91Ó°ÊÓ

The given data is listed below as-

The wavelength of light, $\mathrm{λ}=600\text{nm}$.

$\sin \left({\mathrm{θ}}_2\right)=0.2$

$\sin \left({\mathrm{θ}}_3\right)=0.3$

The condition maxima of the diffraction grating are:

$dsin\mathrm{θ}=m\mathrm{λ}$

Here, d is the distance between adjacent rulings, m is an integer and is the wavelength of light.

Between two adjacent lines, the order difference is equal to unity.

Therefore, its equation is given by –

$d\sin \left({\mathrm{θ}}_1\right)=m\mathrm{λ}$ ......... (1)

And $d\sin \left({\mathrm{θ}}_2\right)=\left(m+1\right)\mathrm{λ}$ .......... (2)

Now, subtract equation (2) from (1)

Therefore,

$d\left[\sin \left({\mathrm{θ}}_2\right)-\sin \left({\mathrm{θ}}_1\right)\right]=\mathrm{λ}$

$d=\frac{\mathrm{λ}}{\sin \left({\mathrm{θ}}_2\right)-\sin \left({\mathrm{θ}}_1\right)}$

Substitute for$\mathrm{λ}=600nm$, $\sin \left({\mathrm{θ}}_2\right)=0.2$ and $\sin \left({\mathrm{θ}}_3\right)=0.3$ in the above equation.

$$\begin{gathered} d=\frac{600×{10}^{-9}\text{m}}{0.3-0.2} \\ d=6.0×{10}^{-6}\text{m} \end{gathered}$$

Thus, the separation between adjacent slits is $6.0×{10}^{-6}\text{m}$

The minima of the diffraction gratingis given by –

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

Here, is the slit width and the fourth order maxima is missing.

Therefore, set $m=4$in equation (1)

So, $d\sin \left(\mathrm{θ}\right)=4\mathrm{λ}$ ............ (4)

And set $m=1$in equation (3)

So, $a\sin \left(\mathrm{θ}\right)=\mathrm{λ}$ .....…(5)

Now, divide equation (4) and (5)

So, $a=\frac{d}{4}$.

For, $d=6.0×{10}^{-6}\text{m}$

$$\begin{gathered} a=\frac{6×{10}^{-6}\text{m}}{4} \\ a=1.5×{10}^{-6}\text{m} \end{gathered}$$

Thus, the smallest width the grating can have is $1.5×{10}^{-6}\text{m}$.

The maxima of the diffraction grating is given by –

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

Put $\sin \mathrm{θ}=1$to find the largest value of m.

So, $m<\frac{d}{\mathrm{λ}}$.

For,$d=6.0×{10}^{-6}\text{m}$and$\mathrm{λ}=600\text{nm}$

$$\begin{gathered} m<\frac{6×{10}^{-6}\text{m}}{600×{10}^{-9}\text{m}} \\ m<10 \end{gathered}$$

Therefore,the largest value of m can be 9.

Thus, the largest value of the order m of the maxima produced by the grating is 9.

$m<10$

Every 8^th order is missed.

Therefore, for second largest value of order,$m=7$

Thus, a second largest value of the order m of the maxima produced by the grating is 7.

$m<10$.

Therefore, for third largest value of order,$m=6$

Thus, a third largest value of the order m of the maxima produced by the grating is 6.