A beam of linearly polarized light is changed into a circularly polari

1.	A beam of linearly polarized light is changed into a circularly polarized light by passing it through a slice of crystal 0.003 cm thick. Calculate the difference in the refractive index of the two rays in the crystal assuming this to be the minimum thickness that will produce the effect and that the wavelength is
Answer» We know that, when we use quantum wave plate then a linear polarized beam change circular polarized beam. t_1/4= \(\cfrac14\)= \(\cfrac{\lambda}{(H_0-H_e)}\) where, H₀ = refractive index of ordinary wave H_e= refractive index of extra ordinary wave. H₀ - H_e = \(\cfrac{\lambda}{4t_{t/4}}\) H₀ - H_e = \(\cfrac{6\times10^{-5}}{4\times0.003}\) H₀ - H_e = \(\cfrac{6\times10^{-2}}{12}\) H₀ - H_e = 0.5 x 10^-2

A beam of linearly polarized light is changed into a circularly polarized light by passing it through a slice of crystal 0.003 cm thick. Calculate the difference in the refractive index of the two rays in the crystal assuming this to be the minimum thickness that will produce the effect and that the wavelength is

Answer»

We know that, when we use quantum wave plate then a linear polarized beam change circular polarized beam.

t_1/4= \(\cfrac14\)= \(\cfrac{\lambda}{(H_0-H_e)}\)

where, H₀ = refractive index of ordinary wave

H_e= refractive index of extra ordinary wave.

H₀ - H_e = \(\cfrac{\lambda}{4t_{t/4}}\)

H₀ - H_e = \(\cfrac{6\times10^{-5}}{4\times0.003}\)

H₀ - H_e = \(\cfrac{6\times10^{-2}}{12}\)

H₀ - H_e = 0.5 x 10^-2

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