# CBSE Important Questions Class 12 Physics Chapter 9

## Important Questions for CBSE Class 12 Physics Chapter 9- Ray Optics and Optical Instruments

With important questions for  Class 12 physics chapter 9, students will get elaborated and authentic solutions to their doubts regarding ray optics and optical instruments.

After preparing chapter 9 class 12 Physics important questions, students will be able to solve CBSE Sample papers, CBSE Past year question papers, and NCERT EXEMPLAR Questions.

In addition, Physics Class 12 chapter 9 important questions also contain important formulas, short derivations and CBSE extra questions that can help students to test their understanding.

### IMPORTANT QUESTIONS CLASS 12 PHYSICS CHAPTER 9

In these important question notes, we will talk about the phenomena of reflection, refraction and dispersion of light, using the ray diagram of light. Using the basic laws of reflection and refraction, we shall understand image formation by plane and spherical reflecting and refracting surfaces.

### 1-MARK QUESTIONS

1. A short pulse of white light is incident from air to a glass slab at normal incidence.

What would the first colour after emergence from a glass slab be?

Ans. Since, V ∝ 𝛌, the light of red colour is of the highest wavelength and therefore, the highest speed.

Hence, red would emerge first.

CONCEPT- When light rays go from one medium to another medium, the frequency of light remains unchanged.

1. The speed of the yellow light in a certain liquid is 2.4×10⁸m/s. Find the refractive index of the liquid.

Ans. Given, cꞌ=2.4×10⁸m/s

We know, c = speed of light =3×10⁸m/s

We have to find, the refractive index “𝜇” of the liquid

Since,𝜇=c/cꞌ

𝜇=3×10⁸/2.4×10⁸

𝜇=1.25

1. A substance has a critical angle of 30° for a yellow light. Find the refractive index.

Ans. We know that refractive index (𝜇)=1sin C

Substituting the values, we have

𝜇=1sin30 =11/2

⇒ 𝜇=2

1. At a given time, it is recorded that in NASA’s Hubble telescope, the focal length of the objective and the eyepieces are 80cm and 10cm respectively. What is its magnification power?

Ans. Given ƒ₀=80cm

ƒₑ=10cm,

We know that magnification(M)= – ƒ₀ƒₑ= – 8010 = -8

Hence, M=-8

CONCEPT – To find tube length (L), Use formula L=ƒ₀+ƒₑ

1. Define power of a lens. What is the S.I. unit of power?

Ans.  The power P of  a lens is defined as the tangent of the angle by which it converges or diverges a beam of light parallel to the principal axis falling at a unit distance from the optical centre.

•                                    P= 1f

The S.I. unit of power is 1/m or dioptre.

CONCEPT-  the power of a lens is positive for a converging lens and negative for a diverging lens.

Two thin lenses of power +4D and -2D are in contact. What is the focal length of the combination?

Ans. The combined power of the two thin lenses is given by the formula

P= P₁ + P₂ = 4 +(-2) = +2D

Since focal length f= 1P

Hence, f = ½  = 0.5 m = 50 cm

1. State the conditions for the phenomenon of total internal reflection to occur.

Ans. Two essential conditions for total internal reflection to occur are-

1. Light should travel from an optically denser medium to an optically rarer medium.
2. The angle of incidence in the denser medium must be greater than the critical angle for the two media.
3. What is the difference between refraction and reflection?

Ans. Refraction can be defined as the process of the shift of light when it passes through a medium leading to the bending of light. Whereas reflection can be simply defined as the bouncing back of light when it strikes a medium on a plane.

### 2-MARK QUESTIONS

1. An object is placed 10cm in front of a concave mirror. If the radius of curvature is 15cm

Find the position, nature, and magnification of the image.

Ans.  Given the radius of curvature(R)=15cm

Hence focal length(ƒ)= – R2= –152= -7.5cm

Given, object distance (u) = -10cm

Applying mirror formula –

1v+1u = 1ƒ

Putting the values,

1v+1-10 = 1-7.5

After solving,ign we get, v= -30cm

Negative sign of v shows that object and image lie on the same side (left side) of the mirror.

