Class 12 Physics
CBSE and UP Board
Ray Optics and Optical Instruments
Chapter-9 Exercise 9.4
NCERT Solutions for Class 12 Physics Chapter 9 Question-4
Figures 9.34(a) and (b) show refraction of a ray in air incident at 60° 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 45s with the normal to a water- glass interface [Fig. 9.34(c)].
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As per the given figure, for the glass – air interface:
Angle of incidence, i = 60°
Angle of refraction, r = 35°
The relative refractive index of glass with respect to air is given by Snell’s law as:
μga= sin i/sin r
= sin 60°/sin 35°= 0.8660/0.5736 = 1.51 ————–Eq-1
As per the given figure, for the air-water interface:
Angle of incidence, i = 60°
Angle of refraction r= 47°
The relative refractive index of water with respect to air is given by Snell’s law as :
μwa= sin i/sin r
= sin 60°/sin 47°= 0.8660/0.7314 = 1.184————–Eq-2
Using (1) and (2), the relative refractive index of glass with respect to water can be obtained as:
μgw = μga/μwa
=1.51/1.184= 1.275
The following figure shows the situation involving the glass – water interface.
Angle of incidence, i = 45°
Angle of refraction = r
From Snell’s law, r can be calculated as:
sin i/sin r= μgw
=>sin 45°/sin r = 1.275
sin r = (1/√2)/1.275 = 0.5546
Therefore ,r = sin⁻¹ (0.5546)
= 38.68°
Hence, the angle of refraction at the water – glass interface is 38.68°.