A beam of light of wavelength 400 nm is incident normally on a right angled prism. It is observed that the
light just grazes along the surface AC after falling on it. Given that the refractive index of the material of the prism
varies with the wavelength as per the relation
R.I. of prism for any wavelength of light= R.I. of prism + b/(lambda)2 where b is a positive constant,
calculate the value of b and the refractive index of the prism material for a wavelength, lamda= 5000 A0
[(Given = Sin-1 (0.625)]
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As the light just grazes along the surface AC after falling on it, so angle of incidence is equal to the critical angle.
ic = thita = sin^ - 1 (0.625)
sin ic = 0.625
meu = 1 / sin ic = 1 / 0.625 = 1.6
hence, for light of wavelength 400nm,
meu = 1.2 + b / (400nm)^2
meu = 0.4 * (400nm)^2
meu = 0.4 * (4 * 10 - 7)^2 m^2 = 6.4 * 10 ^- 14 m^2
for wavelength lammela = 5000A the refractive index of prism material is
meu = 1.2 + 6.4 * 10 ^-14 / (5 * 10 - 7)^2 = 1.2 + 6.4 / 25
meu = 1.456
ic = thita = sin^ - 1 (0.625)
sin ic = 0.625
meu = 1 / sin ic = 1 / 0.625 = 1.6
hence, for light of wavelength 400nm,
meu = 1.2 + b / (400nm)^2
meu = 0.4 * (400nm)^2
meu = 0.4 * (4 * 10 - 7)^2 m^2 = 6.4 * 10 ^- 14 m^2
for wavelength lammela = 5000A the refractive index of prism material is
meu = 1.2 + 6.4 * 10 ^-14 / (5 * 10 - 7)^2 = 1.2 + 6.4 / 25
meu = 1.456
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