In an arrangement shown in the figure, d<<D, where d is the distance of separation of the slits S1 and S2 and D is the distance between the slits and the screen. Source S is emitting monochromatic light of wavelength λ. Then (D = 1m)

strong{font-weight:bold !important} Minima will be formed at O, if SS1+S1O−SS2−S2O=nλ2, n=1,2,……For minimum value of d, n = 1∴λ4=1+d2−11+d21/2−1=λ41+d22−1=λ4 (neglecting smaller terms) ⇒d=λ2; β=Dλd=2λ

In an arrangement shown in the figure, d<<D, where d is the distance of separation of the slits S1 and S2 and D is the distance between the slits and the screen. Source S is emitting monochromatic light of wavelength λ. Then (D = 1m)

strong{font-weight:bold !important} Minima will be formed at O, if SS1+S1O−SS2−S2O=nλ2, n=1,2,……For minimum value of d, n = 1∴λ4=1+d2−11+d21/2−1=λ41+d22−1=λ4 (neglecting smaller terms) ⇒d=λ2; β=Dλd=2λ

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In an arrangement shown in the figure, d<<D, where d is the distance of separation of the slits S₁ and S₂ and D is the distance between the slits and the screen. Source S is emitting monochromatic light of wavelength $λ$ . Then (D = 1m)

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corresponding to minimum value of d, the fringe width is $\sqrt{\frac{λ}{2}}$

minimum value of d for the dark fringe at O is $\sqrt{\frac{λ}{2}}$

corresponding to minimum value of d, the fringe width is $\sqrt{2 λ}$

minimum value of d for the dark fringe at O is $\sqrt{\frac{λ}{3}}$

answer is A, C.

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Detailed Solution

Minima will be formed at $O, if$
${SS}_{1} + S_{1} O - {SS}_{2} - S_{2} O = \frac{nλ}{2}, n = 1, 2, \dots \dots$
For minimum value of d, n = 1
$\begin{array}{l} ∴ \frac{λ}{4} = \sqrt{1 + d^{2}} - 1 \\ {(1 + d^{2})}^{1 / 2} - 1 = \frac{λ}{4} \\ 1 + \frac{d^{2}}{2} - 1 = \frac{λ}{4} (neglecting smaller terms) \\ \Rightarrow d = \sqrt{\frac{λ}{2}}; β = \frac{Dλ}{d} = \sqrt{2 λ} \end{array}$