1.

A ray of light passes from vacuum into amedium of refractive index mu, the angle f incidence is found to be twice the angle of refractive. Then angle of refractive. Then angle of incidences is:

Answer»

`cos^(-1) (MU)/(2)`
`2cos^(-1) (mu)/(2)`
`2 sin^(-1) mu`
`2 sin^(-1) (mu)/(2)`.

SOLUTION :(b) `mu = (sin i)/(sin R), "here" r = i//2`
`thereforemu=(SINI)/(sin""(i)/(2))=(2sin""(i)/(2)cos""(i)/(2))/(sin""(i)/(2))=2cos""(i)/(2)`
`therefore cos""(i)/(2) = (mu)/(2)`
`therefore (i)/(2) = cos^(-1)(mu)/(2)`
`therefore I = 2 cos^(-1)(mu)/(2)`


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