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In the figure A, B and C are three slits each of them individually producing the same intensity I₀ at P when they are illuminated by parallel by parallel beam of light of wavelength $λ$ . It is given that $B P - A P = \frac{λ}{2}$ . Also given that $d ≪ D$ , then wavelength $λ$ and resultant intensity I at P will be

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$λ = \frac{2 d^{2}}{D}$

$I = I_{0}$

$λ = \frac{4 d^{2}}{D}$

$I = 2 I_{0}$

answer is B, D.

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Detailed Solution

$\sqrt{D^{2} + 4 d^{2}} - D = \frac{λ}{2}$

gives, $λ = \frac{4 d^{2}}{D}$

Path difference between CP and AP = $\sqrt{D^{2} + {(3 λ)}^{2}} - D = \frac{9 λ}{8}$

Therefore phase difference between CP and AP = $\frac{2 π}{λ} \times \frac{9 λ}{8} = \frac{9 π}{4} = 360^{0} + 45^{0}$

Resultant intensity at P is due to C only i.e., $I_{R} = I_{0}$

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