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A beam of light of wavelength 400 nm is incident normally on a right angled prism as shown. 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 2 as per the relation.

Manvendra Singh chahar , 10 Years ago
Grade Upto college level
anser 1 Answers
Saurabh Koranglekar

Last Activity: 4 Years ago

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)2m2= 6.4 * 10- 14m2
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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