Question

Figures (a) and (b) show refraction of a ray in air incident at ${60}^{\circ }$ with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is ${45}^{\circ }$ with the normal to a water-glass interface [Figure (c)].

Answer 1

Let us assume the refractive index of air, water and glass are $n_a,n_w$ and $n_g$ respectively. Step1: Find $\frac{n_g}{n_0}$
For the glass-air interface:
Angle of incidence, $i={60}^{\circ }$
Angle of refraction, $r={35}^{\circ }$
From Snell's law we know that the refractive index of the glass with respect to air is given as:
$\frac{n_g}{n_a}=\frac{\sin i}{\sin r}=\frac{\sin {60}^{\circ }}{\sin {35}^{\circ }}=\frac{0.8660}{0.5736}$
$\frac{n_g}{n_a}=n_{ga}=1.51....\left(i\right)$
Step 2: Find $\frac{n_w}{n_a}$
For the air-water interface:
Angle of incidence, $i={60}^{\circ }$
Angle of refraction, $r={47}^{\circ }$
From Snell's law we know that the refractive index of the water with respect to air is given as:
$\frac{n_w}{n_a}=\frac{\sin i}{\sin r}=\frac{\sin {60}^{\circ }}{\sin {47}^{\circ }}=\frac{0.8660}{0.7314}$
$\frac{n_w}{n_a}=n_{wa}=1.184....\left(ii\right)$
Step 3: Find $\frac{{\eta }_g}{{\eta }_w}$ With the help of equations (i) and (ii) and then apply Snell’s law
$\frac{n_g}{n_w}=\frac{n_g}{n_a}\times \frac{n_a}{n_w}$
$\frac{n_g}{n_w}=n_{gw}=\frac{1.51}{1.184}=1.27$
From Snell’s law the relative refractive index of glass with respect to water can be found: Angle of incidence $i={45}^{\circ }$
$\frac{n_g}{n_w}=\frac{\sin \sin {45}^{\circ }}{\sin r}$
$\sin \sin r=\frac{1}{\sqrt{2}}\left(\frac{1}{1.27}\right)$
$r=\left(0.556\right)$
$r={33.83}^{\circ }$
Final Answer : ${33.83}^{\circ }$