Question

AC = CO = D, S₁C = S₂C = d << D

A small transparent slab containing material of µ = 1.5 is placed along AS₂ (Figure). 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?

Answer 1

When we introduce a thin transparent plate in front of one of the slits in YDSE, the fridge pattern shifts toward the side where the plate is present.

`(S_2O^')_("new") = (S_2O^' - t)_("air") + t_("plate") = (S_2O^' - t)_("air") + mut_("air")`

`(S_2O^')_("new") = S_2O^' + [mu(t - 1)]`

Path difference, `Δx = (S_2O^')_("new") - S_1O^' = (S_2O^' - S_1O^') + [mu(t - 1)]`

⇒ `Δx = d sin theta + [mu(t - 1)]`  ......(i)

⇒ Additional path difference = `(mu - 1)t`

Here the separation between the slits = S₁S₂ = 2d. Hence for calculating path difference, equation (i) becomes `Δx = 2d sin theta + [mu(t - 1)]`

For the principal maxima, (path difference is zero)

`Δx = 2d  sin theta_0 + [mu(t - 1)]` = 0

`sin theta_0 = - (L(mu - 1))/(2d) = (-L(0.5))/(2d)`  .....[∵ L = d/4]

or ⇒ `sin theta_0 = (-1)/16`

θ₀ is the angular position corresponding to the principal maxima

⇒ `OP = D tan theta_0 ≈ D sin theta_0 = (_D)/16`

For the first minima, the path difference is `+- λ/2`

`Δx = 2d  sin theta_1  +  0.5 L = +- λ/2`

`sin theta_1 = (+- λ/2 - 0.5 L)/(2d) = (+- λ/2 - d/8)/(2d)`

⇒ `sin theta_1 = (+- λ/2 - λ/8)/(2λ) = +- 1/4 - 1/16`  ......[∵ The diffraction occurs if the wavelength of waves is nearly equal to the side width (d)]

On the positive side `sin theta_1^+ = + 1/4 - 1/16 = 3/16`

On the negative side `sin theta_1^- = 1/4 - 1/16 = - 5/16`

The first principal maxima on the positive side is at distance

`D  tan theta"'"_1^+ = D (sin theta"'"_1^+)/sqrt(1 - sin^2 theta_1^') = D 3/sqrt(16^2 - 3^2) = (3D)/sqrt(247)` above point O

The first principal minima on the negative side is at distance

`D  tan theta_1^+ = (5D)/sqrt(16^2 - 5^2) = (5D)/sqrt(231)` below point O.