Question

A ray of light is incident at the glass-water interface at an angle i = 41.8°. It emerges finally parallel to the surface of the water. What would be the value of the refractive index of the given glass? (Given: refractive index of water = 4/3)


(use the value: sin 41.8° ≈ 0.66)

[0.8 mark]

• A. 7/3
• B. 1.5
• C. 4/3
• D. 1

Answer 1

The correct option is B 1.5

Using Snell’s law at water-air interface,

$\frac{\sin r}{\sin {90}^{\circ }}=\frac{1}{{\mu }_w}\frac{\sin r}{\sin {90}^{\circ }}=\frac{3}{4}$

$\sin r=3/4$

Using Snell’s law at glass-water interface,
$\frac{\sin i}{\sin r}=\frac{{\mu }_w}{{\mu }_g}\frac{\sin 41.8^{\circ }}{3/4}=\frac{(4/3)}{{\mu }_g}$
${\mu }_g=1.5$