Also, magnification (m) = – vu = – -30-10 = -3

The negative sign of magnification shows that the image is magnified, real and inverted.

1. Define the term optical fibre. State one its important uses.

Ans. An optical fibre is a thin, long and transparent rod usually made of glass or plastic through which light can propagate.

Optical fibre work on the principle of total internal reflection and thus, they avoid loss in the transfer of information.

Use of optical fibre –

Optical fibres are often used in medical investigations i.e. one can examine the inside view of the stomach and intestine by a method called endoscopy.

1. A simple microscope has a magnifying power of 3.0 when the image is formed at the near point (25cm) of a normal eye.
1. Find its focal length.
2. What will its magnifying power be if the image is formed at infinity?

(a)

Ans.  Given D = 25cm

m= 3

For simple microscope, m= 1+ Dƒ

Hence f = 12.5 cm

(b)

Ans.  When the image is formed at infinity m= Dƒ ( the adjustment is normal)

m= 2512.5

m= 2.0

So if the magnifying power is 2 if the image is formed at infinity.

1. Explain the mechanism of mirage formation.

Ans. On hot summer days, the air near the ground becomes hotter and less dense

than the air at higher levels.

The refractive index of air increases with its density. If the air currents are small, the optical density at different layers of air increases with height.

As a result, light from a tall object such as a tree passes through a medium whose refractive index decreases towards the ground. Thus, a ray of light from such an object successively

bends away from the normal and undergoes total internal reflection, if the angle of incidence for the air near the ground exceeds the critical angle.

To a distant observer, light appears to be coming from somewhere below the ground. The observer assumes that light is being reflected from the ground by a pool of water near the tall object.

Such inverted images cause an optical illusion to the observer. This phenomenon is called mirage.

1. Why does bluish colour predominate in a clear sky?

Ans. BLUE COLOR OF THE SKY – The scattering of light by the atmosphere is a colour-dependent phenomenon. The scattering of the sky is primarily based on Rayleigh’s Law which shows the intensity of scattered light I ∝ 1𝛌⁴ , where 𝛌 is the wavelength of the light.

Since blue light is scattered much more strongly than red light, the colour of the sky becomes blue. The blue component of light is proportionately more in the light coming from different parts of the sky. This gives the impression of a blue sky.

### 3-MARK QUESTIONS

1. If the focal length of a plane convex lens is 0.3m and the refractive index of the material of the lens is 1.5, find the radius of curvature of the convex surface of a plane convex lens.

Ans. Given that, 𝜇 = 1.5

ƒ= 0.3m

For plane convex lens,

R₂ = – ∞ and let R₁ = R

Substituting these values in the formula for focal length,

1ƒ = (𝜇 – 1) (1R₁ 1R₂)

10.3 = 1.5-1 (1R + 1)

(1R) 0.5 =10.3

⇒ R = 0.15m

Thus, the radius of curvature is R = 0.15m.

1. 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? (Given the refractive index of the material of the prism is 1.524)

Ans.

Given, angle of the prism, ∠A = 60°

Refractive index of the prism, 𝜇 = 1.524

i₁= Incident angle

r₁= Refracted angle

r₂= Angle of incidence at the face AC

e= Emergent angle = 90°

According to Snell’s law, for the face AC, we can have:

sinesin r₂ = 𝜇

r₂= Angle of incidence at the face AC

e= Emergent angle = 90°

According to Snell’s law, for face AC, we can have:

sinesin r₂= 𝜇

Sin r₂ = 1𝜇 × sin 90°

= 11.524×1 = 0.6562

Therefore, r₂ = sin⁻¹ 0.6562 ≈ 41°

It is clear from the figure that angle A = r₁ + r₂

Therefore, r₁ = A – r₂ = 60 – 41 = 19°

According to Snell’s law, we have the relation:

𝜇 = sini₁sin r₁

Sin i₁ = 𝜇 sin r₁

= 1.524 × sin 19° = 0.496

Therefore, i₁ = 29.75°

Hence the angle of incidence is 29.75°.

