# NCERT Solutions Class 12 Physics Chapter 9

## NCERT Solutions Class 12 Physics Chapter 9 – Ray Optics and Optical Instruments

The NCERT Solutions Class 12 Physics Chapter 9 Ray Optics and Optical Instruments area valuable resource. Students preparing for Class 12 annual exams can find these notes on the Extramarks online platform. The important topics of this chapter, such as reflection, refraction, and dispersion of light, are well explained using pictorial representation. Using the Physics class 12 chapter 9 NCERT solutions, students can understand and obtain a firm grip of all concepts included in this chapter. Our NCERT Solutions are based on the latest guidelines for 2022-2023.

In the Ray Optics and Optical Instruments Chapter, students will learn different phenomena of light using the ray diagrams. For example, one will study different images formed when light passes through plane and spherical reflecting and refracting surfaces. The chapter also briefly introduces the optical instruments and their working. The Extramarks NCERT Solutions Class 12 Physics Chapter 9 covers all the concepts, theories, and formulas included in this chapter. The diagrams, charts, and graphs included in the notes make learning engaging and interesting.

The NCERT solutions class 12 Physics chapter 9 aim to help students comprehend each concept easily and attain perfect scores. Extramarks provides chapter-wise NCERT Solutions for all the subjects. Students are recommended to use the NCERT Solutions Class 9, Class 10, and Class 11 for reference to strengthen their base. Using these reference notes will help build a strong foundation for Class 12.

### Key Topics Covered In NCERT Solutions Class 12 Physics Chapter 9

The Extramarks NCERT Solutions Class 12 Physics Chapter 9 provides apt information and knowledge. It is an important resource for the preparation of board exams along with Entrance exams like NEET, JEE (Main and Advanced).

These notes are prepared by the subject elites at Extramarks to give perfect knowledge of complex topics and include all minute detail. The following topics are covered in the NCERT Solutions Class 12 Physics Chapter 9.

 Exercise Topic 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 Introduction Reflection Of Light By Spherical Mirrors Refraction Total Internal Reflection Refraction At Spherical Surfaces And By Lenses Refraction Through A Prism Dispersion By A Prism Some Natural Phenomena Due To Sunlight Optical Instruments

9.1 Introduction

Chapter 9 Physics class 12 is based on the different characteristics of light as it passes through media, namely, the convex lens, concave lens, spherical mirrors, and prism. The chapter begins with a thorough discussion of light processes like reflection, refraction, and dispersion using several ray diagrams. It then describes some natural phenomena that occur due to sunlight, such as formation of rainbow and scattering of light. Students also gain an insight into the working and construction of the human eye and other optical instruments.

The main features of the chapter are

• Rectilinear Propagation of Light
• Characteristics of light
• Laws of Refraction
• Laws of Reflection
• Human Eye

9.2 Reflection Of Light By Spherical Mirrors

This section helps students strengthen their understanding of concepts related to spherical mirrors. Concepts like Sign convention, Focal length of Spherical mirror, and paraxial rays have also been discussed. The NCERT Solutions Class 12 Physics Chapter 9 also gives a step-by-step explanation about the focal length (f) of the spherical mirror with the help of the formula, f = R2, where R= Radius of curvature. Students can get a clear idea of the mirror formula, magnification and lateral inversion included in this section.

9.3 Refraction

This section of class 12 Physics chapter 9 NCERT solutions gives brief information about the refraction of light through a curved surface. Students can comprehend topics like refractive index, critical angle, and relative refractive index. Cauchy’s formula helps students learn the relation between the refractive index of a medium () and the wavelength of light (). The causes and laws of refraction (Snell’s Law) is an important concept in this section of the NCERT Solutions Class 12 Physics Chapter 9.

9.4 Total Internal Reflection

In this section of Chapter 9, students learn about Total Internal Reflection (TIR) in nature and its applications. Concepts like Double refraction and Single refraction from a plane surface and the relation between object distance and image distance have been explained lucidly. This section also includes knowledge about the formation of Mirages, Diamonds, Prisms, and Optical fibres based on TIR.

Students should gain an in-depth understanding of these concepts to solve questions of any difficulty level with ease.

9.5 Refraction At Spherical Surfaces And By Lenses

This section is bifurcated into four parts: Refraction at a Spherical Surface, Refraction by a Lens, Power of a Lens, and Combination of Thin Lenses in Contact. With the help of the NCERT Solutions Class 12 Physics Chapter 9, students will learn to differentiate between a thin lens and a standard lens.

