Question

Light of wavelength 600 nm is incident normally on a diffraction grating. Two adjacent maxima occur at angles given by $\sin \theta =0.2$and$\sin \theta =0.3$ .The fourth order maxima are missing. (a) What is the separation between adjacent slits? (b) What is the smallest slit width this grating can have? For that slit width , what are the (c) largest, (d) second largest, and (e) third largest values of the order of the maxima produced by the grating?

Short answer

a) The separation between adjacent slits is $6.0\times {10}^{-6}\text{m}$.

b) The smallest width the grating can have is$1.5\times {10}^{-6}\text{m}$

c) The largest value of the order m of the maxima produced by the grating is 9.

d) The second largest value of the order m of the maxima produced by the grating is 7.

e) The third largest value of the order m of the maxima produced by the grating is 6.

Step-by-step solution

Identification of the given data

The given data is listed below as-

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

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

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

The condition of the diffraction grating

The condition maxima of the diffraction grating are:

$dsin\theta =m\lambda$

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

To find the separation between adjacent slits(a)

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

Therefore, its equation is given by –

$d\sin \left({\theta }_1\right)=m\lambda$ ......... (1)

And $d\sin \left({\theta }_2\right)=\left(m+1\right)\lambda$ .......... (2)

Now, subtract equation (2) from (1)

Therefore,

$d\left[\sin \left({\theta }_2\right)-\sin \left({\theta }_1\right)\right]=\lambda$

$d=\frac{\lambda }{\sin \left({\theta }_2\right)-\sin \left({\theta }_1\right)}$

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

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

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

To find is the smallest slit width this grating can have(b)

The minima of the diffraction gratingis given by –

$a\sin \left(\theta \right)=m\lambda$ ...........(3)

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

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

So, $d\sin \left(\theta \right)=4\lambda$ ............ (4)

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

So, $a\sin \left(\theta \right)=\lambda$ .....…(5)

Now, divide equation (4) and (5)

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

For, $d=6.0\times {10}^{-6}\text{m}$

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

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

To find largest values of the order m of the maxima produced by the grating(c)

The maxima of the diffraction grating is given by –

$d\sin \left(\theta \right)=m\lambda$

Put $\sin \theta =1$to find the largest value of m.

So, $m<\frac{d}{\lambda }$.

For,$d=6.0\times {10}^{-6}\text{m}$and$\lambda =600\text{nm}$

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

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.

To find second largest values of the order m of the maxima produced by the grating(d)

$m<10$

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.

To find third largest values of the order m of the maxima produced by the grating(e)

$m<10$.

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.