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

The correct option is B Five

For maxima, $dsin\theta =n\lambda$
$\Rightarrow 2\lambda sin\theta =n\lambda$
$2sin\theta =n$
$\therefore sin\theta =\frac{n}{2}$
$\because sin\theta \le 1$
$\Rightarrow \frac{n}{2}\le 1$
$n\le 2$
$n_{max}=2$
$n=0,1,2,$
$Somaximumnumberofpossiblemaxima=5$
$n=0,\pm 1,\pm 2$