Let maximum focal length of the convex lens = f_max

For real image on the wall, the maximum focal length is given as:

\begin{array}{l}{\text{f}}_{\text{max}}\text{=}\frac{\text{s}}{\text{4}}\\ \text{=}\frac{\text{3 m}}{\text{4}}\\ \text{= 0}\text{.75}\text{m}\\ \text{Therefore, the maximum focal length of the lens required for this purpose = 0}\text{.75 m}\end{array}

Q.20 A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.

Ans.

Distance between object and the screen, D = 90 cm

Separation between two locations of the convex lens, d = 20 cm

Let focal length of the lens = f

The relation connecting focal length to d and D is given as:

\begin{array}{l}\text{f =}\frac{{\text{D}}^{\text{2}}{\text{– d}}^{\text{2}}}{\text{4D}}\\ \text{=}\frac{{\left(\text{90 cm}\right)}^{\text{2}}\text{–}{\left(\text{20 cm}\right)}^{\text{2}}}{\text{4×90 cm}}\\ \text{=}\frac{\text{770 cm}}{\text{36}}\\ \text{= 21}\text{.39}\text{cm}\\ \text{Therefore, focal length of the convex lens = 21}\text{.39 cm}\end{array}

Q.21 (a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?
(b) An object 1.5 cm in size is placed on the side of the convex lens. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.

Ans.

Here, focal length of the convex lens, f₁ = 30 cm
Focal length of the concave lens, f₂ = −20 cm
Distance between the two lenses, d = 8.0 cm

(a) When the parallel beam of light is incident on the convex lens first:
For convex lens:
Here, object distance, u₁= ∞
Let image distance=v₁
From the lens formula, we have:

