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Light of wavelength 4500 Åin 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. |
Answer» Solution : Angle of incidence ` i=90^@ - 30^@ =60^@ .`Asthe raybendstowardsthe normal,itdeviatesby anangle `i-r =15^@ `( given ) ` thereforer=45^@ ` Applyingsnell.slaw `mu_("sir") sin i=mu_("MED") sin r ` ` therefore1 xx SIN60^@= muxx sin45^@` ` thereforemu = ( sin 60^@)/(sin 45^@) = ( sqrt( 3) //2)/( 1//2)(or)mu = sqrt(1.5 )` in termsof 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` |
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