Q.
A ray of light is incident at the glass-water interface at an angle $i$it emerges finally parallel to the surface of water, then the value of $\mu_{g}$ would be
A) $(4/3)\ \sin i$
B) $1/\sin i$
C) 4/3
D) 1
Explanation
(b) For glass-water interface $\ _{g}\mu_{w} = \frac{\sin i}{\sin r}$
For water air interface $\ _{w}\mu_{a}\ = \ \frac{\sin r}{\sin{90{^\circ}}}$
$\Rightarrow \ _{g}\mu_{w} \times \ _{w}\mu_{a} = \ \frac{\sin i}{\sin r} \times \frac{\sin r}{\sin{90{^\circ}}} = \sin i$
or $\frac{\mu_{w}}{\mu_{g}} \times \frac{\mu_{a}}{\mu_{w}} = \sin i$
$\Rightarrow \mu_{g} = \frac{1}{\sin i}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.