1. A myopic person has been using spectacles of power -1.0 dioptre for distant vision.

During old age, he also needs to use a separate reading glass of power +2.0 dioptres.

Explain what may have happened.

Ans.

The power of the spectacles used by the myopic person, P = -1.0 D

Focal length of the spectacles, ƒ = 1p = 1-1✕10⁻² = -100 cm

Hence, the far point of the person is 100 cm. He might have a normal near point of 25 cm.

When he uses the spectacles, the objects placed at infinity produce virtual images at 100 cm.

He uses the ability of accommodation of the eye lens to see the objects placed between 100 cm and 25 cm.

During old age, the person uses reading glasses of power, Pꞌ = +2D

The ability to accommodation is lost in old age. This defect is called presbyopia. As a result, he is unable to see clearly the objects placed at 25 cm.

1. There is a prism made up of two different materials such as crown glass and flint glass with a wide variety of angles. What can be the combinations of prisms which will:
2. Deviate a pencil of white light without many dispersions.
3. Disperse and displace a pencil of white light without much deviation.

Ans. a. Place the two given prisms next to each other.

Make sure that their bases are on the opposite sides of the incident white light, with their faces touching each other.

When the white light is incident on the first prism, it will get dispersed. This acts as the incident light for the second prism and the dispersed light this time will recombine to give white light as a result of the combination of the two prisms.

Ans. b. Take the system of the two prisms as suggested in answer (a).

Adjust (increase) the angle of the flint glass prism so that the deviations due to the combination of the prisms become equal.

This combination will now disperse the pencil of white light without much deviation.

1. What are the factors that affect the refractive index of a medium?

Ans.  Following are the factors that affect the refractive index of a medium –

1. The refractive index of the medium will depend on the nature and temperature of the medium. It also depends on the color of the light ray.
2. Refractive index is an optical property therefore any impurity added to the medium will alter the refractive index of the medium.
3. The absolute refractive index of the medium is the ratio of the velocity of light in air or vacuum to that in the given medium. The velocity of light is maximum in vacuum. The velocity in any other medium is less than the value in air. Thus the absolute refractive index of the medium is always greater than the unity.

### 5-MARK QUESTIONS

1. Briefly explain the formation of a rainbow.

Ans.  The rainbow is an example of the dispersion of sunlight by the water drops in the atmosphere. This a phenomenon due to the combined effect of dispersion, refraction, and reflection.

The conditions for observing a rainbow are that the sun should be shining in one part of the sky, while it should be raining in the opposite part.

MECHANISM –

Consider figure A, in which sunlight is first refracted, which causes the different wavelengths of white light to separate. Longer wavelengths of light (RED) are bent the least while the shorter wavelength (VIOLET) is bent the most.

If the refracted angle is greater than the critical angle, then the light will get internally reflected.

It is found that violet light emerges at an angle of 40° while red light emerges at an angle of 42°.

Consider figure B, in which the formation of the primary rainbow takes place. The observer sees a rainbow with red color on the top and violet on the bottom which is a result of three-step processes i.e., refraction, reflection, and refraction.

When light rays undergo two internal reflections inside a raindrop, instead of one as in the primary rainbow, a secondary rainbow is formed, as shown in figure C .

It is due to a four-step process.

The intensity of light is reduced. Hence, the secondary rainbow is fainter.

1. A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom 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.

Actual depth of the needle in water, h₁ = 12.5 cm

Apparent depth of the needle in water, h₂ = 9.4 cm

Refractive index of water = 𝜇

The value of 𝜇 can be obtained as follows:

𝜇 = h₁h₂

= 12.5 / 9.4 = 1.33

Hence, 1.33 is the refractive index of water.

Now water is replaced by a liquid of refractive index 1.63

The actual depth of the needle remains the same, but its apparent depth changes.

Let x be the new apparent depth of the needle.

𝜇ꞌ =h₁x

Therefore, y = h₁𝜇ꞌe

= 12.5 / 1.63 = 7.67 cm

The new apparent depth of the needle is 7.67 cm. It is observed that the value is less than h₂,

therefore, the needle needs to be moved up to the focus again.

