Question

Calculate the thickness of double
refracting plate capable of producing
a path difference of N4 between
ordinary and extraordinary ray. Given
ue =1.532, PO =1.544, 1=6000 Å.
O 1.25 x 10^-2 cm
O 1.25 x 10^3 cm
1.25 x 10^-3 cm
O 1.25 x 10^-4 cm​

Answer 1

Answer:

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Answer 2

Answer:

The double refracting frame which produces a path difference of λ/4 between O-ray and E-ray are called quarter wave plate

Let μ∘ = refractive index of crystal with respect to O- ray.

And μE = refractive index of crystal with respect to E- ray.

Here, O and E represents the ordinary and extraordinay light ray.

And if t represents the thickness of the crystal, then for negative crystal

pathdifference =μ∘t−μEt

(μ∘−μE)t=λ/4

t=λ/4(μ∘−μE)

∴t=λ/4(μ∘−μE)

The above equation gives the thickness of quarter wave plate for the negative crystal.

Also ,for positive crystal,

(μE−μ∘)t=λ/4

∴t=λ/4(μE−μ∘)