NCERT Exemplar for Class 12 Physics Chapter 9

NCERT Exemplar for Class 12 Physics is an incredible way to increase your knowledge about the chapters discussed in the class 12 science syllabus. Through this exemplar students can gain in-depth knowledge on the related topic of Ray Optics and Optical Instruments. The NCERT exemplar prepared by the top subject matter experts at SimplyAcad allows students to learn and cover all the sections presented in the chapter 9 of the physics textbook. This provided exemplar contains questions based on different marks such as 1, 2,3 and 5. It will benefit the students as they become more familiar with these dynamic patterns of short, long and very long answer type questions. Students can easily access this NCERT exemplar for class 12 physics in this article below to perform incredibly well in their upcoming 12th board examinations. Along with this, there are several NCERT exemplar for class 12 science of all the chapters provided in a detailed manner.

Access the NCERT Exemplar Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

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.

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

3. 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

4. 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

5. 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.

6. 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

7. 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.

2. 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.
2. Find its focal length.
3. 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.

4. 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.

5. 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.

2. 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°.

3. 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.

4. 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.

2.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

3. 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 = 1ƒ – 1ƒ₁

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 )

4. 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

NCERT Exemplar For Class 12 Science

Students must practise these additional questions for their own benefits, the ncert exemplar are curated by the best subject-matter experts to boost your knowledge on the presented topic. Students can easily access the ncert exemplar for class 12 science by visiting our website SimplyAcad and solve all the questions listed to secure maximum marks.

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