In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other…

Q: In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other, then in the interference pattern

(A) the intensities of both the maxima and the minima increases

(B) the intensity of maxima increases and the minima has zero intensity

(C) the intensity of maxima decreases and that of minima increases

(D) the intensity of maxima decreases and the minima has zero intensity

Ans: (A)

Sol: In normal conditions , when widths of both the slits are equal.

I1 = I2 = I

$\large I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 = 4 I$

$\large I_{min} = (\sqrt{I_1} – \sqrt{I_2})^2 = 0$

When width of one of the slit is increased , Intensity due to that slit will be increases while that of other will remain same .

I1 = n I & I2 = I (here n> 1)

$\large I_{max} = I(\sqrt{n}+1)^2 > 4 I $

$\large I_{min} = I(\sqrt{n}-1)^2 > 0 $

intensities of both the maxima and the minima increases .

Author: Rajesh Jha

QuantumStudy.com