$\begin{array}{l}\frac{1}{{\mathrm{v}}_1}–\frac{1}{{\mathrm{u}}_1}=\frac{1}{{\mathrm{f}}_1}\\ \frac{1}{{\mathrm{v}}_1}=\frac{1}{{\mathrm{f}}_1}–\frac{1}{{\mathrm{u}}_1}\\ =\frac{1}{30}–\frac{1}{\mathrm{\infty }}\\ \therefore {\mathrm{v}}_1=30\mathrm{cm}\\ \text{For concave lens:}\\ \text{In this case, the image will act as a virtual object.}\\ {\text{Here, object distance, u}}_{\text{2}}\text{=}\left(\text{30 – d}\right)\text{=30 – 8=22 cm}\\ {\text{Let image distance=v}}_{\text{2}}\\ \text{From lens formula for the concave lens, we obtain:}\\ \frac{\text{1}}{{\text{v}}_{\text{2}}}\text{–}\frac{\text{1}}{{\text{u}}_{\text{2}}}\text{=}\frac{\text{1}}{{\text{f}}_{\text{2}}}\\ \therefore \frac{\text{1}}{{\text{v}}_{\text{2}}}\text{=}\frac{\text{1}}{{\text{f}}_{\text{2}}}\text{–}\frac{\text{1}}{{\text{u}}_{\text{2}}}\\ \text{=}\frac{\text{1}}{\text{22 cm}}\text{–}\frac{\text{1}}{\text{20 cm}}\text{=}\frac{\text{10 – 11}}{\text{220 cm}}\\ \therefore {\text{v}}_{\text{2}}\text{= -220\hspace{0.33em}cm}\\ \text{Therefore, the parallel incident beam appears to diverge from a point that is}\left(\text{220-4}\right)\text{216 cm from}\\ \text{the centre of the system of the two lenses.}\\ \left(\text{ii}\right)\text{When the parallel beam of light is incident from the left on the concave lens first:}\\ \text{For concave lens:}\\ {\text{Here, object distance, u}}_{\text{2}}\text{=-}\mathrm{\infty }\\ {\text{Let image distance=v}}_{\text{2}}\\ \text{From the lens formula, we have:}\\ \frac{\text{1}}{{\text{v}}_{\text{2}}}\text{–}\frac{\text{1}}{{\text{u}}_{\text{2}}}\text{=}\frac{\text{1}}{{\text{f}}_{\text{2}}}\\ \therefore \frac{\text{1}}{{\text{v}}_{\text{2}}}\text{=}\frac{\text{1}}{{\text{f}}_{\text{2}}}\text{+}\frac{\text{1}}{{\text{u}}_{\text{2}}}\\ \text{=}\frac{\text{1}}{\text{-20 cm}}\text{–}\frac{\text{1}}{\text{–}\mathrm{\infty }}\text{=-}\frac{\text{1}}{\text{20 cm}}\\ \therefore {\text{v}}_{\text{2}}\text{= -20\hspace{0.33em}cm}\\ \text{For convex lens:}\\ {\text{Here, object distance, u}}_{\text{1}}\text{=-}\left(\text{20 cm + 8 cm}\right)\text{= -28 cm}\\ {\text{Let image distance = v}}_{\text{2}}\\ \text{From lens formula for the convex lens, we obtain:}\\ \frac{\text{1}}{{\text{v}}_{\text{1}}}\text{–}\frac{\text{1}}{{\text{u}}_{\text{1}}}\text{=}\frac{\text{1}}{{\text{f}}_{\text{1}}}\\ \therefore \frac{\text{1}}{{\text{v}}_{\text{1}}}\text{=}\frac{\text{1}}{{\text{f}}_{\text{1}}}\text{+}\frac{\text{1}}{{\text{u}}_{\text{1}}}\\ \text{=}\frac{\text{1}}{\text{30 cm}}\text{+}\frac{\text{1}}{\text{-28 cm}}\\ \text{=}\frac{\text{14-15}}{\text{420 cm}}\text{=-}\frac{\text{1}}{\text{420 cm}}\\ \therefore {\text{v}}_{\text{2}}\text{= -420\hspace{0.33em}cm}\\ \text{Therefore, the parallel beam of incident light appears to diverge from a point 420-4=416 cm}\\ \text{from the centre of the two lens system.}\\ \left(\text{b}\right){\text{Given, height of the object, h}}_{\text{1}}\text{= 1.5 cm}\\ {\text{Object distance for convex lens, u}}_{\text{1}}\text{=- 40 cm}\\ \text{From the lens formula, we obtain:}\\ \frac{\text{1}}{{\text{v}}_{\text{1}}}\text{–}\frac{\text{1}}{{\text{u}}_{\text{1}}}\text{=}\frac{\text{1}}{{\text{f}}_{\text{1}}}\\ \therefore \frac{\text{1}}{{\text{v}}_{\text{1}}}\text{=}\frac{\text{1}}{{\text{f}}_{\text{1}}}\text{+}\frac{\text{1}}{{\text{u}}_{\text{1}}}\\ \text{=}\frac{\text{1}}{\text{30 cm}}\text{–}\frac{\text{1}}{\text{40 cm}}\text{=}\frac{\text{4-3}}{\text{120}}\\ \text{=}\frac{\text{1}}{\text{120}}\\ \therefore {\text{v}}_{\text{1}}\text{=120\hspace{0.33em}cm}\\ \text{Magnitude\hspace{0.33em}of\hspace{0.33em}magnification\hspace{0.33em}produced\hspace{0.33em}by\hspace{0.33em}convex\hspace{0.33em}lens\hspace{0.33em}is\hspace{0.33em}}\\ \text{given\hspace{0.33em}as:}\\ {\text{m}}_{\text{1}}\text{=}\frac{{\text{v}}_{\text{1}}}{\left|{\text{u}}_{\text{1}}\right|}\text{=}\frac{\text{120 cm}}{\text{40 cm}}\text{= 3}\\ \text{Therefore, the magnitude of magnification produced by convex lens = 3}\\ \text{The image formed by the convex lens acts as an object for the concave lens.}\\ \text{For concave lens:}\\ \text{Object distance,}\\ {\text{u}}_{\text{2}}\text{=120 cm – 8 cm = 112 cm}\\ \text{From the lens formula,}\\ \text{we obtain:}\\ \frac{\text{1}}{{\text{v}}_{\text{2}}}\text{–}\frac{\text{1}}{{\text{u}}_{\text{2}}}\text{=}\frac{\text{1}}{{\text{f}}_{\text{2}}}\\ \therefore \frac{\text{1}}{{\text{v}}_{\text{2}}}\text{=}\frac{\text{1}}{{\text{f}}_{\text{2}}}\text{+}\frac{\text{1}}{{\text{u}}_{\text{2}}}\\ \text{=}\frac{\text{1}}{\text{-20 cm}}\text{+}\frac{\text{1}}{\text{112 cm}}\\ \text{=}\frac{\text{-112+20}}{\text{2240 cm}}\\ \text{=}\frac{\text{-92}}{\text{2240 cm}}\\ \therefore {\text{v}}_{\text{2}}\text{=}\frac{\text{-2240}}{\text{92}}\text{\hspace{0.33em}cm}\\ \text{Magnitude\hspace{0.33em}of\hspace{0.33em}magnification\hspace{0.33em}produced\hspace{0.33em}by\hspace{0.33em}concave\hspace{0.33em}lens\hspace{0.33em}is\hspace{0.33em}}\\ \text{given\hspace{0.33em}as:}\\ {\text{m}}_2\text{=}\left|\frac{{\text{v}}_2}{{\text{u}}_2}\right|\\ \text{=}\frac{\text{2240}}{\text{92}}\text{×}\frac{\text{1}}{\text{112}}\text{=}\frac{\text{20}}{\text{92}}\\ {\text{Net\hspace{0.33em}magnification\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}lens\hspace{0.33em}combination,\hspace{0.33em}m=m}}_{\text{1}}{\text{×m}}_{\text{2}}\\ \text{=3×}\frac{\text{20}}{\text{92}}\text{=0.652}\\ \therefore {\text{Height\hspace{0.33em}of\hspace{0.33em}image,\hspace{0.33em}h}}_{\text{2}}{\text{= emh}}_{\text{1}}\\ \text{= 0.652×1.5 = 0.98\hspace{0.33em}cm}\end{array}$