Some of the main formulas explained in this section are as follows:

• Lens Formula
• Mirror Formula
• Lens makers formula
• Power of a Lens
• The focal length of lens Combination
• Power of Lens Combination
• Linear magnification
• The focal length of the convex lens using the displacement method

The NCERT Solutions Class 12 Physics Chapter 9 includes the derivation of the equations that helps students get an in-depth understanding of these concepts.

9.6 Refraction Through A Prism

In this section of chapter 9 Physics class 12, students learn about the prism. It covers various features and the importance of a prism. Topics like Angle of Deviation and Prism formula are very well explained with the help of the prism diagram in the NCERT Solutions Class 12 Physics Chapter 9. Moreover, students can synopsize with the help of the simple explanation given about the condition of no emergence and dispersive power along with the equations.

9.7 Dispersion By A Prism

Based on the above topic, students learn the theory of dispersion and deviation of light by a prism. In NCERT Solutions Class 12 Physics Chapter 9, students can acquire accurate information of the angular dispersion and dispersive power within this section.

It is an important topic from the Physics class 12 chapter 9 NCERT solutions and has been asked several times in CBSE class 12 board exams.

9.8 Some Natural Phenomena Due To Sunlight

A brief discussion on the scattering of light, Rainbow, Blue Sky, and the reddish appearance of the sun in full moon near horizon has been given in NCERT Solutions Class 12 Physics Chapter 9. Furthermore, the Rayleigh’slaw of scattering is explained as the relationship between the amount of scattering and the wavelength of light.

9.9 Optical Instruments

The working and construction of Optical Instruments such as the Human Eye, Telescope, and Microscope have been explained with the help of diagrams. In addition, NCERT Solutions Class 12 Physics Chapter 9 also includes detailed explanations of eye defects such as myopia or near-sightedness, Hypermetropia or far-sightedness, Astigmatism, Colour Blindness, and Cataract using the ray diagrams. This segment will help the students to understand the ray diagrams and magnifying power of the Telescope and simple Microscope.

Students are advised to use the chapter 9 Physics class 12 NCERT solutions to significantly speed up their learning process.

### List of NCERT Solutions Class 12 Physics Chapter 9 Exercise & Answer Solutions

Extramarks provides study materials like the NCERT Solutions Class 12 Physics Chapter 9, CBSE previous year question papers, sample papers, NCERT Solutions from Class 1 to 12 for the benefit of the students. These notes ensure a thorough understanding of all essential topics included in the chapter. With the help of these notes, students can understand and revise the formulas, equations, and ray diagrams quickly to solve all questions based on them. Refer to the below content to get more details on NCERT Solutions Class 12 Physics Chapter 9:

Students can also click on the below links to get access to the NCERT Solutions based on new guidelines:

• NCERT Solutions Class 1
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### NCERT Exemplar Class 12 Physics

Students can start their preparation for Class 12 Physics Chapter 9 using the NCERT Exemplar Solutions available to the Extramarks platform. They can find solutions to all subjective, objective, in-text, and exercise questions included in the NCERT books. In addition, there are also many extra questions based on important topics mentioned in the  NCERT Solutions Class 12 Physics Chapter 9. With the help of these notes, students can prepare for the CBSE school level exams and many other national level entrance exams.

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### Key Features of NCERT Solutions Class 12 Physics Chapter 9

The key features of NCERT Solutions Class 12 Physics Chapter 9 include

• The solutions include all important concepts, formulas, equations, and solved examples in one place.
• Students will learn to analyze themselves based on their performance and know which topic they should practice more.
• Experts curate the NCERT Solutions Class 12 Physics Chapter 9 based on the latest NCERT textbooks.
• It is considered the one-stop solution for students’ every need.
• The Notes help students clear any doubts and understand each concept well to attempt questions based on them easily.

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

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

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

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

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:

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

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.

