In two-slit interference, if the slit separation is$\mathbf{14}\mathbf{}\mathit{\mu }\mathit{m}$and the slit widths are each 2.0$\mathit{\mu }$m, (a) how many two-slit maxima are in the central peak of the diffraction envelope and (b) how many are in either of the first side peak of the diffraction envelope?

The given data can be listed below,

• The slit separation is,$d=14\mu m$
• The slit width is,$a=2.0\text{mm}$

A wave bends in the direction of the obstacles when it travels through a tiny slit whose width is similar to the wave's wavelength. The screen's minima and maxima are then created by superimposing this twisted wave. The term for this occurrence is diffraction.

(a)

The initial diffraction minimum's angle,${\theta }_1={\sin }^{-1}\left(\frac{\lambda }{a}\right)$ , is the range that the central diffraction envelope covers$-{\theta }_1<\theta <+{\theta }_1$ -1 b +1. The double-slit pattern's maxima are located at is given by,

${\theta }_m={\sin }^{-1}\left(\frac{m\lambda }{d}\right)$

So, the range of diffraction is given by,

$-{\sin }^{-1}\left(\frac{\lambda }{a}\right)<{\sin }^{-1}\left(\frac{m\lambda }{d}\right)<+{\sin }^{-1}\left(\frac{\lambda }{a}\right)$

Rearrange the range of the diffraction is given by,

$-\frac{d}{a}<m<+\frac{d}{a}$

The ratio of the slit separation and slit width is given by,

$$\begin{gathered} \frac{d}{a}=\frac{14\mu m}{2\mu m} \\ =7 \end{gathered}$$

Substitute all the values in the above,

$-7<m<+7$

Thus, the range shows there will be 13 distinct values of m means 13 maxima’s in central envelope.

(b)

The range within one of the envelope is given by,

$-{\sin }^{-1}\left(\frac{\lambda }{a}\right)<{\sin }^{-1}\left(\frac{m\lambda }{d}\right)<+{\sin }^{-1}\left(\frac{2\lambda }{a}\right)$

This results as$\frac{-d}{a}<m<\frac{d}{a}$ gives $7<m<14$

Since m is an integer, this indicates that$8<m<13$ and that there are 6 different values of m contained within that one envelope.

Thus, there are 6 first side peaks in a diffraction envelope.