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(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 quantity μ for 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
    μ1 sin 1 = μ2 sin 2 = μ3 sin 3 =  μ4 sin 4 = constant
In general, μ sinθ  = constant 

1277_A series of transparent layers of different refractive indices.jpg

Fig 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
μ1 sin θ1 = μ2 sin θ2
When a light ray passes from denser to rarer medium it bends away from the normal as shown in the fig. b above

For a given point object, the image formed by refraction at plane surface is illustrated by the following diagrams.

888_case 1 object in denser medium.jpg

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 = μ1/ μ2
•    If there are a number of slabs with different refractive indices placed between the observer and the object.
Total apparent shift = ΣΔyi

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?

2138_fig 1.jpg

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.
           285_fig 1.jpg

For refraction at C : sin r / sin i  = μ1 = 1.5

sin r = AD / AC = 10/ √( 102 + h2), where h is height of vessel


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