A ray of light is incident at an angle of 30º on a glass prism of refractive index 1.5. Calculate the angle through which the ray is minimally deviated in the prism. (The medium surrounding the prism is air)
10.5o
19.5o
21.1o
38.9o
40.5o
Explanation
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Discussions (7)

Required is Minimum Deviation, Dm ( with 'm' written as subscript).
In any case the answer is only half inaccurate as,
Dm = 2(i - r) ..................recall that at min deviation, for a glass prism experiencing double refraction,
i = ½ [A + Dm] for A = 2r or, r = ½A
putting these together proves that ,
Dm = 2(i - r) with i = 30 deg and r as found overleaf
already = 19.47 deg
= 2 ( 30 - 19.47) deg
= 21.1 deg (approx) matching with OPTION C
THANK YOU FOR LEARNING

Is this another new formula to solve this because the formula I know is refractive index=sin1/2(A+D)/sin1/2A and D= 2i-A
Please review and correct me if I am wrong


