Question

In Young’s double slit experiment, light from two identical sources are superimposing on a screen. The path difference between the two lights reaching at a point on the screen is \(\frac{7\lambda}{4}\). The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is : 

(1) \(\frac{1}{2}\)

(2) \(\frac{3}{4}\)

(3) \(\frac{1}{3}\)

(4) \(\frac{1}{4}\)

Answer 1

The correct option is (1) \(\frac{1}{2}\).

\(\Delta x = \frac{7\lambda}{4}\)

\(\therefore \Delta \phi = \frac{2\pi}{\lambda} \Delta x = \frac{7\pi}{2}\)

Now intensity at given point is \(I = 4I_0 cos^2 \frac{\Delta \phi}{2}\) \(= 4I_0 cos^2 \frac{7\pi}{4} = 2I_0\)

\(\therefore \frac{I}{4I_0} = \frac{1}{2}\)