1.

The speed of light in media M_(1) and M_(2) are .5 xx 10^(8) ms^(-1) and 2 xx 10^(8)ms^(-1) respectively. A ray travels from medium M_(1) to the medium M_(2) with an angle of incidence theta. The ray suffers total internal reflection. Then the value of the angle of incidence theta is :

Answer»

`sin^(-1)((3)/(4))`
`LT sin^(-1)((3)/(4))`
`=sin^(-1)((3)/(4))`
`lesin^(-1)((3)/(4))`. '

Solution :(a) For TOTAL internal reflection, the angle of incidence of light in denser medium must be greater than the critical angle for the pair of MEDIA in contact, i.e.,
`I gt C`
We have, `(1)/(sin C) = (v_(2))/(v_(1))`
Given `v_(1) = M_(1) = 1.5 xx 10^(8) ms^(-1)`
`v_(2) = M_(2) = 2 xx 10^(8) ms^(-1)`
Here `C = sin^(-1) ((M_(1))/(M_(2)))`
`THEREFORE "" i gt sin^(-1)((M_(-1))/(M_(2)))`
`rArr "" i gt sin^(-1) ((1.5 xx 10^(8))/(2 xx 10^(8)))`
`i gt sin^(-1) ((3)/(4))`.


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