Consider a two slit interference arrangements (figure) such that the distance of the screen from the slits is. half the distance between the slits. Obtain the value of D in terms of X such that the first minima on the screen falls at a distance D from the centre O.

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Vishal Baghel | Contributor-Level 10

7 months ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- T₂P= T₂O+OP= D+x

T₁P= T₁O-OP= D-x

S₁P $\frac{λ}{2}$ = $\sqrt[]{{{(S}_{1} T_{1})}^{2} + {(P T_{1})}^{2}} = \sqrt[]{D^{2} + {(d - x)}^{2}}$

S₂P= $\sqrt[]{{{(S}_{2} T_{2})}^{2} + {(P T_{2})}^{2}} = \sqrt[]{D^{2} + {(d + x)}^{2}}$

The minima will occur when S₂P- S₁P= (2n-1) $\frac{λ}{2}$

$\sqrt[]{D^{2} + {(d + x)}^{2}}$ - $\sqrt[]{D^{2} + {(d - x)}^{2}}$ = $\frac{λ}{2}$

$x = D$

[D²+4D²]^1/2-[D²+0]^1/2= $\frac{λ}{2}$

[5D²]^1/2- [D²]^1/2= $\frac{λ}{2}$

$\sqrt[]{5}$ D-D= $\frac{λ}{2}$

$a f t e r s o l v i n g$ we get D= 0.404 $λ$

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In Young’s double slit experiment the two slits are 0.6mm distance apart. Interference pattern is observed on a screen at a distance 80cm from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light will be__________nm.

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3d = 0.6mm

D = 80cm

= 800mm

Path difference is given by

BP – Andhra Pradesh = Dx

$\Rightarrow \frac{d y}{D} = (2 n + 1) \frac{λ}{2}$ [for Dark fringe at P]

n = 0, for first dark fringe

$\frac{d y}{D} = \frac{λ}{2}$

$λ = \frac{2 d}{D} y$

$= \frac{2 d}{D} \times \frac{d}{2} [y = \frac{d}{2}, G i v e n$ first dark fringe is observed on the screen directly opposite to one of the slits]

$λ = \frac{2 \times 0.6 \times 0.6}{800 \times 2}$

$λ = 450 m m$

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