1.

A ray of light covers distance .d. in time t, in air and distance 10 d in time t_2, in medium, then critical angle for medium is .....

Answer»

`sin^(-1)((t_1)/(t_2))`
`sin^(-1)((10t_1)/(t_2))`
`tan^(-1)((t_1)/(t_2))`
`tan^(-1)((10t_1)/(t_2))`

Solution :ABSOLUTE refractive index n = `c/v` and
sinC=`1/n=v/c`
`therefore` sinC=`(d_m//t_2)/(d_a//t_1)`
[`because d_m` and `d_a` are distances COVERED in air and medium]
`thereforesinC=(d_mt_1)/(d_at_2)`
`=(10d xx t_1)/(d xx t_2)`
`(10t_1)/(t_2)`
`therefore C=sin^(-1)((10t_1)/(t_2))`


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