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)

For possible interference maxima on the screen, the condition is
$d\sin \theta =n\lambda$ ...(i)
Given : $d=$ slit - width $=2\lambda$
$\therefore 2\lambda \sin \theta =n\lambda$
$\Rightarrow 2\sin \theta =n$
The maximum value of $\sin \theta$ is 1 , hence, $n=2\times 1=2$
Thus, Eq. (i) must be satisfied by 5 integer values ie, $-2,-1,0,1,2$. Hence, the maximum number of possible interference maxima is $5.$