Q.22 At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

Ans.

The given figure shows the incident, refracted and emergent rays associated with the glass prism ABC.

Here, angle of prism, A = 60°

Refractive index of the prism, μ = 1.524

As shown in the given figure, a ray of light is incident at the face AB of the prism, gets refracted and strikes the face AC of the prism. Here, i₁ and r₁ represent the angle of incidence and refraction at AB respectively. r₂ and e represent angle of incidence and angle of emergence at face AC respectively.

As per Snell’s law, we have:

\begin{array}{l}\frac{\sin e}{\sin r_2}=\mu \\ \therefore \sin r_2=\frac{\sin e}{\mu }\\ \text{For total internal reflection at face AC}:\\ \text{e}=\text{9}{0}^{\text{o}}\\ \therefore \sin r_2=\frac{\sin {90}^o}{\mu }\\ =\frac{\sin {90}^o}{1.524}\\ =\frac{1}{1.524}\\ =0.6562\\ \therefore r_2={\sin }^{-1}\left(0.6562\right)\approx {41}^o\\ \text{Since}\text{angle}\text{of}\text{prism}\text{is}\text{given}\text{as}:\\ A=r_1+r_2\\ \therefore r_1=A-r_2\\ ={60}^{\text{o}}-{41}^o\\ ={19}^o\\ \text{For}\text{face}AB:\\ \text{From}\text{Snell’s}\text{law,}\text{we}\text{have}:\\ \mu =\frac{\sin i_1}{\sin r_1}\\ \therefore \sin i_1=\mu \sin r_1\\ =1.524\times \sin {19}^o\\ =0.496\\ \therefore i_1={\sin }^{-1}\left(0.496\right)\\ ={29.75}^o\\ \therefore \text{Angle}\text{of}\text{incidence}={29.75}^o\end{array}

Q.23 A card sheet divided into squares each of size 1 mm² is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye.
(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.

Ans.

