Q. When a glass slab is placed on a cross made on a sheet, the cross appears raised by 1 cm. The thickness of the glass is 3 cm. The critical angle for glass is
A) $\sin^{- 1}{(0.33)}$
B) $\sin^{- 1}{(0.5)}$
C) $\sin^{- 1}{(0.67)}$
D) $\sin^{- 1}{(\sqrt{3}/2)}$
Explanation
(c) $\mu = 1 = 3\left( 1 - \frac{1}{\mu} \right)$
Or $1 - \frac{1}{\mu} = \frac{1}{3}\ \ $or$\ \ \frac{1}{\mu} = 1 - \frac{1}{3} = \frac{2}{3}\ $ o$r\ \mu = \frac{2}{3}$
Now, $\ \frac{1}{\sin i_{c}} = \frac{3}{2}$
Or ${\ sin}i_{c} = \frac{2}{3}$ or$\ i_{c} = \sin^{- 1}\left( \frac{2}{3} \right)$
Or $i_{c} = \sin^{- 1}(0.67)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.