Question

The maximum number of possible interference maxima for slit separation equal to twice the wavelength in youngs double slit experiment is.

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

Answer 1

The correct option is B Five

$-d\sin \theta =n\lambda$
$d=2\lambda$
$2\lambda \sin \theta =n\lambda$
$2\sin \theta =n$
When $\sin \theta =1,n=2\times 1=2$
The interference maxima will be formed when $n=0,\pm 1,\pm 2$
Therefore the maximum number of possible interference maxima =5