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For a prism, angle of prism is 60^@ and it.s refractive index is 1.5, find (1) angle of incidence corresponding to the angle of minimum deviation and (2) angle of emergence for angle of maximum deviation. |
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Answer» Solution :For MINIMUM deviation, `r_1 = r_2 and A = r_1 + r_2` `thereforeA = 2r_1` `r_1 = A/2 = 60 /2 = 30^@` Now, n = 1.5 and `n = (sin i)/(sin r_1)` `therefore n sin r_1 = sin i` `therefore 1. 5 xx sin 30^@ = sin i` `therefore 1.5 xx 0.5 = sin i ` `therefore i = 48^@ 35` (2) For maximum deviation , `i = 90^@` ` therefore1.5 = (sin 90^@)/(sin r_1) thereforer_1 = 41^@ 48.` `therefore r_2 = A - r_1 = 60 - 41^@ 48. = 18^@ 12.` `( because r_1 + r_2 =A)` `therefore1.5 sin r_2 sin e (because n sin r_2 = sin e)` `thereforesin e = 0.4685` `thereforee = 27.9^@` |
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