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

Hint: In this question, we will directly use the formula for the maxima, observed in a Young’s double slit experiment. Now, by substituting the given values in the equation will help us to get our answer. Further, we will study the basics of Young’s double slit experiment for our better understanding.

Formula used:
$d\sin \theta = n\lambda $

Complete step by step solution:
As we know, for maxima
$d\sin \theta = n\lambda $

Now, substituting the value of d in above equation:
$2\lambda \sin \theta = n\lambda $
$ \Rightarrow 2\sin \theta = n$
$ \Rightarrow \sin \theta = \dfrac{n}{2}$
$ \Rightarrow \sin \theta \leqslant 1$
$ \Rightarrow \dfrac{n}{2} \leqslant 1$
$ \Rightarrow n \leqslant 2$
From above observation we get:
${n_{\max }} = 2$
So, $n = 0,1,2$
Now, the maximum number of possible maxima is 5.
$\therefore n = 0, \pm 1, \pm 2$

Therefore, the correct option is B.

Additional information:
As we know, Young’s double-slit experiment uses two coherent sources of light placed at a small distance apart. These two sources of light are a few orders of magnitude greater than the wavelength of light is used. Young’s double-slit experiment helps us in understanding the superposition of light waves, also it tells about the type of interference taking place.
Constructive and destructive are the two types of inference, which can be observed at the screen which is placed at a few distances. A screen or photo detector is placed at a large distance ’D’ away from the slits.
Due to the distance between the two light sources, there is some path difference observed. This is the path difference between two meetings at a point on the screen. Due to this path difference in Young’s double slit experiment, some points on the screen are bright and some points are dark.
Therefore, we finally observe the constructive and destructive fringes on the screen.
Further, intensity is defined as the power radiated per unit area. Here the area is measured on the plane perpendicular to the direction of propagation of the energy. The S.I unit of intensity is measured in watts per square meter.

Note:
We should remember that in Young’s-Double slit experiment the light source is coherent. Lasers can be used as the light source because of its coherence property. Also, we should know that the intensity can be calculated as the square of the amplitude.