Optics Question 456
Question: A rectangular glass slab ABCD, of refractive index $ n _{1} $ is immersed in water of refractive index $ n _{2}(n _{1}>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
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:
[a] Ray comes out from CD, means rays after refraction form AB get, total internally reflected at AD $ \frac{n _{1}}{n _{2}}=\frac{\sin {a _{\max }}}{\sin r _{1}}\Rightarrow {a _{\max }}={{\sin }^{-1}}[ \frac{n _{1}}{n _{2}}\sin r _{1} ] $ …(i)
Also $ r _{1}+r _{2}=90{}^\circ \Rightarrow r _{1}=90-r _{2}=90-c $
$ r _{1}=90-{{\sin }^{-1}}( \frac{1}{2{\mu _{2}}} )\Rightarrow r _{1}=90-{{\sin }^{-1}}( \frac{n _{2}}{n _{1}} ) $ …(ii)
Hence form equations (i) and (ii) $ {a _{\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}} ) ] $