(a) Area of each square, A = 1 mm²
Object distance, u = −9 cm
Since if we take the focal length of the converging lens equal to 9 cm, the value of image distance will become infinity, which is not sensible,
Therefore, we take the focal length of the converging lens equal to 10 cm.
Focal length of the converging lens, f = 10 cm
Let image distance=v
From the lens formula,

\begin{array}{l}\frac{\text{1}}{\text{f}}\text{=}\frac{\text{1}}{\text{v}}\text{–}\frac{\text{1}}{\text{u}}\\ \therefore \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{f}}\text{+}\frac{\text{1}}{\text{u}}\\ \text{=}\frac{\text{1}}{\text{10 cm}}\text{+}\frac{\text{1}}{\text{-9 cm}}\\ \text{= –}\frac{\text{1}}{\text{90 cm}}\\ \therefore \text{v = -90}\text{cm}\\ \text{Magnification}\text{of}\text{the}\text{converging}\text{lens}\text{is}\text{given}\text{as:}\\ \text{m =}\frac{\text{v}}{\text{u}}\text{=}\frac{\text{-90}}{\text{-9}}\text{=10}\\ \text{Since}\text{area}\text{of}\text{each}\text{square,}\text{A = 1}{\text{mm}}^{\text{2}}\\ \therefore \text{Area}\text{of}\text{each}\text{square}\text{in}\text{the}\text{virtual}{\text{image = m}}^{\text{2}}\text{A}\\ \text{=}{\left(\text{10}\right)}^{\text{2}}\text{×1}{\text{mm}}^{\text{2}}\\ \text{=100}{\text{mm}}^{\text{2}}\\ \text{=1}{\text{cm}}^{\text{2}}\\ \text{(b)}\text{Magnifying}\text{power}\text{of}\text{the}\text{converging}\text{lens}\text{is}\text{given}\text{as:}\\ {\text{m}}_{\text{c}}\text{=}\frac{\text{d}}{\left|\text{u}\right|}\\ \text{=}\frac{\text{25 cm}}{\text{9 cm}}\text{= 2}\text{.8}\\ \text{(c) The magnification in (a) is not equal to the magnifying power in (b)}\text{. The magnification can be equal to}\\ \text{the magnifying power only if the image is formed at the least distance of distinct vision}\text{.}\end{array}

Q.24 (a) At what distance should the lens be held from the figure in Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case? Explain.

Ans.

\begin{array}{l}\left(\text{a}\right)\text{The maximum possible magnification is obtained when the image is formed at the least}\\ \text{distance of distinct vision}\left(\text{d = 25 cm}\right)\text{.}\\ \text{Here, image distance, v = -d = -25 cm}\\ \text{Focal length of the lens, f = 10 cm}\\ \text{Let object distance = u}\\ \text{From the lens formula, we obtain}\\ \frac{\text{1}}{\text{v}}\text{–}\frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{f}}\\ \therefore \frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{v}}\text{–}\frac{\text{1}}{\text{f}}\\ \frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{-25 cm}}\text{–}\frac{\text{1}}{\text{10 cm}}\text{= –}\frac{\text{7}}{\text{50 cm}}\\ \therefore \text{u = -7}\text{.14}\text{cm}\\ \text{Therefore, in order to view the squares distinctly, the distance of lens from them = 7}\text{.14 cm}\\ \left(\text{b}\right)\text{Magnification is given as}\\ \text{m =}\left|\frac{\text{v}}{\text{u}}\right|\\ \text{m =}\frac{\text{25 cm}}{\text{7}\text{.14 cm}}\text{= 3}\text{.5}\\ \left(\text{c}\right)\text{Magnifying power is given as}\\ \text{Magnifying}{\text{power, m}}_{\text{P}}\text{=}\frac{\text{d}}{\left|\text{u}\right|}\\ {\text{m}}_{\text{P}}\text{=}\frac{\text{25 cm}}{\text{7}\text{.14 cm}}\text{= 3}\text{.5}\\ \text{Since, the image is formed at the least distance of distinct vision}\left(\text{25 cm}\right)\text{, therefore,}\\ \text{the magnifying power is equal to the magnitude of magnification in this case}\text{.}\end{array}

Q.25 What should be the distance between the object in Exercise 9.24 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm². Would you be able to see the squares distinctly with your eyes very close to the magnifier?

