Q. Wavelength of given light waves in air and in a medium are $6000\ Å$ and $4000\ Å$ respectively. The critical angle is
A) $\sin^{- 1}\begin{array}{r}
\left( \frac{3}{2} \right)
\end{array}$
B) $\sin^{- 1}\begin{array}{r}
\left( \frac{2}{3} \right)
\end{array}$
C) $\tan^{- 1}\left( \frac{3}{2} \right)$
D) $\tan^{- 1}\left( \frac{2}{3} \right)$
Explanation
(b) $\frac{n₂}{n₁} = \frac{1}{\sin C}$
$\frac{1}{\sin C} = \ \frac{\lambda ₁}{\lambda ₂}\ = \ \frac{6000}{4000} = \frac{3}{2}$
$C = \sin^{- 1}\left( \frac{2}{3} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.