Question

The maximum number of possible interference maxima for slit separation is equal to twice the wavelength in Young's double-slit experiment, which is

• A. Infinite
• B. Five
• C. Three
• D. Zero

Answer 1

The correct option is B

Five

1. In the Young double-slit experiment, the condition for maxima is$d\sin \theta =n\lambda .$
2. Where$d=$ separation between the slits and$\lambda$ is the wavelength of light used.
3. The maxima that are farthest from the slits are infinitely up or infinitely down the screen and correspond to$\theta =\pi /2.$
4. So, $n=d\sin (90°)/\lambda =d/\lambda =2$.
5. The maximum number of possible maxima observed is$2+1+2=5.$
6. We have two maxima on the screen on either side of the central maxima.

The maximum number of possible maxima is$5.$

Hence, option B is the correct answer.