Chapter 24, Problem 14Q

To determine

To Arrange: The materials according to their index of refraction, least to great.

Answer to Problem 14Q

The ranking of the materials according to their refractive index: ${\text{n}}_{\text{1}}{\text{ < n}}_{\text{3}}{\text{ < n}}_{\text{2}}\text{.}$

Explanation of Solution

Introduction:The ratio of light’s speed in a vacuum to the light’s speed in material is called refractive index. The refractive index affects the focal length of any lens.

The direction and relative amount that the light rays bend at each interface.

When the ray travels first material to second material, it will bend towards normal. Due to this, ray will slow down. So, the speed: $\left({\text{n}}_1{\text{ < n}}_{\text{2}}\right)\text{. }$

When the ray travels second material to third material, it will bend away from normal. Due to this, ray will speed up. So, the speed: $\left({\text{n}}_{\text{2}}{\text{ > n}}_{\text{3}}\right)\text{. }$

In the third material, the ray will not bend. So, the speed: $\left({\text{n}}_1{\text{ < n}}_3\right)\text{. }$

Thus, the overall ranking is ${\text{n}}_1{\text{ < n}}_{\text{3}}{\text{ < n}}_2\text{.}$

Where,

  ${\text{n}}_{\text{1}}$ is refractive index of first material

  ${\text{n}}_2$ is refractive index of second material,

  ${\text{n}}_{\text{3}}$ is refractive index of third material.

Conclusion:The ranking of the materials according to their refractive index: ${\text{n}}_{\text{1}}{\text{ < n}}_{\text{3}}{\text{ < n}}_{\text{2}}\text{.}$