NCERT Questions for Class 12 Physics Chapter 9 – Ray Optics and Optical Instruments

Important Questions with Solutions of Class 12 Physics Chapter 9 – Ray Optics and Optical Instruments

Short Answer Questions

1) Does myopia or hypermetropia necessarily indicate that the eye has partially diminished its capacity for accommodation? What could potentially cause these visual impairments?

Ans – Common eye defects include myopia and hypermetropia.

An individual with myopia or hyperopia does not inevitably experience a diminished capacity for ocular accommodation.

Myopia arises when the eyeballs elongate from front to rear, and hypermetropia occurs when the eyeballs shorten.

Conversely, when the eye lens entirely loses its capacity for accommodation, the condition is termed presbyopia.

2) Describe optical fibres. Provide one application of it.

Ans – Optical fibres are long, thin strands of high-quality glass or quartz that have been lightly coated with a substance that has a lower refractive index than the strands themselves. They avoid any loss in information transfer since they operate on the tenet of complete internal reflection.

Applications – Optical fibres are frequently employed in medical examinations; for example, endoscopy allows one to observe the inside of the stomach and intestine.

3) A converging lens and a diverging lens with identical focal lengths are positioned coaxially in contact. Determine the focal length and power of the optical combination.

Ans – The lens formula is given as follows,

For converging lens the focal length is f₁ = +f

For diverging lens the focal length is f₂ = -f

We know,

4) A newspaper reflects light, which is why you read it. If so, why don’t you see a faint picture of yourself in the text?

Ans – We are aware that typical light reflection creates images. However, there is diffused (irregular) light reflection when we read a newspaper, making it impossible for us to detect even a faint image of ourselves there.

5) When violet light illuminates on a thin converging lens, its focal length is (f). Explain why the focal length of the lens will change if red light is used instead of violet light.

Ans – According to the formula

where n is the refractive index, R₁ and R₂ are the radii of curvature of the two surfaces of the lens and f is the focal length.

We know that n for violet is greater than that of red, and n∝f. This means that when red light replaces violet light, the focal length of the lens will decrease.

3 Mark Questions

1) Determine the radius of curvature of the convex surface of a plano-convex lens, given a focal length of 0.3 m and a refractive index of 1.5 for the lens material.

Ans – Given,

μ = 1.5

f = 0.3m

For plane convex lens,

R₂ = −∞ and let R₁ = R

Formula for focal length is,

Substituting the values we get,

⇒ R = 0.15m

2) Prove that the limiting value of the prism angle is double its critical angle. In turn, define the crucial angle.

Ans – We know that angle of the prism is given by A = r₁ + r₂

In a triangular prism scenario where i₁ = i₂ = 90°, the angle of refraction is expressed as r₁ = r₂ = C, where C is the critical angle.

A = r₁ + r₂

⇒ A = C + C

⇒ A = 2C

∴ The angle of incidence at which the angle of refraction equals 90° is referred to as the crucial angle.

3) A convex lens with a focal length of 0.2 m, composed of glass (μ = 1.50), is submerged in water (μ = 1.33). Determine the variation in the focal length of the lens.

Ans – Here f_a = 0.2m and ^aμ_g = 1.50

We can get the value of ^wμ_g by

^wμ_g = ^aμ_g / ^aμ_w = 1.5/1.33

= 1.128

Now the focal length of the lens when submerged in water,

f_w = 1/1.28 = 0.78m

Thus, change in focal length = f_w – f_a

= 0.78 – 0.20 = 0.58m

4) The concave reflector of a reflecting type telescope has a radius of curvature of 120 cm. Calculate the eye piece’s focal length to reach a 20x magnification.

Ans – Given M = 20 and R = 120cm

f_o = R/2 = (-120)/2 = -60cm

M = f_o / f_e

⇒ -20 = (-60)/f_e

⇒ f_e = 3cm

As a result, the eyepiece’s focal length is 3 cm.

5 Marks Questions

1) a. The vertical lines are easier to see than the horizontal ones when a person is staring at a mesh of crossed wires. What causes the effect? How can a visual impairment like this be fixed?

Ans – The individual can see vertical lines more clearly than horizontal lines in the example. This indicates that the cornea and eye lens, which make up the eye’s refracting mechanism, are not functioning similarly in various planes. Astigmatism is the name given to this condition.

The eye’s curvature in the vertical plane is adequate. But the curvature of the horizontal plane is not enough. As a result, the retina produces distinct vertical lines but appears to blur horizontal lines. Cylindrical lenses can be used to correct this issue.

b. A man with a standard near point of 25 cm reads a book with small print using a magnifying glass, which is a thin convex lens with a focal length of 5 cm.
(i) What are the nearest and farthest distances at which he can read the book while using the magnifying glass?
(ii) What are the maximum and minimum angular magnifications achievable with the aforementioned basic microscope?

Ans – (i) Given focal length f = 5 cm and normal near point d = 25 cm.

Distance to the nearest object = u

Image distance, v = -d = -25 cm

According to the lens formula,

Distance to the farthest object = u’

Image distance, v’ = ∞

Hence nearest and farthest distances at which he can read the book while using the magnifying glass will be 4.167cm and 5cm respectively.

(ii) Maximum angular magnification is given by,

Minimum angular magnification is given by,

As a result, with the provided basic microscope, the maximum angular magnification (magnifying power) is 6 and the minimum is 5.