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

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:

$\begin{array}{l}\mu =\frac{\text{Real}\text{\hspace{0.17em}}\text{depth}}{\text{Apparent}\text{\hspace{0.17em}}\text{depth}}=\frac{12.5}{9.4}\\ =1.33\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \text{For}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{second}\text{\hspace{0.17em}}\text{case:}\\ \text{Refractive}\text{\hspace{0.17em}}\text{index}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{liquid},\text{\hspace{0.17em}}\mu =1.63\\ \text{Real}\text{\hspace{0.17em}}\text{depth}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{needle}=12.5\text{\hspace{0.17em}}cm\\ \text{Let}\text{\hspace{0.17em}}\text{apparent}\text{\hspace{0.17em}}\text{depth}=y\\ \text{The}\text{\hspace{0.17em}}\text{relation}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}\text{refractive}\text{\hspace{0.17em}}\text{index}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{as:}\\ \mu =\frac{\text{Real}\text{\hspace{0.17em}}\text{depth}}{\text{Apparent}\text{\hspace{0.17em}}\text{depth}}\\ \therefore 1.63=\frac{12.5}{y}\\ \therefore y=\frac{12.5}{1.63}\\ =7.67\text{\hspace{0.17em}}\text{cm}\end{array}$

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 45o 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:

$μ air glass = sin60 o sin35 o = 0.8660 0.5736 = 1.51 →(i)$

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:

$μ air water = sin60 o sin47 o = 0.8660 0.7314 = 1.184 →(ii)$

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

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

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

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

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,

$Area of the surface through which the light from the bulb can emerge out =Area of circle of radius, r = NP 2 = ON = OP For angle of incidence, i = Critical angle = C Angle of refraction, r = 90 o According to Snell’s law, the relation for refractive index of water is given as: μ= sinr sini ∴1.33 = sin90 o sinC = 1 sinC ∴sin C = 1 1.33 = 0.75 ∴ C = sin -1 0.75 = 48 .6 o From the given figure, in ΔOPM, we have: tani = OP OM ∴OP = OM tan48 .6 o = 0.8 × 1.1345 = 0.907 m Area of the surface through which the light from the bulb can emerge out = π r 2 = 3.14 OP 2 = 3.14 0.907 2 = 2.59 m 2$

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 connecting refractive index of the material }\\ \text{of the prism and angle of deviation is given 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 index of the material of the prism=1.532}\\ \text{When the placed in water:}\\ {\text{Let new minimum angle of deviation=δ}}_{\text{m}}\text{‘}\\ \text{Refractive index of the material of the prism with respect }\\ \text{to water is given 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:

$1 f = μ – 1 1 R 1 – 1 R 2 1 20 cm = 1.55 – 1 1 R + 1 R 1 20 cm = 0.55 × 2 R ∴ R = 0.55 × 2 × 20 cm = 22 cm$

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:

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

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

$1 v – 1 u = 1 f ∴ 1 v = 1 f + 1 u 1 v = – 1 16 cm + 1 12 cm 1 v = -3 + 4 48 cm = 1 48 ∴v = 48 cm ∴The image is formed at 48 cm to the right of the lens.$

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.

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

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

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:

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

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.

$Here, focal length of the objective lens, f o = 2.0 cm Focal length of the eyepiece, f e = 6.25 cm Distance between the objective lens and the eyepiece, d = 15 cm ( a ) Least distance of distinct vision, d’ = 25 cm Image distance for the eyepiece, v e = -25 cm Let object distance for the eyepiece = u e From the lens formula, we have 1 v e – 1 u e = 1 f e ∴ 1 u e = 1 v e – 1 f e 1 u e = 1 -25 cm – 1 6.25 cm = -1 5 cm ∴ u e = -5 cm Image distance for the objective lens is given as v o = d + u e = 15 cm – 5 cm = 10 cm Object distance for the objective lens = u o From the lens formula, we have 1 v o – 1 u o = 1 f o ∴ 1 u o = 1 v o – 1 f o 1 u o = 1 10 cm – 1 2 cm = -2 5 cm ∴ u o = -2.5 cm Magnifying power of the compound microscope is given as: m = v o | u o | ( 1+ d’ f e ) = 10 cm 2.5 cm ( 1+ 25 cm 6.25 cm ) = 20 ∴Magnifying power of the compound microscope = 20$

$(b) Since the image is formed at infinity, ∴Image distance for the eyepiece, v e = ∞ Let object distance for the eyepiece be u e From lens formula, we have 1 v e – 1 u e = 1 f e 1 ∞ – 1 u e = 1 6.25 ∴ u e =-6.25 cm Image distance for the objective lens is given as v o = d + u e v o = 15 cm – 6.25 cm = 8.75 cm Let object distance for the objective lens be u o From lens formula, we have 1 v o – 1 u o = 1 f o ∴ 1 u o = 1 v o – 1 f o 1 u o = 1 8.75 cm – 1 2 cm ∴ u o =- 17.5 6.75 cm = -2.59 cm Magnifying power of the microscope is given as m = v o | u o | × v e | u e | m = 8.75 cm 2.59 cm × 25 cm 6.25 cm = 13.5 Therefore, magnifying power of the microscope is 13.51.$

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.

