A small transparent slab containing material of μ =1.5 is placed along AS₂ (Fig.10.6). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab.

AC = CO = D, S₁C = S₂C = d << D

Question

A small transparent slab containing material of μ =1.5 is placed along AS₂ (Fig.10.6). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab.

AC = CO = D, S₁C = S₂C = d << D

Philoid · Accepted Answer

Whenever a transparent slab of refractive index μ and thickness t is inserted in the path of the ray the fringes on the screen shifts by (μ-1) t towards the slab.

Given AC = CO = D, S₁C = S₂C = d << D

L=d/4, μ=1.5

∴ the optical path difference is given by

For principal maxima OPD=0, i.e.

The distance from O to principal maxima is,

The path difference for first minima is ± (λ/2)

i.e.

Since diffraction occurs when slit width is equals to the wavelength i.e. λ =d

Considering the positive side,

Considering the negative side,

The distance of first principal minima from O on the positive side is,

The distance of first principal minima from O on the negative side is,

A small transparent slab containing material of μ =1.5 is placed along AS2 (Fig.10.6). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab.AC = CO = D, S1C = S2C = d << D

A small transparent slab containing material of μ =1.5 is placed along AS₂ (Fig.10.6). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab.

AC = CO = D, S₁C = S₂C = d << D