Distance to be moved to focus = 9.4 – 7.67 = 1.73 cm

1. The figure depicted below shows a biconvex 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 45.0 cm. The liquid is removed, and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?

Ans.

The focal length of the convex lens, f₁ = 30 cm

The liquid acts as a mirror. The focal length of the liquid = f₂

Focal length of the system ( convex lens + liquid ) f = 45cm

The equivalent focal length of the pair of optical systems that are in contact are given as:

1ƒ = 1ƒ₁ + 1ƒ₂

1f₂ = 1ƒ1ƒ₁

145 130

190

Therefore, f₂ = -90 cm

Let 𝜇₁ be the refractive index, and the radius of curvature of one surface be R.

Hence, -R will be the radius of the curvature of the other surface.

R can be obtained by using the relation:

1ƒ₁ = ( 𝜇₁ -1 ) ( 1R + 1-R

1. Draw a ray diagram showing the image formation by a compound microscope. Then obtain the expression for total magnification when the image is formed at infinity.

Ans. Ray diagram:

(a) Ray diagram of a compound microscope: A schematic diagram of a compound microscope is shown in the figure. The lens nearest the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens, the eyepiece, which functions essentially like a simple microscope or magnifier, and produces the final image, which is enlarged and virtual. The first inverted image is thus near (at or within) the focal plane of the eyepiece, at a distance appropriate for final image formation at infinity, or a little closer for image formation at the near point. Clearly, the final image is inverted with respect to the original object.

Magnification due to a compound microscope.

The ray diagram shows that the (linear) magnification due to the objective, namely h’/h, equals

Here h’ is the size of the first image, the object size being h and f0 being the focal length of the object. The first image is formed near the focal point of the eyepiece. The distance L, i.e., the distance between the second focal point of the objective and the first focal point of the eyepiece (focal length fe), is called the tube length of the compound microscope.

As the first inverted image is near the focal point of the eyepiece, we use for a simple microscope to obtain the (angular) magnification me due to it when the final image is formed at the near point, is

When the final image is formed at infinity, the angular magnification due to the eyepiece, me = (D//e)

Thus, the total magnification from equation (i) and (iii), when the image is formed at infinity, is

### RELEVANCE OF RAY OPTICS IN FUTURE

Geometrical optics or ray optics is a model that describes light propagation in terms of rays.

Ray optics is a sub-branch that is significantly important in the field of fiber optics. Optical fiber is used as a medium for telecommunication and computer networking.

It is especially advantageous for long-distance communications.

Ray optics is also used in the field of astronomy. All microscopes, telescopes, and other optical instruments work under the phenomena of reflection and refraction.

Recently, a black hole was discovered by NASA was found by the James Webb telescope.

We can say that ray optics is an important area of study for the generations to come because of its significant role in ARTIFICIAL INTELLIGENCE and quantum computing.

### WHY IS IT IMPORTANT TO STUDY IMPORTANT QUESTIONS FOR RAY OPTICS CLASS 12?

Benefits of preparing NCERT Chapter 9 Ray Optics and optical instruments from Important questions for RAY OPTICS CLASS 12.

• These physics Class 12 chapter 9 important question notes of NCERT give us a wide coverage of the topic keeping in mind the CBSE syllabus.
• Important questions for Ray optics Class 12 cover a wide range of CBSE extra questions and also the CBSE past year question paper.
• The questions have been set up  keeping in mind the CBSE Sample papers, and also the CBSE revision notes have been taken into consideration.
• The important question for the chapter Ray Optics covers all the important formulas, derivations, and necessary diagrams.
• Class 12 Physics chapter 9 important questions will benefit students in covering this whole chapter in a lucid and understanding manner.

HOW WILL STUDYING FROM EXTRAMARKS IMPORTANT QUESTION FOR RAY OPTICS CLASS 12 CBSE HELP STUDENT?

Extramarks Class 12 physics chapter 12 important questions for Ray Optics Class 12 are very significant in preparing for CBSE Class 12 Board exam and also other competitive exams. These important questions consist of NCERT Solutions, CBSE sample paper questions, and CBSE past year questions also.