$Here, least distance of distinct vision, d = 25 cm Focal length of the objective of the compound microscope, fo = 8 mm = 0.8 cm Focal length of the eyepiece of the compound microscope, fe = 2.5 cm Object distance,u o = -9.0 mm = -0.9 cm From the lens formula, we have: 1 v e – 1 u e = 1 f e ∴ 1 u e = 1 v e – 1 f e Here, image distance, v e = -d = -25 cm ∴ 1 u e = 1 -25 cm – 1 2.5 cm = -1 – 10 25 cm = -11 25 cm u e = -25 11 = -2.27 cm In case of objective lens: From lens formula, we have: 1 v o – 1 u o = 1 f o ∴ 1 v o = 1 f o + 1 u o = 1 0.8 cm – 1 0.9 cm 1 v o = 0.9 – 0.8 0.72 cm = 0.1 0.72 ∴ v o = 0.72 0.1 = 7.2 cm Thus, image distance for objective, v o =7.2 cm ∴Distance between the two lenses = u e +v o 2.27 cm + 7.2 cm = 9.47 cm Magnifying power of the microscope is given as: m = v o u o 1+ d f e m = 7.2 cm 0.9 cm 1+ 25 cm 2.5 cm =88 ∴Magnifying power of the microscope = 88$

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, fo = 144 cm

Focal length of eyepiece, fe = 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.

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

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:

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

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.

$Refractive index of glass, μ = Real depth Apparent depth ∴μ = d d’ ∴d’ = d μ = 15 cm 1.5 =10 cm ∴Distance through which the pin appears to be raised = Real depth – Apparent depth = 15 cm – 10 cm = 5 cm$

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

(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 = fmax

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

$f max = s 4 = 3 m 4 = 0.75 m Therefore, the maximum focal length of the lens required for this purpose = 0.75 m$

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:

$f = D 2 – d 2 4D = 90 cm 2 – 20 cm 2 4×90 cm = 770 cm 36 = 21.39 cm Therefore, focal length of the convex lens = 21.39 cm$

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\text{ }\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 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 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 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 cm}\\ \text{Magnitude of magnification produced by convex lens is }\\ \text{given 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{ cm}\\ \text{Magnitude of magnification produced by concave lens is }\\ \text{given 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 magnification of the lens combination, m=m}}_{\text{1}}{\text{×m}}_{\text{2}}\\ \text{=3×}\frac{\text{20}}{\text{92}}\text{=0.652}\\ \therefore {\text{Height of image, h}}_{\text{2}}{\text{= emh}}_{\text{1}}\\ \text{= 0.652×1.5 = 0.98 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:

$\begin{array}{l}\frac{\mathrm{sin}e}{\mathrm{sin}{r}_{2}}=\mu \\ \therefore \mathrm{sin}{r}_{2}=\frac{\mathrm{sin}e}{\mu }\\ \text{For total internal reflection at face AC}:\\ \text{e}=\text{9}{0}^{\text{o}}\\ \therefore \mathrm{sin}{r}_{2}=\frac{\mathrm{sin}{90}^{o}}{\mu }\\ =\frac{\text{\hspace{0.17em}}\mathrm{sin}{90}^{o}}{1.524}\\ =\frac{1}{1.524}\\ =0.6562\\ \therefore {r}_{2}={\mathrm{sin}}^{-1}\left(0.6562\right)\approx {41}^{o}\\ \text{Since}\text{\hspace{0.17em}}\text{angle}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{prism}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{as}:\text{\hspace{0.17em}}\\ A={r}_{1}+{r}_{2}\\ \therefore {r}_{1}=A-{r}_{2}\\ ={60}^{\text{o}}-{41}^{o}\\ ={19}^{o}\\ \text{For}\text{\hspace{0.17em}}\text{face}\text{\hspace{0.17em}}AB:\\ \text{From}\text{\hspace{0.17em}}\text{Snell’s}\text{\hspace{0.17em}}\text{law,}\text{\hspace{0.17em}}\text{we}\text{\hspace{0.17em}}\text{have}:\\ \mu =\frac{\mathrm{sin}{i}_{1}}{\mathrm{sin}{r}_{1}}\\ \therefore \mathrm{sin}{i}_{1}=\mu \mathrm{sin}{r}_{1}\\ =1.524×\mathrm{sin}{19}^{o}\\ =0.496\\ \therefore {i}_{1}={\mathrm{sin}}^{-1}\left(0.496\right)\\ ={29.75}^{o}\\ \therefore \text{Angle}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{incidence}={29.75}^{o}\end{array}$

