Assertion: In Young's double slit experiment, | vedclass

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(B) Let the two slits be $S_1$ and $S_2$. The point $P$ is directly opposite to slit $S_1$. Thus,$S_1P = D$.The distance $S_2P = \sqrt{D^2 + d^2} = D(

Vedclass · Accepted Answer

If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.

Vedclass · Answer

(B) Let the two slits be $S_1$ and $S_2$. The point $P$ is directly opposite to slit $S_1$. Thus,$S_1P = D$.The distance $S_2P = \sqrt{D^2 + d^2} = D(1 + \frac{d^2}{D^2})^{1/2}$.Using the binomial theorem for $d The path difference $\Delta x = S_2P - S_1P = (D + \frac{d^2}{2D}) - D = \frac{d^2}{2D}$.For a dark fringe,the path difference must be an odd multiple of $\lambda/2$. For the first dark fringe,$\Delta x = \frac{\lambda}{2}$.Equating the two,$\frac{d^2}{2D} = \frac{\lambda}{2}$,which gives $\lambda = \frac{d^2}{D}$.Thus,$\lambda \propto d^2$. The Assertion is correct.The Reason states that for a dark fringe,the intensity is zero,which is a correct statement,but it does not explain why the wavelength is proportional to $d^2$ in this specific geometric configuration. Therefore,the Reason is not the correct explanation of the Assertion.

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$A$ double slit experiment is immersed in water of refractive index $1.33$. The slit separation is $1 \,mm$, and the distance between the slit and the screen is $1.33 \,m$. The slits are illuminated by light of wavelength $6300 \,Å$. The fringe width is:

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