Chapter 27, Problem 60P

(a)

To determine

The angle of the second order ray in diffraction grating.

Answer to Problem 60P

The angle of the second order ray in the diffraction grating that has $400$ grooves/mm is $25.6°$.

Explanation of Solution

Given info: The number of grooves per $\text{mm}$ in the diffraction grating is 400, the incident wavelength is $541\text{nm}$

Formula to calculate angle of $m^{th}$ order ray is ,

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

Here,

$d$ is the grid spacing.

$\theta$ angle of the $m^{th}$ order ray.

$\lambda$ wavelength of the incident light.

First calculate the grid spacing, formula to calculate grid spacing is,

$d=\frac{1}{\text{number}\text{of}\text{grooves}\text{per}\text{metre}}$ (2)

Substitute $400\text{mm}$ for $\text{number}\text{of}\text{grooves}\text{per}\text{metre}$ in equation (2)

$\begin{array}{c}d=\frac{1}{\text{number}\text{of}\text{grooves}\text{per}\text{metre}}\\ =\frac{1}{400\text{mm}}\\ =\frac{1}{400\text{mm}\times \frac{1\text{m}}{{10}^{-3}\text{mm}}}\\ =2.50\times {10}^{-6}\text{m}\end{array}$ (3)

Substitute $2.50\times {10}^{-6}\text{m}$ for $d$, $2$ for $m$ and $541\text{nm}$ for $\lambda$ (1),

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{m\lambda }{d}\right)\\ ={\sin }^{-1}\left(\frac{2\times 541\text{nm}}{2.50\times {10}^{-6}\text{m}}\right)\\ ={\sin }^{-1}\left(\frac{2\times 541\text{nm}\times \frac{1\text{m}}{{10}^9\text{nm}}}{2.50\times {10}^{-6}\text{m}}\right)\\ =25.6°\end{array}$

Conclusion:

Therefore, The angle of the second order ray in the diffraction grating that has $400$ grooves/mm is $25.6°$.

(b)

To determine

The change in the angle of the second order ray in the diffraction grating if the apparatus is in water.

Answer to Problem 60P

If the apparatus is submerged in the water the angle of the second order ray in the diffraction grating that has $400\text{grooves}/\text{mm}$ is $18.9°$

Explanation of Solution

Given info: The number of grooves per $\text{mm}$ in the diffraction grating is 400, the incident wavelength is $541\text{nm}$ The refractive index of the water is $1.33$

From equation (1) formula to calculate angle of $m^{th}$ order ray in water is

$\theta ={\sin }^{-1}\left(\frac{m{\lambda }^{\prime }}{d}\right)$ (4)

Here,

${\lambda }^{\prime }$ is the wavelength of the light in water

The whole apparatus is now submerged into water therefore the wavelength of the incident light will be reduced to factor of the refractive index of water.

The new incident wavelength is

${\lambda }^{\prime }=\frac{\lambda }{n}$ (5)

Here,

$\lambda$ is the wavelength of the incident light.

$n$ is the refractive index of water.

Substitute $1.33$ for $n$, $541\text{nm}$ for $\lambda$ in equation (5),

$\begin{array}{c}{\lambda }^{\prime }=\frac{\lambda }{n}\\ =\frac{541\text{nm}\times \frac{1\text{m}}{{10}^9\text{nm}}}{1.33}\\ =4.06\times {10}^{-7}\text{m}\end{array}$ (6)

From equation (6), substitute $4.06\times {10}^{-7}\text{m}$ for $\lambda$ $2.50\times {10}^{-6}\text{m}$ for $d$, $2$ for $m$, in equation (4).

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{m\lambda }{d}\right)\\ ={\sin }^{-1}\left(\frac{2\times 4.06\times {10}^{-7}\text{m}}{2.50\times {10}^{-6}\text{m}}\right)\\ =18.9°\end{array}$

Conclusion:

Therefore, if the apparatus is submerged in the water the angle of the second order ray in the diffraction grating that has $400$ grooves/mm is $18.9°$

(c)

To determine

To show: The two diffracted rays from part (a) and part (b) are related through the law of refraction.

Answer to Problem 60P

The relation between the diffracted rays for the part (a) and part b is $n_{2}d\sin \theta =n_{1}d\sin \theta$ satisfying the law of refraction.

Explanation of Solution

 for the part (a) from equation (1),

$\begin{array}{l}\theta ={\sin }^{-1}\left(\frac{m\lambda }{d}\right)\\ d\sin \theta =m\lambda \end{array}$ (7)

For second order ray substitute $2$ for $m$ in equation (7)

$\begin{array}{l}d\sin \theta =m\lambda \\ d\sin \theta =2\lambda \end{array}$ (8)

From equation (4) for the second part when the apparatus is submerged in water,

$d\sin \theta =m{\lambda }^{\prime }$ (9)

Form equation (5) Substitute $\frac{\lambda }{n}$ for ${\lambda }^{\prime }$ in equation (9)

$\begin{array}{l}d\sin \theta =m{\lambda }^{\prime }\\ d\sin \theta =\frac{m\lambda }{n}\\ nd\sin \theta =m\lambda \end{array}$

For second order, ray substitute $2$ for $m$ in the above equation.

$nd\sin \theta =2\lambda$ (10)

Compare equation (8) and (10).

$nd\sin \theta =d\sin \theta$

Substitute $n_2$ for $n$ in left hand side of the equation $n_1$ for $1$ in the right hand side of the above equation.

$\begin{array}{l}nd\sin \theta =d\sin \theta \\ n_{2}d\sin \theta =1\times d\sin \theta \\ n_{2}d\sin \theta =n_{1}d\sin \theta \end{array}$ (11)

Here,

$n_2$ is the refractive index of the material

$n_1$ is the refractive index of the air.

From equation (11), it is evident that it can be expressed in terms of law of refraction.

Conclusion:

Therefore, the two diffracted rays of part (a) and part (b) is related through law of refraction.