Question

The maximum number of possible interference maxima when slit separation is equal to $4$ times the wavelength of light used in a double slit experiment is

• A. $\infty$
• B. 9
• C. 8
• D. 4

Answer 1

Answer: (B)

$d\sin \theta =n\lambda$
$\sin \theta =\frac{n}{4}$
maximum values of $\sin \theta$ can not exceed \pm 1 So possible value os $n$ is $0,\pm 1,\pm 2,\pm 3,\pm 4$ answer is 9