Important Questions Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

Important Questions Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

Ray Optics and Optical Instruments is one of the longest chapters of Class 12 Physics. It covers reflection by spherical mirrors, refraction and Snell's law, total internal reflection, refraction at spherical surfaces and through lenses, the lens maker's formula, power of a lens, prisms and dispersion, and optical instruments such as the microscope and the telescope.

Below are 12 important questions from Ray Optics and Optical Instruments with step-by-step answers from Extramarks experts. The questions include refraction through a sphere, the compound microscope, lens combinations, the prism formula, total internal reflection and ray diagrams. Draw the ray diagrams yourself while reading the answers.

Q1. The focal lengths of the objective and the eye-piece of a compound microscope are 1.0 cm and 2.5 cm respectively. Find the tube length of the microscope for obtaining a magnification of 300.

Answer: Here, 𝑓₀ = 1.0 cm, 𝑓𝑒 = 2.5 cm, 𝑚 = 300, 𝐷 = 25 cm, Magnifying power,

∴ 𝐿 = 30 cm

Q2. Describe the construction and working of an optical fibre. State its uses

Answer:

Core - Transparent fibre of higher refractive index 𝜇₁

Cladding - Transparent fibre of lower refractive index 𝜇₂

Sheath - Hard, tough, protecitve and opaque covering of fibre. Ray of light incident on the interface of cladding and cone at angle more than critical angle so the ray suffers, total internal reflection on each incidence.

It is used for long distance transmission of signal with minimum loss of energy

Q3.

beam of light is incident on the lens. Trace the path of emergent rays when

Answer:

(i) for 𝑛₁ = 𝑛₂ ⇒ 𝑓 = ∞

(ii) for 𝑛₁ < 𝑛₂ ⇒ 𝑓 > 0

(iii) for 𝑛₁ > 𝑛₂ ⇒ 𝑓 < 0

The path of rays in three cases is shown in fig.

Q4. When light travels from an optically denser medium to a rarer medium, why does the critical angle of incidence depend on the colour of light?

Answer:

for different colour of light.

Q5. Draw the ray diagram for convex mirror producing real image

Answer:

Q6. A school plans to buy an astronomical telescope. One student recommends a very short focal-length objective because the telescope should then be compact, while another recommends a large-aperture, long focal-length objective used with a suitable eyepiece. Decide which proposal better suits the observation of faint, distant objects and justify.

Answer: The second proposal is better. A large objective aperture collects more light and produces a brighter image of a faint object. A long focal- length objective forms a larger real image near its focal plane, while the short focal-length eyepiece provides useful angular magnification. The objective and eyepiece therefore perform different functions.

Q7. A swimmer under water sees a rescue ring held above the surface. At the same time, a person on the bank sees the swimmer closer to the surface than the swimmer actually is. Explain both observations using refraction and assess why judging the true positions by sight alone can be misleading.

Answer: Light from the ring bends towards the normal on entering water, so its apparent direction differs from its actual position. Light from the swimmer bends away from the normal on emerging into air, making the swimmer appear raised. In both cases the observer traces the refracted rays backward in straight lines and therefore locates a virtual image at an incorrect position.

Q8. A ray of light 'PQ' is incident normally on the face 'AB' of a triangular prism of refracting angle of made of a

60°

(i) Calculate angle of emergence.

(ii) Define critical angle and trace the path of the ray as it passes through the prism.

Answer:

So, the angle of incidence at the face 'AC' of prism is = ( Angle of incidence at face 'AC' = critical angle)

(ii) Critical angle: The angle of incidence corresponding to an angle of refraction 90° is called the critical angle for the given pair of media. Here, PQ is the incident ray, and after final refraction, the light ray emerging out at with the normal..

Q9. A reflecting telescope has a concave mirror of radius of curvature 120 cm as objective and a convex lens of focal length 5 cm as eyepiece. Calculate the magnifying power of the telescope when the final image is formed at infinity. Also find the separation between the mirror and the eyepiece.

Answer:

Magnifying power for final image at infinity:For a reflecting telescope, when final image is at infinity:

The negative sign indicates inverted image. The magnitude of magnifying power is 12. Separation between mirror and eyepiece:For normal adjustment (final image at infinity), the intermediate image formed by the objective must be at the focal point of the eyepiece.

Q10. Define absolute and relative refractive index. A light ray travels from water (n = 1.33) to glass (n = 1.5). Calculate the relative refractive index of glass with respect to water and explain its physical significance.

Answer: Absolute refractive index of a medium is the ratio of speed of light in vacuum to the speed of light in that medium and is given by

Relative refractive index of medium 2 with respect to medium 1 is given by

For glass with respect to water we have

This means light travels 1.128 times slower in glass compared to water. The relative refractive index determines the amount of bending when light passes from water to glass.

that use multiple media.

Q11. An object in water is at a depth of 10 m. If the refractive index of water is 1.33, what will be the apparent depth of the object when seen from the air? Will there be any change in the apparent depth if the object is seen inside the water?

Answer: Here, or,

When the same object is viewed by an observer who is also inside the water, there is no change in apparent depth. It is because: Both the object and observer are in the same medium (water). Light travels from the object to the observer's eye entirely within water. No refraction occurs at any interface during this light path. Therefore, the object appears at its actual depth of 10 m.

Q12. Derive an expression for the critical angle in terms of refractive indices. Calculate the critical angle for a diamond-air interface (refractive index of diamond = 2.42).

Answer: Critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90°.

This small critical angle explains why diamonds sparkle brilliantly - most light rays inside undergo total internal reflection, emerging only from specific faces after multiple reflections.

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