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)

five

For interferance maxima in YDSE
$dsin\theta =n\lambda$
Here $d=2\lambda$
$\therefore sin\theta =\frac{4}{2}$
So, $n$ can take $5$ integral values
i,e. $(-2,-1,0,1,2)$ because $sin\theta$ varies from $-1$ to $+1$.
So, therefore $5$ possible interference maxima.