Question

Q.3. The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment, is-
(A) infinite
(B) five
(C) three
(D) zero

Answer 1

The condition for maxima in Young's double slit experiment is $\text{dsinθ=nλ}$ where $\text{d}$ is the separation between the slits and λ is the wavelength of light used.

The maxima that is farthest from the slits is infinitely up or infinitely down the screen and corresponds to $\text{θ=}\frac{\pi }{2}$

​So, with the given condition, we have, $\text{n=}\frac{dsin{90}^0}{\lambda }=\frac{d}{\lambda }$

Thus, we have two maxima on the screen on either side of the central maxima. Thus, the maximum number of possible maxima observed are 2+1+2=5.