Question

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

Answer: (B)

Five

The condition for maxima in Young's double slit experiment is $d\sin \theta =n\lambda$ where $d$ is the separation between the slits and $\lambda$ 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 $\theta =\frac{\pi }{2}$.

So, with the given condition, we have, $n=\frac{d\sin \left(90°\right)}{\lambda }=\frac{d}{\lambda }=2$

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