[Note: Exercises 9.23 to 9.25 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]

Ans.

\begin{array}{l}\begin{array}{l}\text{Here, areal magnification of each square, A = 6}{\text{.25 mm}}^{\text{2}}\hfill \\ \text{Therefore, the linear magnification is given as:}\hfill \end{array}\\ \text{m=}\sqrt{\text{6}{\text{.25 mm}}^{\text{2}}}\text{=2}\text{.5 mm}\\ \text{Since}\text{magnitude}\text{of}\text{magnification}\text{is}\text{given}\text{as:}\\ \text{m =}\frac{\text{v}}{\text{u}}\\ \therefore \text{v = mu}\\ \text{v = 2}\text{.5u}\\ \text{Given,}\text{focal}\text{length}\text{of}\text{magnifying}\text{glass,}\text{f=10}\text{cm}\\ \text{From}\text{the}\text{lens}\text{formula,}\text{we}\text{obtain:}\\ \frac{\text{1}}{\text{f}}\text{=}\frac{\text{1}}{\text{v}}\text{–}\frac{\text{1}}{\text{u}}\\ \therefore \frac{\text{1}}{\text{10}}\text{=}\frac{\text{1}}{\text{2}\text{.5u}}\text{–}\frac{\text{1}}{\text{u}}\\ \text{=}\frac{\text{1}}{\text{u}}\left(\frac{\text{1}}{\text{2}\text{.5}}\text{-1}\right)\\ \text{=}\frac{\text{1}}{\text{u}}\left(\frac{\text{-1}\text{.5}}{\text{2}\text{.5}}\right)\\ \therefore \text{u =}\frac{\text{-1}\text{.5 ×10 cm}}{\text{2}\text{.5}}\text{= -6}\text{cm}\\ \text{Since,}\text{v = 2}\text{.5u}\\ \therefore \text{v = 2}\text{.5×-6 cm = -15}\text{cm}\\ \text{Since the virtual image is formed at 15 cm, which is less than the least distance of distinct vision for a normal eye}\left(\text{25 cm}\right)\text{,}\\ \text{therefore, the squares cannot be seen distinctly}\text{.}\end{array}

Q.26 Answer the following questions:
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?

Ans.

(a) Although the virtual image produced by the magnifying glass is of larger size, but the angular size of the image is equal to the angular size of the object. With the help of magnifying glass, we place the object much closer than the least distance of distinct vision. The magnifying glass helps us to see objects placed at the distance less than least distance of distinct vision. Since the closer object will have the larger angular magnification as compared to the same object at least distance of distinct vision, therefore, the magnification is achieved in this sense.

(b) Yes, the angular magnification changes when distance between the lens and a magnifying glass is increased, the angular magnification decreases a little. This is beacuse the angle subtended at the eye is slightly less than the angle subtended at the lens. The image distance does not have any effect on angular magnification.

(c) Since it is very difficult to grind a convex lens of such a small focal length, therefore, it is very difficult to produce a convex lens of very small focal length. Moreover, when we decrease the focal length of the convex lens by increasing its thickness at the middle, both the spherical and chromatic aberrations increase as well. Therefore, we cannot produce a convex lens of very small focal length.

(d) The angular magnification of the eye piece of the compound microscope is given as:

${\mathrm{m}}_{\mathrm{e}}=1+\frac{\mathrm{d}}{{\mathrm{f}}_{\mathrm{e}}}$

Here, f_e = Focal length of the eye piece
From the above relation, we observe that in order to obtain the higher magnifying power of the compound microscope, the focal length of the eye piece should be small.
The angular magnification of the objective of the compound microscope is given as:

${\mathrm{m}}_{\mathrm{o}}=\frac{{\mathrm{v}}_{\mathrm{o}}}{\left|{\mathrm{u}}_{\mathrm{o}}\right|}$

