Optics Question 86
Question: A rectangular glass slab ABCD, of refractive index $n _{1}$, is immersed in water of refractive index $n _{2}$ ($n _{1} \gt n _{2}$). A ray of light in incident at the surface AB of the slab as shown. The maximum value of the angle of incidence $\alpha _{max}$, such that the ray comes out only from the other surface CD is given by
[IIT-JEE (Screening) 2000]
Options:
A) $ {{\sin }^{-1}}[ \frac{n _{1}}{n _{2}}\cos ( {{\sin }^{-1}}\frac{n _{2}}{n _{1}} ) ] $
B) $ {{\sin }^{-1}}[ n _{1}\cos ( {{\sin }^{-1}}\frac{1}{n _{2}} ) ] $
C) $ {{\sin }^{-1}}( \frac{n _{1}}{n _{2}} ) $
D) $ {{\sin }^{-1}}( \frac{n _{2}}{n _{1}} ) $
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Answer:
Correct Answer: A
Solution:
Ray comes out from CD, means rays after refraction from AB get, total internally reflected at AD $ \frac{n _{1}}{n _{2}}=\frac{\sin {\alpha _{\max }}}{\sin r _{1}}\Rightarrow {\alpha _{\max }}={{\sin }^{-1}}[ \frac{n _{1}}{n _{2}}\sin r _{1} ] $ -(i)
Also $ r _{1}+r _{2}=90^{o}\Rightarrow r _{1}=90-r _{2}=90-C $
$ \Rightarrow $ $ r _{1}=90-{{\sin }^{-1}}( \frac{1}{ _{2}{\mu _{1}}} )\Rightarrow r _{1}=90-{{\sin }^{-1}}( \frac{n _{2}}{n _{1}} ) $ …(ii)
Hence from equation (i) and (ii) $ {\alpha _{\max }}={{\sin }^{-1}}[ \frac{n _{1}}{n _{2}}\sin { 90-{{\sin }^{-1}}\frac{n _{2}}{n _{1}} } ] $ = $ {{\sin }^{-1}}[ \frac{n _{1}}{n _{2}}\cos ( {{\sin }^{-1}}\frac{n _{2}}{n _{1}} ) ] $
BETA
