Optics Question 733
Question: In the adjoining diagram, a wavefront AB, moving in air is incident on a plane glass surface XY. Its position CD after refraction through a glass slab is shown also along with the normals drawn at A and D. The refractive index of glass with respect to air ( $ \mu =1 $ ) will be equal to
[CPMT 1988; DPMT 1999]
Options:
A) $ \frac{\sin \theta }{\sin \theta ‘} $
B) $ \frac{\sin \theta }{\sin \varphi ‘} $
C) $ \frac{\sin \varphi ‘}{\sin \theta } $
D) $ \frac{AB}{CD} $
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Answer:
Correct Answer: B
Solution:
In the case of refraction if CD is the refracted wave front and v1 and v2 are the speed of light in the two media, then in the time the wavelets from B reaches C, the wavelet from A will reach D, such that $ t=\frac{BC}{v _{a}}=\frac{AD}{v _{g}} $
$ \Rightarrow \frac{BC}{AD}=\frac{v _{a}}{v _{g}} $ …..(i) But in $ \Delta ACB, $
$ BC=AC\sin \theta $ …..(ii) while in $ \Delta ACD, $
$ AD=AC\sin {\varphi }’ $ …..(iii) From equations (i), (ii) and (iii) $ \frac{v _{a}}{v _{g}}=\frac{\sin \theta }{\sin {\varphi }’} $ .
Also $ \mu \propto \frac{1}{v}\Rightarrow \frac{v _{a}}{v _{g}}=\frac{{\mu _{g}}}{{\mu _{a}}}=\frac{\sin \theta }{\sin {\varphi }’} $
$ \Rightarrow {\mu _{g}}=\frac{\sin \theta }{\sin {\varphi }’} $
BETA
