A) \[>{{\sin }^{-1}}\left( \frac{3}{4} \right)\]
B) \[<{{\sin }^{-1}}\left( \frac{3}{4} \right)\]
C) \[={{\sin }^{-1}}\left( \frac{4}{3} \right)\]
D) \[\le {{\sin }^{-1}}\left( \frac{4}{3} \right)\]
Correct Answer: A
Solution :
: Given, \[{{\upsilon }_{1}}=1.5\times {{10}^{8}}\,m\,{{s}^{-1}}\] \[{{\upsilon }_{2}}=2.0\times {{10}^{8}}\,m\,{{s}^{-1}}\] Refractive index for medium \[{{M}_{1}}\] is \[{{\mu }_{1}}=\frac{c}{{{\upsilon }_{1}}}=\frac{3\times {{10}^{8}}\,m\,{{s}^{-1}}}{1.5\times {{10}^{8}}m\,{{s}^{-1}}}=2\] Refractive index for medium \[{{\mu }_{2}}=\frac{c}{{{\upsilon }_{2}}}=\frac{3\times {{10}^{8}}\,m\,{{s}^{-1}}}{2.0\times {{10}^{8}}m\,{{s}^{-1}}}=\frac{3}{2}\] is If \[\theta \] is the angle of incidence and C is the critical angle, then for total internal reflection \[\sin \theta .\sin C\] But \[\sin C=\frac{{{\mu }_{2}}}{{{\mu }_{1}}}\] \[\therefore \] \[\sin \theta >\frac{{{\mu }_{2}}}{{{\mu }_{1}}}>\frac{3/2}{2}\] or \[\theta >{{\sin }^{-1}}\left( \frac{3}{4} \right)\]You need to login to perform this action.
You will be redirected in
3 sec