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प्रश्न
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?
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उत्तर
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.
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