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)



Answer :

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 :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
RELATED QUESTIONS :