Question

Explain why.

Tutorial question. This question consists of two independent parts, A and B. Consider a plane wave of monochromatic light, which is shone on two identical narrow slits (the widths of the slits are much less than the wavelength of the light). The slits are separated by a distance d. The interference pattern at right is observed on a distant screen. i. If point P marks the center of the screen, what is the path-length-difference between the light from each slits at points Q and S? How would your answer change if point R was the center of the screen?

Answer 1

In youngs double slit experiment, if the center fring is brigth then the path difference for the dark fring is obtained by the follwing equation

$dsin(\theta) = 2(n + 1)\frac{\lambda }{2}$

where n is number of the first dark fring, second dark fring and so on...

Now in the above case, path difference for the first dark fring is (n=1)

$P.D = 2(1 + 1)\frac{\lambda }{2} = 2\lambda$

path difference for the second dark fring is (n=2)

$P.D = 2(2 + 1)\frac{\lambda }{2} = 3\lambda$

Now, path difference between first and second dark fring is:

$3\lambda - 2\lambda = \lambda$

Answer: Path difference between Q and S is equal to the wavelength of light