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,

$1 f = 1 v – 1 u ∴ 1 v = 1 f + 1 u = 1 10 cm + 1 -9 cm = – 1 90 cm ∴v = -90 cm Magnification of the converging lens is given as: m = v u = -90 -9 =10 Since area of each square, A = 1 mm 2 ∴Area of each square in the virtual image = m 2 A = 10 2 ×1 mm 2 =100 mm 2 =1 cm 2 (b) Magnifying power of the converging lens is given as: m c = d u = 25 cm 9 cm = 2.8 (c) The magnification in (a) is not equal to the magnifying power in (b). The magnification can be equal to the magnifying power only if the image is formed at the least distance of distinct vision.$

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.

$( a ) The maximum possible magnification is obtained when the image is formed at the least distance of distinct vision ( d = 25 cm ). Here, image distance, v = -d = -25 cm Focal length of the lens, f = 10 cm Let object distance = u From the lens formula, we obtain 1 v – 1 u = 1 f ∴ 1 u = 1 v – 1 f 1 u = 1 -25 cm – 1 10 cm = – 7 50 cm ∴u = -7.14 cm Therefore, in order to view the squares distinctly, the distance of lens from them = 7.14 cm ( b ) Magnification is given as m = | v u | m = 25 cm 7.14 cm = 3.5 ( c ) Magnifying power is given as Magnifying power, m P = d | u | m P = 25 cm 7.14 cm = 3.5 Since, the image is formed at the least distance of distinct vision ( 25 cm ), therefore, the magnifying power is equal to the magnitude of magnification in this case.$

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.

$Here, areal magnification of each square, A = 6 .25 mm 2 Therefore, the linear magnification is given as: m= 6 .25 mm 2 =2.5 mm Since magnitude of magnification is given as: m = v u ∴v = mu v = 2.5u Given, focal length of magnifying glass, f=10 cm From the lens formula, we obtain: 1 f = 1 v – 1 u ∴ 1 10 = 1 2.5u – 1 u = 1 u 1 2.5 -1 = 1 u -1.5 2.5 ∴u = -1.5 ×10 cm 2.5 = -6 cm Since, v = 2.5u ∴v = 2.5×-6 cm = -15 cm Since the virtual image is formed at 15 cm, which is less than the least distance of distinct vision for a normal eye 25 cm , therefore, the squares cannot be seen distinctly.$

(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, fe = 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:

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

In order to increase the magnifying power of the compound microscope, we should increase vo and decrease uo.
Maximum value of vo = Length of microscope tube = L
Minimum value of uo = Focal length of objective of the compound microscope = fo
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, fo= 1.25 cm

Focal length of the eyepiece, fe = 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:

$m e = 1+ d f e = 1+ 25 cm 5 cm =6 Angular magnification of the compound microscope is given as: m= m e m o ∴30=6× m o ∴ m o = 30 6 =5 ∴Angular magnification of the objective lens=5 Let object distance for object lens= u o Let image distance for object lens= v o Relation for the angular magnification of the objective is given as: m o =− v o u o ∴5=− v o u o ∴ v o =−5 u o From the lens formula for objective lens, we obtain: 1 v o − 1 u o = 1 f o ∴ 1 −5 u o − 1 u o = 1 1.25 − 1 u o 1 5 +1 = 1 1.25 − 6 5 u o = 1 1.25 cm ∴ u o = 1.25 cm 1 × −6 5 =−1.5 cm ∴Object distance for objective lens, u o =−1.5 cm→(i) Since, v o =−5 u o →(ii) From equation (i) and (ii), we obtain: v o =−5×−1.5 =7.5 cm Let image distance for eye piece= v e Let object distance for eye piece= u e From lens formula for eye piece, we obtain: 1 v e − 1 u e = 1 f e Image distance for eye piece, v e =−d=−25 cm ∴ 1 u e = 1 v e − 1 f e = −1 25 cm − 1 5 cm =− 6 25 ∴ u e =− 25 6 =− 4.17 cm Distance between objective lens and eye piece= u e + v e =4.17 cm+7.5 cm =11.67 cm ∴Distance between objective lens and eye piece=11.67 cm$

