Assertion : In Young's double slit experiment the two slits are a… - Vidyadip

Question

Assertion : In Young's double slit experiment the two slits are at distance $d$ apart. Interference pattern is observed on a screen at distance $D$ from the slits. At a point onthe screen when it is directly opposite to one of the slits, a dark fringe is observed. Then, the wavelength of wave is proportional to square of distance of two slits

Reason : For a dark fringe intensity is zero.

✓

Answer

When dark fringe is obtained at the point opposite to one of the slits then

and $\mathrm{S}_{2} \mathrm{P}=\sqrt{\mathrm{D}^{2}+\mathrm{d}^{2}}=\mathrm{D}\left(1+\frac{\mathrm{d}^{2}}{\mathrm{D}^{2}}\right)^{1 / 2}$

$=\mathrm{D}\left(1+\frac{\mathrm{d}^{2}}{2 \mathrm{D}^{2}}\right) \quad$ (By binomial theorem)

Path difference $=\mathrm{S}_{2} \mathrm{P}-\mathrm{S}_{1} \mathrm{P}$

$=D\left(1+\frac{d^{2}}{2 D^{2}}\right)-D=\frac{d^{2}}{2 D}=\frac{\lambda}{2}$

or $\lambda=\frac{\mathrm{d}^{2}}{\mathrm{D}} \Rightarrow \lambda \propto \mathrm{d}^{2}$

Now, intensity of a dark fringe is zero.

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