1.

A ray of light is incident on a glass slab at the polarising angle of58^(@) . Calculate the percentage change in the speed of light in glass.

Answer»

Solution :Data : ` i_(P) = 58^(@)`
` n = tan i_(P)`
` = tan 58^(@) = 1.6003`
` n = lambda_(a)/lambda_(G)`
` :.lambda_(g)/lambda_(a) = 1/n = 1/(1.6003)`
` :. (lambda_(a)-lambda_(g))/lambda_(a) = (1.6003 - 1)/(1.6003) = (0.6003)/(1.6003)`
` = 0.3753`
`{:(log 0.6003,," "bar(1).7784),(log 1.6003,,UL(-0.2041)),(,,ul(" "bar(1).5743)),(ALbar(1).5743,=," "0.3753):}`
` :. ` The percentage change in the velocity of the light in glass is
`(lambda_(a)-lambda_(g))/lambda_(a) XX 100 = 37.53 %`


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