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Let the refractive index of a denser medium with respect to a rare medium be N12 and its critical angleθC. At an angle of incidence A when light is travelling from denser medium to rarer medium, a part of the light is reflected and the rest is refracted and the angle between reflected and refracted rays is 90. Angle A is given by:
(A) tan1(sinθC)
(B) 1tan1(sinθC)
(C) 1cos1(sinθC)
(D) cos1(sinθC)

Answer
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Hint: To find the angle of incidence we have to learn about the total internal reflection and its condition and about the critical angle we should also have knowledge about the Snell’s law to relate the critical angle with the angle of refraction and incidence.

Complete step by step answer
Angle of incidence is defined as the angle between a ray incident on a surface and the line perpendicular to the surface at the point of incidence.
When a ray of light hits a smooth polished surface, the light ray bounces back, it is called the reflection. The angle at which it bounce back is known as the angle of reflection
When light waves move from one medium to another, there is a change in the direction of waves due to the obstacles, so they bend and continue to pass it is known as refraction. The angle at which it bends is known as angle of refraction.
The relation between the angle of incidence and angle of refraction is given by Snell’s law
Snell's law states that the ratio of the sines of the angles of incidence to angle of refraction is equivalent to the ratio of phase velocities in the two media, or to the reciprocal of the ratio of the refractive indices
sin isin r=vivr or μrμi
Where,
Sin i is the angle of incidence
Sin r is the angle of refraction
μi is the refractive index of the incident medium
μr is the refractive index of the refractive medium
This relationship between the angles of incidence and refraction and the refractive indices of the two media is known as Snell's Law.
The critical angle is the angle of incidence beyond which light rays passing through a denser medium to the surface of a less dense medium are no longer refracted but totally reflected.
The critical angle is given by
sinθc=μrμi
Here sinθc is the critical angle.
Hence here the total internal reflection takes place.
Given,
The refractive index of a denser medium with respect to a rare medium be N12
The critical angle is θC
The angle between refracted and reflected ray is 90
The angle of incidence is A

From the diagram
The angle of refraction is 90A
Critical angle is sinθc=μrμi
We have seen that sin isin r=μrμi
The critical angle becomes
sinθc=sin isin r
Substitute the known values
sinθc=sin Asin (90 - A)
sinθc=sin Acos A
sinθc=tanA
A=tan1sinθc

Hence the correct answer is option (A) tan1sinθc

Note In the diagram we have noted an angle as A other than the incident angle it is the angle of reflection. The angle of reflection is also A because the angle of reflection is equal to the angle of incidence in total internal reflection.
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