A ray of light travels from an optically denser to rarer medium. The c

1.	A ray of light travels from an optically denser to rarer medium. The critical angle of the twomedia is ‘c’. What is the maximum possible deviation of the ray?
Answer» Answer: Given: A ray of light TRAVELS from an OPTICALLY denser to rarer medium. The critical angle for the two media is C. To find the maximum POSSIBLE deviation of the ray Solution: Critical angle is the angle of incidence for which the angle of refraction is 2 π for a given pair of media. LET θ be the angle of incidence and ϕ the angle of refraction. When the ray passes from denser to rarer medium, the deviation is δ=ϕ−θ. If θ=C then ϕ= 2 π and deviation δ= 2 π −C. This is the maximum deviation. When total internal reflection occurs, the deviation is given by δ=π−2θ. Here δ is maximum when θ is minimum. The minimum value of θ is C. Hence maximum deviation δ=π−2C. solution

A ray of light travels from an optically denser to rarer medium. The critical angle of the twomedia is ‘c’. What is the maximum possible deviation of the ray?

Answer»

Answer:

Given: A ray of light TRAVELS from an OPTICALLY denser to rarer medium. The critical angle for the two media is C.

To find the maximum POSSIBLE deviation of the ray

Solution:

Critical angle is the angle of incidence for which the angle of refraction is

for a given pair of media.

LET θ be the angle of incidence and ϕ the angle of refraction.

When the ray passes from denser to rarer medium, the deviation is δ=ϕ−θ. If θ=C then ϕ=

and deviation δ=

−C. This is the maximum deviation. When total internal reflection occurs, the deviation is given by δ=π−2θ. Here δ is maximum when θ is minimum. The minimum value of θ is C.

Hence maximum deviation δ=π−2C.

solution

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