Extramarks provides important questions for the chapter Ray Optics which widely cover the CBSE syllabus; this would prove to be very beneficial for the students. Extramarks important questions are compiled using important formulas, derivations, and explanations that go in very lucid and understandable patterns.

HOW TO STUDY FOR RAY OPTICS FROM EXTRAMARKS NOTES?

Students are advised to first go through the basic text for the chapter on ray optics using the NCERT Textbook Class 12. After glancing through the chapter, they should refer to the Extramarks Important Question for Ray Optics. These important questions widely cover the CBSE Syllabus, CBSE Past year question papers, etc., which would help students to know the kind of questions asked from the topic of ray optics.

Students can first refer to the important questions and try solving them on their own and, after that, go through the detailed solutions provided by Extramarks subject experts.

### INTRODUCTION TO RAY OPTICS

A lightwave can be considered to travel from one point to another along a straight line joining them. The path is called a ray of light, and a bundle of such rays constitutes a beam of light.

The branch of Physics that deals with the nature, properties, sources, and effects of light is called optics. Optics is broadly divided into two branches namely physical optics, which is the study of the wave-like nature of light and the interactions between light and matter.

Ray optics is the study of simple properties of light and optical instruments by assuming that light travels in a straight line.

Ray optics deals with the geometry of light. In ray optics, we study the image formed by mirrors, lenses, and prisms.

#### A ray is a part of the line having one fixed point, and the other point does not have an end.

In optics, a ray is an idealized geometrical model of light obtained by choosing a curve, i.e., is perpendicular to the wavefronts of the actual light and that points in the direction of energy flow.

### CONCLUSION –

Based on the property of reflection and refraction, many optical instruments have been designed to understand the behavior of light better. Based on the total internal reflection, the phenomenon like mirages,  transmission through optical fiber, and the property of diamonds could be well understood. Likewise, other optical instruments like microscopes, telescopes, etc., have been designed to help in the advancement of technology and research studies.

Q1-Two thin lenses P1 and P2 are made in contact to get the focal length of 60 cm. If the focal length of one of the lens is 20 cm, then the power of the other lens is

opt-

a–3.34 D

b–1 D

c-3.34 D

d-6.64 D

ans-3.34 D

Q2-The refractive indices of glass and water are 1.62 and 1.32 respectively. The critical angle for a ray going from glass to water is

opt-

A-

B

ans

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### 1. What are the defects of the human eye?

The defects of the human eye are as follows –

Myopia – It is also known as short-sightedness in which nearby objects appear to be seen clearly, and far objects appear to be seen blurred. It can be corrected by a concave lens.

Hypermetropia – It is also called farsightedness in which far away appear to be seen clearly and nearby objects appear to be blurred. It can be corrected using the convex lens.

Astigmatism – This defect causes impaired vision due to which some parts of the picture may appear clear, whereas some parts may appear blurred. It can be corrected with the help of spherical or cylindrical lenses.

### 2. What is illuminance?

Illuminance is defined as luminous flux incident per unit area on a surface.

The illuminance E produced by a source of luminous intensity I is given by

E = I / R²

Where R is the normal distance of the surface from the source.

Its S.I. unit is lux.

### 3. Why are diamonds known for their spectacular brilliance?

The critical angle for diamond – air interface is very small, therefore, once light enters a diamond, it is very likely to undergo Total Internal Reflection inside it.

Hence, their brilliance is mainly due to total internal reflection.

### 4. State the advantages of a reflecting type telescope over a refracting type of telescope.

Following are the advantages of a reflecting type telescope over a refracting type of telescope –

1. Due to the large aperture of the mirror used in the reflecting telescopes, it has high resolving power.
2. The reflecting type of telescope is free from chromatic aberration (formation of the coloured image of a white object).
3. The use of paraboloidal mirrors reduces spherical aberration (formation of a nonpoint, blurred image of a point object).
4. The image formed by a reflecting telescope is brighter than a refracting telescope.