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Two light rays 1 and 2 are incident on t...

Two light rays 1 and 2 are incident on two faces AB and AC on an isosceles prism as shown in the figure. The rays emerge from the side BC. Then, `

A

minimum deviation of ray `1gt`minimum deviation of `ray 2``

B

minimum deviation of ray `1lt`minimum deviation of ray `2`

C

minimum deviation of ray `1=`minimum deviation of ray `2`

D

cannot be determined

Text Solution

Verified by Experts

The correct Answer is:
C
`mu=sin(A+delta_m/2)/sin(A//2)` ,
`mu` and A for both rays are same. Hence, value of `delta_m`
is also same for both rays.
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