Two beakers A and B contain kerosene oil and water respectively. The speed of light in water is 2.25 x 108 ms-1 and in kerosene oil 2.15 x 108 ms-1.
(a) In which liquid will the angle of refraction be more, when light ray strikes the surface of both liquids at the same angle of incidence?
(b) Calculate the ratio of refractive indices of water to kerosene oil.
(c) If liquid A is heated, what change will you expect in the path of the refracted ray?
Show SolutionHide Solution↓
(a) The angle of refraction will be more in water (beaker B).
Reason — The refractive index of a medium is:
$\text{μ} = \frac{\text c}{\text v}$
Since the speed of light in water is greater than that in kerosene oil,
μ water $\lt$ μ kerosene
For the same angle of incidence, light bends less towards the normal in the optically rarer medium. Therefore, the angle of refraction is greater in water (beaker B).
(b) Given,
- Speed of light in water (vw) = 2.25 x 108 m s-1
- Speed of light in kerosene oil (vk) = 2.15 x 108 m s-1
As, refractive index of a medium = $\frac{\text{speed of light in air}}{\text{speed of light in the medium}} = \frac{\text c}{\text v}$
Then,
$$\begin{aligned}\frac{\text{μ}_\text w}{\text{μ}_\text k} = \frac{\frac{\text c}{\text v_\text w}}{\frac{\text c}{\text v_\text k}} \\[1em] = \frac{\text v_\text k}{\text v_\text w} \\[1em] = \frac{2.15 \times 10^8}{2.25 \times 10^8} \\[1em] = \frac{215}{225} \\[1em] = \frac{43}{45}\end{aligned}$$
Hence, the ratio of the refractive indices of water to kerosene oil is 43 : 45.
(c) Liquid A is kerosene oil. On heating, the refractive index of kerosene oil decreases. Therefore the light ray bends less towards the normal, i.e., the angle of refraction increases and the refracted ray shifts away from the normal.