Question

The maximum numbers 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

The condition of interference maxima is
$d\sin \theta =n\lambda$
$\sin \theta =\frac{n\lambda }{d}$
Given, $d=2\lambda$
$\sin \theta =\frac{n\lambda }{2\lambda }=n/2$
The magnitude of $\sin \theta$ lies between $0$ and $1$
When $n=0,\sin \theta =0\Rightarrow \theta =0$
When $n=1,\sin \theta =1/2\Rightarrow \theta ={30}^o$
When $n=2,\sin \theta =1\Rightarrow \theta ={90}^o$
Thus, there is central maximum $(\theta =0^o)$, on other side of it maxima lie at $\theta ={30}^o$ and $\theta ={90}^o$, so maximum number of possible interference maxima is $5$.