Consider a Young’s double slit experiment as shown in figure. What should be the slit separation d in terms of wavelength λ such that the first minima occurs directly in front of the slit (S1)? (a) (λ)/(2(5 - √(2))) (b) (λ)/(2(√(5) - 2)) (c) (λ)/((5 - √(2))) (d) (λ)/((√(5) - 2))

Answer: (b). S1S2P2d√5 ddD = 2d P is directly opposite S₁ with screen at 2d: S₁P = 2d. S₂P = √((2d)² + d²) = √(5) d (exact, since D is not d). First minimum: (√(5) - 2)d = (λ)/(2) ⇒ d = (λ)/(2(√(5) - 2)) So the answer is (b) (λ)/(2(√(5) - 2)).

Consider a Young’s double slit experiment as shown in figure. What… | JEE Main 2019 (10 Jan, Shift 2)

Q61JEE Main 2019 · 10 JAN · SHIFT 2Medium

YDSE: Film, Medium & Two Wavelengths

Consider a Young’s double slit experiment as shown in figure. What should be the slit separation d in terms of wavelength $\lambda$ such that the first minima occurs directly in front of the slit (S1)?

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Answer: (b)

P is directly opposite $S_1$ with screen at $2d$ : $S_1P = 2d$ .
$S_2P = \sqrt{(2d)^2 + d^2} = \sqrt{5}\,d$ (exact, since $D$ is not $\gg d$ ).
First minimum: $(\sqrt{5} - 2)d = \dfrac{\lambda}{2} \Rightarrow d = \dfrac{\lambda}{2(\sqrt{5} - 2)}$
So the answer is (b) $\frac{\lambda}{2(\sqrt{5} - 2)}$ .

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