Light of wavelength 4500 A^@ in air in incident on a plane boundary between air and another medium at an angle 30^@ with the plane boundary. As if 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?

Light of wavelength 4500 A^@ in air in incident on a plane boundary be

1.	Light of wavelength 4500 A^@ in air in incident on a plane boundary between air and another medium at an angle 30^@ with the plane boundary. As if 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?
Answer» Solution :Angle of INCIDENCE `i=90^@-30^@=60^@` As the ra bends towards the normal, it deviates by an angle `i-r=15^@` (given) `therefore r=45^@` Applying snell.s LAW `mu_(air) SIN I=mu_(med) sin r` `therefore 1 TIMES sin 60^@=mu times sin 45^@` `therefore mu = (sin 60^@)/(sin 45^@) =(sqrt3//2)/(1//sqrt2)(or) mu =sqrt1.5` In terms of wavelength `mu=sqrt1.5=lamda_(air)/lamda_(med) (or) lamda_(med)= lamda_(air)/sqrt1.5=4500/sqrt1.5=3674 A^@`