# Physics Class 12 NCERT Solutions: Chapter 9 Ray Optics and Optical Instruments Part 2 (For CBSE, ICSE, IAS, NET, NRA 2022)

Glide to success with Doorsteptutor material for CBSE/Class-12 : get questions, notes, tests, video lectures and more- for all subjects of CBSE/Class-12.

Q: 4. Figures (a) and (b) show refraction of a ray in air incident at 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 with the normal to a water-glass interface [Figure (c) ] .

Answer:

As per the given figure, for the glass – air interface:

Angle of incidence,

Angle of refraction,

The relative refractive index of glass with respect to air is given by Snell՚s law as:

As per the given figure, for the air – water interface:

Angle of incidence,

Angle of refraction,

The relative refractive index of water with respect to air is given by Snell՚s law as:

Using (1) and (2) , the relative refractive index of glass with respect to water can be obtained as:

The following figure shows the situation involving the glass – water interface.

Angle of incidence,

Angle of refraction

From Snell՚s law, can be calculated as:

Hence, the angle of refraction at the water – glass interface is .

Q: 5. A small bulb is placed at the bottom of a tank containing water to a depth of . What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is . (Consider the bulb to be a point source.)

Answer:

Actual depth of the bulb in water,

Refractive index of water,

The given situation is shown in the following figure:

Where,

Since the bulb is a point source, the emergent light can be considered as a circle of radius,

Using Snell ′ law, we can write the relation for the refractive index of water as:

Using the given figure, we have the relation:

∴ Area of the surface of water

Hence, the area of the surface of water through which the light from the bulb can emerge is approximately .

Developed by: