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Question 27
Explain the refraction of light through a triangular glass prism using a labelled ray diagram. Hence define the angle of deviation.


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Solution

A sheet of white paper is fixed on a drawing board using drawing pins. A glass prism is placed on it in such a way that it rests on its triangular base. The outline of the prism is traced using a pencil. A straight line PE is drawn inclined to one of the refracting surfaces, AB, of the prism. Two pins are fixed at points P and Q, on the line PE as shown. The images of the pins, fixed at P and Q are looked through the other face AC. Two more pins are fixed at points R and S, such that the pins at R and S and the images of the pins at P and Q lie on the same straight line.

The pins and the glass prism are removed. The points P and Q are joined and the line is extended till it meets the nearest boundary of the prism. Similarly, the points R and S are joined and produced. The lines should meet the boundary of the prism at E and F respectively. E and F are joined. Perpendiculars are drawn to the refracting surfaces AB and AC of the prism at points E and F respectively. The angle of incidence ( i), the angle of refraction ( r) and the angle of emergence ( e) are marked.
It is seen that a ray of light enters from air to glass at the first surface AB. This light ray on refraction bends towards the normal. At the second surface AC, the light ray enters the air from glass and bends away from normal. These rays obey the laws of refraction of light. But the emergent ray is not parallel to the incident ray.
The angle between the direction of the incident ray and that of the emergent ray is called angle of deviation ( D).


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