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Question

(B) In this configuration,the path difference between the waves from the two slits at a point $P$ on the screen is given by $\Delta x = d \cos 	heta$

Vedclass · Accepted Answer

$\sqrt{8} D$

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(B) In this configuration,the path difference between the waves from the two slits at a point $P$ on the screen is given by $\Delta x = d \cos 	heta$,where $d = 3 \lambda$ is the separation between the slits and $	heta$ is the angle with the axis of the slits.For a maxima,the path difference must be an integer multiple of the wavelength: $\Delta x = n \lambda$.Since we want the smallest distance $x$ (where $x$ corresponds to $y$ in the diagram),we consider the first possible maxima after the central axis,which corresponds to $n = 1$ (or $n=2$ depending on geometry,but here $n=1$ gives the first non-zero maxima).Thus,$1 \cdot \lambda = 3 \lambda \cos 	heta \Rightarrow \cos 	heta = \frac{1}{3}$.From the geometry of the setup,$\cos 	heta = \frac{D}{\sqrt{D^2 + x^2}}$.Equating the two expressions: $\frac{D}{\sqrt{D^2 + x^2}} = \frac{1}{3}$.Squaring both sides: $\frac{D^2}{D^2 + x^2} = \frac{1}{9} \Rightarrow 9 D^2 = D^2 + x^2$.$x^2 = 8 D^2 \Rightarrow x = \sqrt{8} D$.

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