In order to increase the magnifying power of the compound microscope, we should increase v_o and decrease u_o.
Maximum value of v_o = Length of microscope tube = L
Minimum value of u_o = Focal length of objective of the compound microscope = f_o
Therefore, magnifying power of objective of compound microscope is given as:

${\mathrm{m}}_{\mathrm{o}}=\frac{\mathrm{L}}{{\mathrm{f}}_{\mathrm{o}}}$

From the above relation, we observe that in order to obtain the higher value of the magnifying power of the compound microscope, the focal length of the objective should be small.
Thus, we conclude that, in order to obtain the higher value of magnifying power of the compound microscope, the focal length of both the eye piece and the objective of the compound microscope should be small.

(e) The image of the objective lens of the compound microscope formed by the eye piece is called as the eye ring. All the rays of light refracted by the objective lens pass through the eye ring. If the eye is placed very close to the eye piece, then the field of view of the eye will decrease and all the rays of light coming through the eye will not be collected by the eye. The ideal position for the eye to view through the compound microscope is the eye ring. The exact position of the eye ring depends on the distance between the objective and the eye piece and focal length of the eye piece as well.

Q.27 An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?

Ans.

Given, focal length of the objective lens, f_o= 1.25 cm

Focal length of the eyepiece, f_e = 5 cm

Least distance of distinct vision, d = 25 cm

Since angular magnification of the compound microscope = 30X

Therefore, total magnifying power of the compound microscope, m = 30

Angular magnification of the eyepiece is given as:

\begin{array}{l}{m}_e=\left(1+\frac{d}{f_e}\right)\\ =\left(1+\frac{25\text{cm}}{5\text{cm}}\right)\\ =6\\ \text{Angular}\text{magnification}\text{of}\text{the}\text{compound}\text{microscope}\text{is}\\ \text{given}\text{as}:\\ m=m_e{m}_o\\ \therefore 30=6\times m_o\\ \therefore m_o=\frac{30}{6}\\ =5\\ \therefore \text{Angular}\text{magnification}\text{of}\text{the}\text{objective}\text{lens}=5\\ \text{Let}\text{object}\text{distance}\text{for}\text{object}\text{lens}=u_o\\ \text{Let}\text{image}\text{distance}\text{for}\text{object}\text{lens}=v_o\\ \text{Relation}\text{for}\text{the}\text{angular}\text{magnification}\text{of}\text{the}\text{objective}\\ \text{is}\text{given}\text{as}:\\ m_o=-\frac{v_o}{u_o}\\ \therefore 5=-\frac{v_o}{u_o}\\ \therefore v_o=-5u_o\\ \text{From}\text{the}\text{lens}\text{formula}\text{for}\text{objective}\text{lens,}\text{we}\text{obtain}:\\ \frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_o}\\ \therefore \frac{1}{-5u_o}-\frac{1}{u_o}=\frac{1}{1.25}\\ -\frac{1}{u_o}\left(\frac{1}{5}+1\right)=\frac{1}{1.25}\\ -\frac{6}{5u_o}=\frac{1}{\mathrm{1.25\; cm}}\\ \therefore u_o=\frac{\mathrm{1.25\; cm}}{1}\times \frac{-6}{5}\\ =-1.5\text{\hspace{0.33em}cm}\\ \therefore \text{Object}\text{distance}\text{for}\text{objective}\text{lens,}{u}_o=-1.5\mathrm{cm}\to (i)\\ Since,v_o=-5u_o\to (ii)\\ \text{From}\text{equation}\text{(i)}\text{and}\text{(ii),}\text{we}\text{obtain}:\\ v_o=-5\times -1.5\\ =7.5cm\\ \text{Let}\text{image}\text{distance}\text{for}\text{eye}\text{piece}=v_e\\ \text{Let}\text{object}\text{distance}\text{for}\text{eye}\text{piece}=u_e\\ \text{From}\text{lens}\text{formula}\text{for}\text{eye}\text{piece,}\text{we}\text{obtain}:\\ \frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}\\ \text{Image}\text{distance}\text{for}\text{eye}\text{piece,}{v}_e=-d=-25cm\\ \therefore \frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}\\ =\frac{-1}{25cm}-\frac{1}{5\text{cm}}\\ =-\frac{6}{25}\\ \therefore u_e=-\frac{25}{6}\\ =-\mathrm{4.17\; cm}\\ \text{Distance}\text{between}\text{objective}\text{lens}\text{and}\text{eye}\text{piece}=\left|u_e\right|+\left|v_e\right|\\ =4.17\text{cm}+7.5\text{cm}\\ =\mathrm{11.67\; cm}\\ \therefore \text{Distance}\text{between}\text{objective}\text{lens}\text{and}\text{eye}\text{piece}\\ =\mathrm{11.67\; cm}\end{array}

