REFRACTION AT A PLANE SURFACE

(i) The incident ray, the refracted ray and the normal to the refracting surface at the point of incidence all lie in the same plane.

(ii) The ratio of the sines of the angle of incidence (i) and of the angle of refraction (r) is a constant quantityμfor two given media, which is called the refractive index of the second medium with respect to the first.(sin i / sin r) = constant = μ

When light propagates through a series of layers of different medium as shown in the figure, then the Snell’s law may be written as

μ_{1}sin 1 = μ_{2}sin 2 = μ_{3}sin 3 = μ_{4}sin 4 = constant

In general, μ sinθ = constantFig a (series of transparent layers of different refractive indices), Fig b (A light ray passing from air to water bends toward the normal)

When light passes from rarer to denser medium it bends toward the normal as shown in the fig.

According to Snell’s law

μ_{1}sin θ1 = μ_{2}sin θ_{2}

When a light ray passes from denser to rarer medium it bends away from the normal as shown in the fig. b aboveFor a given point object, the image formed by refraction at plane surface is illustrated by the following diagrams.

The same result is obtained for the other case also. The image distance from the refracting surface is also known as Apparent depth or height.

Apparent Shift

Apparent shift = Object distance from refracting surface – image distance from refracting surface.Δy (apparent shift) t = 1 - μ

^{-1}where t is the object distance and μ = μ_{1}/ μ_{2}

• If there are a number of slabs with different refractive indices placed between the observer and the object.

Total apparent shift = ΣΔy_{i}

Illustration 3: A person looking through a telescope T just sees the point A on the rim at the bottom of a cylindrical vessel when the vessel is empty. When the vessel is completely filled with a liquid (? = 1.5), he observes a mark at the centre, B, of the vessel. What is the height of the vessel if the diameter of its cross-section is 10cm?

Solution:It is mentioned in the problem that on filling the vessel with the liquid, point B is observed for the same setting; this means that the images of point B, is observed at A, because of refraction of the ray at C.

For refraction at C :

sin r / sin i= μ_{1}= 1.5

sin r =AD / AC = 10/ √( 10, where h is height of vessel^{2}+ h^{2})