# Physics Class 12 NCERT Solutions: Chapter 9 Ray Optics and Optical Instruments Part 2

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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)].

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.)

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.

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