Q.28 A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e., when the final image is at infinity)?
(b) the final image is formed at the least distance of distinct vision (25 cm)?

Ans.

Given, focal length of the objective lens, f_o= 140 cm

Focal length of the eyepiece, f_e = 5 cm

Least distance of distinct vision, d = 25 cm

(a) In normal adjustment:

Magnifying power of the telescope is given as:

\begin{array}{l}m=\frac{f_o}{f_e}\\ =\frac{140\text{cm}}{5cm}\\ =28\\ (b)\text{When}\text{final}\text{image}\text{is}\text{formed}\text{at}\text{least}\text{distance}\text{of}\text{distinct}\\ \text{vision:}\\ \text{Magnifying}\text{power}\text{of}\text{the}\text{telescope}\text{is}\text{given}\text{as}:\\ m=\frac{f_o}{f_e}\left(1+\frac{f_e}{d}\right)\\ =\frac{140\text{cm}}{5\text{cm}}\left(1+\frac{5\text{cm}}{25\text{cm}}\right)\\ =28\left(1+0.2\right)\\ =28\times 1.2\\ =33.6\end{array}

Q.29 (a) For the telescope described in Exercise 9.28 (a), what is the separation between the objective lens and the eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?
(c) What is the height of the final image of the tower if it is formed at 25 cm?

Ans.

Here, focal length of the objective lens, f_o = 140 cm
Focal length of the eyepiece, f_e = 5 cm

(a) For normal adjustment:
Distance between the objective lens and the eyepiece
= f_o + f_e = 140 + 5 = 145 cm

(b) Here, height of the tower, h₁ = 100 m
Separation between the tower and the telescope, u = 3 km = 3000 m
Angle subtended by 100 m tall tower at the telescope is given as:

$\begin{array}{l}\mathrm{\theta }=\frac{{\mathrm{h}}_{\mathrm{t}}}{\mathrm{u}}\\ =\frac{100}{3000}\\ =\frac{1}{30}\mathrm{rad}\to \left(\mathrm{i}\right)\\ \text{Let\hspace{0.33em}height\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}image\hspace{0.33em}formed\hspace{0.33em}by\hspace{0.33em}the\hspace{0.33em}objective}={\mathrm{h}}_{\mathrm{i}}\\ \text{Angle\hspace{0.33em}subtended\hspace{0.33em}by\hspace{0.33em}the\hspace{0.33em}image\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}tower\hspace{0.33em}formed\hspace{0.33em}}\\ \text{by\hspace{0.33em}the\hspace{0.33em}objective\hspace{0.33em}lens\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}as}:\\ \mathrm{\theta }=\frac{{\mathrm{h}}_{\mathrm{i}}}{{\mathrm{f}}_{\mathrm{o}}}\\ =\frac{{\mathrm{h}}_{\mathrm{i}}}{140}\mathrm{rad}\to \left(\mathrm{ii}\right)\\ \text{From\hspace{0.33em}equation\hspace{0.33em}}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right),\text{\hspace{0.33em}we\hspace{0.33em}obtain}:\\ \frac{1}{30}=\frac{{\mathrm{h}}_{\mathrm{i}}}{140}\\ \therefore {\mathrm{h}}_{\mathrm{i}}=\frac{140}{30}=4.7\mathrm{cm}\\ \left(\mathrm{c}\right)\text{\hspace{0.33em}Here,\hspace{0.33em}final\hspace{0.33em}image\hspace{0.33em}distance,\hspace{0.33em}d=25\hspace{0.33em}cm}\\ \text{Magnification\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}eye\hspace{0.33em}piece\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}as}:\\ \mathrm{m}=1+\frac{\mathrm{d}}{{\mathrm{f}}_{\mathrm{e}}}\\ =1+\frac{25\mathrm{cm}}{5\text{cm}}\\ =6\\ \text{Height\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}final\hspace{0.33em}image\hspace{0.33em}of\hspace{0.33em}tower}={\mathrm{mh}}_{\mathrm{i}}\\ =6\times 4.7\text{cm}\\ =28.2\mathrm{cm}\end{array}$

Q.30 A Cassegrain telescope uses two mirrors as shown in Fig. 9.30. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?

Ans.

\begin{array}{l}\text{Given, distance between the objective mirror and the secondary mirror, d = 20 mm}\\ {\text{Radius of curvature of objective mirror, R}}_{\text{1}}\text{= 220 mm}\\ \text{Therefore, focal length of the objective mirror is given as:}\\ {\text{f}}_{\text{1}}\text{=}\frac{{\text{R}}_{\text{1}}}{\text{2}}\text{=}\frac{\text{220 mm}}{\text{2}}\text{= 110}\text{mm}\\ {\text{Radius of curvature of secondary mirror, R}}_{\text{1}}\text{= 140 mm}\\ \text{Therefore, focal length of the secondary mirror is given as:}\\ {\text{f}}_{\text{2}}\text{=}\frac{{\text{R}}_{\text{2}}}{\text{2}}\text{=}\frac{\text{140 mm}}{\text{2}}\text{= 70}\text{mm}\\ \text{When the object is placed at infinity, the parallel rays of light coming from the object at infinity}\\ \text{will collect at the focus of the object lens and act as the imaginary object for the secondary mirror}\text{.}\\ \therefore \text{For}\text{secondary}\text{mirror,}\text{object}\text{distance,}{\text{u = f}}_{\text{o}}\text{– d}\\ \text{u = 110 mm – 20 mm = 90}\text{mm}\\ \text{Let}\text{image}\text{distance}\text{for}\text{secondary}\text{mirror = v}\\ \text{From}\text{the}\text{mirror}\text{formula}\text{for}\text{secondary}\text{mirror,}\text{we}\text{obtain:}\\ \frac{\text{1}}{\text{v}}\text{+}\frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{{\text{f}}_{\text{2}}}\\ \Rightarrow \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{{\text{f}}_{\text{2}}}\text{–}\frac{\text{1}}{\text{u}}\\ \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{70 mm}}\text{–}\frac{\text{1}}{\text{90 mm}}\\ \frac{\text{1}}{\text{v}}\text{=}\frac{\text{2}}{\text{630 mm}}\\ \Rightarrow \text{v =}\frac{\text{630 mm}}{\text{2}}\text{= 315}\text{mm}\\ \therefore \text{Final}\text{image}\text{distance}\text{from}\text{secondary}\text{mirror = 315}\text{mm}\end{array}

Q.31 Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.36. A current in the coil produces a deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?

Ans.

Angle of deflection of mirror, θ = 3.5°

Distance of the screen from the mirror, d = 1.5 m

Since the mirror turns by an angle of amount θ , therefore the reflected rays get deflected by twice the angle of deflection of the mirror i.e., 2θ =7.0°

Let displacement of the reflected spot of light on the screen=s

Displacement of the reflected spot of light on the screen is given by the relation:

\begin{array}{l}\tan 2\theta =\frac{s}{1.5m}\\ \therefore s\; =1.5\text{m}\times \tan 7^o\\ =0.184m\\ =18.4cm\\ \therefore \text{Displacement}\text{of}\text{the}\text{reflected}\text{spot}\text{of}\text{light}=18.4cm\end{array}