Question

A ray of light is incident at the glass-water interface at an angle $i$ as shown in the figure. It finally emerges parallel to water-air interface. The value of ${\mu }_g$ would be (Refractive index of water, ${\mu }_w=\frac{4}{3}$ )

• A. $\frac{4}{3}\sin i$
• B. $\frac{1}{\sin i}$
• C. $\frac{4}{3}$
• D. none of these

Answer 1

Answer: (B)

Apply Snell's law at glass-water interface, we get
${\mu }_g\sin i={\mu }_w\sin r$
Apply Snell's law at water air-interface, we get
${\mu }_{air}\sin 90^{\circ }={\mu }_w\sin r$

$\sin r=\frac{1}{{\mu }_w}=\frac{3}{4}$
Thus, ${\mu }_g\sin i=\frac{4}{3}\times \frac{3}{4}=1$
or ${\mu }_g=\frac{1}{\sin i}$