Light of wavelength 4500 Å in air is incident on a plane boundary between air and another medium at an angle 30° with the plane of boundary. As it enters from air into the other medium, it deviates by `15^@` towards the normal. Find refractive index of the medium and also the wavelength of given light in the medium.

Explanation

Angle of incidence ` i=90^@ - 30^@ =60^@ .`As the ray bends towards the normal ,it deviates by an angle
`i-r =15^@ `( given )
` therefore r=45^@ ` Applying snell.s law
`mu_("sir") sin i=mu_("med") sin r `
` therefore 1 xx sin 60^@ = mu xx sin 45^@`
` therefore mu = ( sin 60^@)/(sin 45^@) = ( sqrt( 3) //2)/( 1//2) (or) mu = sqrt(1.5 )`
in terms of wavelengths ,
`mu = sqrt(1.5 ) = ( lamda_("air"))/( lamda_("med")) (or) lamda_("med") = ( lamda_( "air")/( sqrt(1.5 )) = ( 4500 )/( sqrt( 1.5 ))`
` lamda_("med")= 3674 Å`