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, fo= 140 cm

Focal length of the eyepiece, fe = 5 cm

Least distance of distinct vision, d = 25 cm

Magnifying power of the telescope is given as:

$m= f o f e = 140 cm 5 cm =28 (b) When final image is formed at least distance of distinct vision: Magnifying power of the telescope is given as: m= f o f e 1+ f e d = 140 cm 5 cm 1+ 5 cm 25 cm =28 1+0.2 =28×1.2 =33.6$

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, fo = 140 cm
Focal length of the eyepiece, fe = 5 cm

Distance between the objective lens and the eyepiece
= fo + fe = 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:

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

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.

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

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:

$tan2θ= s 1.5 m ∴s =1.5 m×tan 7 o =0.184 m =18.4 cm ∴Displacement of the reflected spot of light=18.4 cm$ Hence, the displacement of the reflected spot of light is 18.4 cm.

Q.32 Figure 9.34 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0cm. What is the refractive index of the liquid? Ans.

$\begin{array}{l}\text{Given, refractive index of equiconvex lens, = 1.50}\\ \text{Focal length of equiconvex lens, f = 30 cm}\\ \text{Here, the liquid in contact with the equiconvex lens acts as the mirror.}\\ \text{Let refractive index of liquid in contact with the equiconvex lens = f}\\ \text{For the combination of the equiconvex lens and the liquid in contact with}\\ \text{the equiconvex lens, the equivalent focal length is given as:}\\ \frac{1}{\mathrm{f}}=\frac{1}{{\mathrm{f}}_{1}}+\frac{1}{{\mathrm{f}}_{2}}\text{ }\\ \therefore \frac{1}{45\text{}\mathrm{cm}}=\frac{1}{30\text{}\mathrm{cm}}+\frac{1}{{\mathrm{f}}_{2}}\\ \therefore \frac{1}{{\mathrm{f}}_{2}}=\frac{1}{45\text{}\mathrm{cm}}-\frac{1}{30\text{}\mathrm{cm}}\\ =-\frac{1}{90}\mathrm{cm}\\ \therefore {\mathrm{f}}_{2}=-90\text{ }\mathrm{cm}\\ \text{For the equiconvex lens}:\\ \text{Let radius of curvature of one surface}=\mathrm{R}\\ \therefore \text{Radius of curvature of the other surface}=-\mathrm{R}\\ \text{Refrative index of equiconvex lens},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{\mu }=1.50\\ \text{From lens maker’s formula for equiconvex lens, we get}\\ \frac{1}{{\mathrm{f}}_{1}}=\left(\mathrm{\mu }-1\right)\left(\frac{1}{\mathrm{R}}-\frac{1}{-\mathrm{R}}\right)\\ \therefore \frac{1}{30\text{}\mathrm{cm}}=\left(1.5-1\right)\left(\frac{2}{\mathrm{R}}\right)\\ \therefore \mathrm{R}=\frac{30\text{}\mathrm{cm}}{0.5×2}\\ =30\text{ }\mathrm{cm}\\ \text{Let refractive index of liquid}={\mathrm{\mu }}_{1}\\ \text{Radius of curvature of the liquid on the side of equiconvex}\\ \text{lens}=-30\text{ }\mathrm{cm}\\ \text{Radius of curvature of the liquid on the side of plane mirror}\\ =\mathrm{\infty }\\ \text{From the lens maker’s formula for the liquid, we obtain}:\\ \frac{1}{{\mathrm{f}}_{2}}=\left({\mathrm{\mu }}_{1}-1\right)\left(\frac{1}{-\mathrm{R}}-\frac{1}{\mathrm{\infty }}\right)\\ \therefore \frac{-1}{90\text{}\mathrm{cm}}=\left({\mathrm{\mu }}_{1}-1\right)\left(\frac{1}{30\text{}\mathrm{cm}}-0\right)\\ \therefore {\mathrm{\mu }}_{1}-1=\frac{1}{3}\\ \therefore {\mathrm{\mu }}_{1}=\frac{4}{3}\\ =1.33\\ \therefore \text{Refractive index of the liquid}=1.33\end{array}$ 