In standard YDSE (Ddλ) with identical slits S1 and S2 light reaching a point A on the screen opposite to slit S2 has an intensity I. It was also found that when only one of the two slits S1 and S2 was illuminated by same light beam, the intensity at A is still I. Now when a third slit S3 of four times the slit width is made as shown (S1S2 = S2S3 = d) and all the three slits are illuminated, then intensity of light reaching A is nI where n =

Question

In standard YDSE (D>>d>>λ) with identical slits S1 and S2 light reaching a point A on the screen opposite to slit S2 has an intensity I. It was also found that when only one of the two slits S1 and S2 was illuminated by same light beam, the intensity at A is still I. Now when a third slit S3 of four times the slit width is made as shown (S1S2 = S2S3 = d) and all the three slits are illuminated, then intensity of light reaching A is nI where n =

Infinity Learn · Accepted Answer

If we denote light leaching A team each slit S₁and S₂ phase of amplitude E₀ . Then phase difference for light from S₁and S₂ should be 120° and S₁ should be in phase with phase of S₃ being 2E₀. So resultant is
$\begin{array}{l} \sqrt{E_{0}^{2} + {(3 E_{0})}^{2} + 2 E_{0} (3 E_{0}) \cos 120^{0}} \\ = \sqrt{(9 - 2) E_{0}^{2}} = \sqrt{7} E_{0} \\ \Rightarrow intensity = 7 I \end{array}$

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