Q28 of 32 Page 60

Show that for a material with refractive index light incident at any angle shall be guided along a length perpendicular to the incident face.

The ray travels along the length of the medium MC, when the value of the sine is maximum and angle “a” is minimum.


Given:


The refractive index of the material is. To show that the light refracted is parallel to the medium of the surface by showing relationship between angle of incidence, refraction and refractive index of the material.


Formula used:


Snell’s law states the relationship between the sine values of incidence, refractive angles and refractive index of the material.



where


i is the angle of incidence, and r is the refractive angle and is the refractive index of the material. is the refractive index of the air


Explanation:


The path of refraction at critical angle is MC



For reflection to stay on the path MC



From triangle MNO








Using Snell’s law, we get



Squaring both the sides we get



When the value of angle of incidence is maximum then angle of refraction is maximum and the value of angle of a is minimum.



Let us take the value of i to be maximum that is





Therefore, it is proved that if refractive index is , then ray of light travels along the length perpendicular to the incidence.


More from this chapter

All 32 →
26

A jar of height h is filled with a transparent liquid of refractive index μ (Fig. 9.6). At the center of the jar on the bottom surface is a dot. Find the minimum diameter of a disc, such that when placed on the top surface symmetrically about the center, the dot is invisible.


 27

A myopic adult has a far point at 0.1 m. His power of accommodation is 4 diopters. (i) What power lenses are required to see distant objects? (ii) What is his near point without glasses? (iii) What is his near point with glasses? (Take the image distance from the lens of the eye to the retina to be 2 cm.)

 29

The mixture a pure liquid and a solution in a long vertical column (i.e., horizontal dimensions << vertical dimensions) produces diffusion of solute particles and hence a refractive index gradient along the vertical dimension. A ray of light entering the column at right angles to the vertical is deviated from its original path. Find the deviation in travelling a horizontal distance d << h, the height of the column.

 30

If light passes near a massive object, the gravitational interaction causes a bending of the ray. This can be thought of as happening due to a change in the effective refractive index of the medium given by


Where r is the distance of the point of consideration from the center of the mass of the massive body, G is the universal gravitational constant, M the mass of the body and c the speed of light in vacuum. Considering a spherical object find the deviation of the ray from the original path as it grazes the object.