Q. 33

# Consider the situation shown in figure (17-E6). The two slits S1 and S2 placed symmetrically around the central line are illuminated by a monochromatic light of wavelength λ. The separation between the slits is d. The light transmitted by the slits falls on a screen ∑1 placed at a distance D from the slits. The slit S3 is at the central line and the slit S4 is at a distance z from S3. Another screen ∑2 is placed a further distance D away from ∑1. Find the ratio of the maximum to minimum intensity observed on ∑2 if z is equal to(a) (b) (c)  Let the intensity of slits and be . The amplitude through each slit be .

For , the phase difference is zero as it lies in the center of the screen. So here, the intensity( ) would be maximum. For , the path difference is And the corresponding phase difference is (a) For , This is the condition of destructive interference, so the intensity at is zero. Hence, only one of the slits is illuminated and so the pattern on Σ₂ would be uniform with no maxima or minima. As maxima and minima are same, the ratio is one.

(b)For , In this case, constructive interference is taking place at . So the intensity is maximum and is equal to that is .

In this case, slits and will form an interference pattern with alternate dark and bright fringes. Let the maximum intensity be while the minimum will be zero for a dark fringe.

Hence the ratio is .

(c)For , The intensity at is thus .

And intensity at is Let the amplitude at be and at be .

Therefore,  And  So maximum amplitude at Σ₂ would be and minimum amplitude would be .

Hence, the ratio of intensities is  Rationalizing, Rate this question :

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