91Ó°ÊÓ

The given data is listed below as-

Ratio of $d$tocauses diffraction to eliminate the fourth bright side fringe.

The two phenomena in the double-slit experiment are:

(1) Due to the difference in the two paths, there is interference and

(2) The diffraction is there in the single silt.

(a)

The condition for diffraction is:

$a\sin \left(\mathrm{θ}\right)=\mathrm{λ}$ (1)

Now, the angular locations of the bright fringes are:

$d\sin \left(\mathrm{θ}\right)=m_2\mathrm{λ}$ (2)

Now, combining the equations (1) and (2).

$m_2=\frac{d}{a}$

Here, $a$ is the width of silt and $d$ is the separation between the silts.

Now, to substitute $m_2=4$ to eliminate the fourth bright side fringe.

Therefore,

$$\begin{gathered} 4=\frac{d}{a} \\ d=4a \end{gathered}$$

Thus, largest ratio of $d$ to that causes diffraction to eliminate the fourth bright side fringe is 4.

(b)

The minima in the diffraction pattern is given by-

$a\sin \left(\mathrm{θ}\right)=m_1\mathrm{λ}$….. (3)

Now, divide (2) over (3) to obtain:

$\frac{m_2}{m_1}=\frac{d}{a}$…… (4)

Also, substituting $\frac{d}{a}=4$in above equation,

Therefore, $m_2=4m_1$

Thus, the fringes that are missing are 4^th , 8^th, 12^th and so on.

(c)

Substitute $m_2=4$in equation (4) to eliminate the fourth bright fringe:

As, $\frac{m_2}{m_1}=\frac{d}{a}$

Therefore,

$\frac{d}{a}=\frac{4}{m_1}$

Now, according to above relation, there are four values of $m_1$ at which $d>a$

Thus, to eliminate the exactly fourth maximum there are four ratios.