Saved Bookmarks
| 1. |
A ray of light is incident normally on one side of an isosceles right-angled prism and is totally reflected from the other side. (i) What is the value of minimum refractive index of the material of the prism? (ii) If the prism is immersed in water, draw the diagram showing the directing of the emergent ray. In the diagram, point out the values of the angles, mu of water =(4)/(3). |
|
Answer» Solution :(i) ABC is an isosceles right - angled prism and its sides AB = BC. The light ray PQ is incident on the face AB normally and enters the prism. The ray is incident at R on the face AC [Fig. 2.64(a)]. It is EVIDENT from the figure that angle of incidence of the ray at R is `45^(@)`. Now if the ray is to be totally reflected from R then this angle of incidence should be greater than the critical angle of the material of the prism i.e. maximum critical angle will be `theta_(c) = 45^(@)`. If `mu` be the minimum refractive index of the material of the prism then, `sintheta_(c) = (1)/(mu) or, sin45^(@) = (1)/(mu) or, mu = sqrt(2) = 1.414`.  If the prism is immersed in water, then refractive index of glass relative to water is, `W^(mu)G = (mu_(g))/(mu_(w)) = (sqrt(2))/((4)/(3)) = (3sqrt(2))/(4)` If the critical angle between the prism and water is `theta_(c)`. then, `theta_(c). = sin^(-1) = (1)/(w^(mu)g) = sin^(-1)((4)/(3sqrt(2))) = 70.53^(@)` But the angle of incidence at R is `45^(@)` [Fig. 2.64(b)] and it is less than the critical angle. So at R, light ray will not be totally reflected. It will be refracted and will enter water. If the angle of refraction is r, then `sqrt(2) sin45^(@) = (4)/(3) SINR` `therefore "" sinr = sqrt(2) XX (3)/(4) xx (1)/(sqrt(2)) = 0.75 = sin 48.59^(@)` `or, "" r = 48.59^(@)` |
|