Chapter 40, Problem 39SP

A spectrum of white light is obtained with a grating ruled with 2500 lines/cm. Compute the angular separation between the violet $\left({\lambda }_u=400\text{ nm}\right)$ and red $\left({\lambda }_r=700\text{ nm}\right)$ in the (a) first order and (b) second order. (c) Does yellow $\left({\lambda }_y=600\text{ nm}\right)$ in the third order overlap the violet in the fourth order?

To determine

The angular separation between the first order image formed on the screen due to the violet and red wavelength if the spectrum of white light is obtained with a grating rule with $2500\text{ lines/cm}$.

Answer to Problem 39SP

Solution:

$4.33°$

Explanation of Solution

Given data:

The wavelength of the red light ${\lambda }_r$ is $700\text{ nm}$.

The wavelength of the violet light ${\lambda }_u$ is $400\text{ nm}$.

The number of lines in the diffraction grating is $2500{\text{ cm}}^{-1}$.

The order of the spectrum formed on the screen is $1$.

Formula used:

The expression for diffraction grating is written as:

$a\sin {\theta }_m=m\lambda$

Here, $a$ is the width of the slit, $\lambda$ is the wavelength of the light, and ${\theta }_m$ is the angle formed by the $m^{th}$ order of image.

Rearrange for ${\theta }_m$.

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

The value of $m$ varies as $\left(1,2,3,4,\mathrm{...}\right)$.

Explanation:

The width of the slit in the diffraction grating is equal to the inverse of the number of lines per unit. Hence, the value of $a$ is equal to $\frac{1}{2000{\text{ cm}}^{-1}}$.

Rewrite the expression for diffraction grating for the violet light.

${\theta }_1={\sin }^{-1}\left(\frac{m{\lambda }_u}{a}\right)$

Here, ${\lambda }_u$ is the wavelength of the violet light and ${\theta }_1$ is the angle formed by the first image due to the violet light.

Substitute $1$ for $m$, $\frac{1}{2500{\text{ cm}}^{-1}}$ for $a$ and $400\text{ nm}$ for ${\lambda }_u$

$\begin{array}{c}{\theta }_1={\sin }^{-1}\left(\frac{\left(1\right)\left(400\text{ nm}\right)}{\frac{1}{2500{\text{ cm}}^{-1}}}\right)\\ ={\sin }^{-1}\left(\frac{\left(1\right)\left(400\text{ nm}\left(\frac{{10}^{-9}\text{m}}{1\text{ nm}}\right)\right)}{\frac{1}{2500{\text{ cm}}^{-1}}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)}\right)\\ =5.74°\end{array}$

Rewrite the expression for diffraction grating for the red light.

${\theta }_2={\sin }^{-1}\left(\frac{m{\lambda }_r}{a}\right)$

Here, ${\lambda }_r$ is the wavelength of thered light and ${\theta }_2$ is the angle formed by the first image due to thered light.

Substitute $1$ for $m$, $\frac{1}{2500{\text{ cm}}^{-1}}$ for $a$ and $700\text{ nm}$ for ${\lambda }_r$

$\begin{array}{c}{\theta }_2={\sin }^{-1}\left(\frac{\left(1\right)\left(700\text{}\text{ nm}\right)}{\frac{1}{2500{\text{ cm}}^{-1}}}\right)\\ ={\sin }^{-1}\left(\frac{\left(1\right)\left(700\text{}\text{ nm}\left(\frac{{10}^{-9}\text{ m}}{1\text{ nm}}\right)\right)}{\frac{1}{2500{\text{ cm}}^{-1}}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)}\right)\\ ={\sin }^{-1}\left(0.175\right)\\ =10.07°\end{array}$

The angular separation between the first order image due to the violet and red lights is written as:

$\Delta \theta ={\theta }_2-{\theta }_1$

Here, $\Delta \theta$ is the angular separation.

Substitute $10.07°$ for ${\theta }_2$ and $5.74°$ for ${\theta }_1$

$\begin{array}{c}\Delta \theta =\left(10.07°\right)-\left(5.74°\right)\\ =4.33°\end{array}$

Conclusion:

Therefore, the angular separation between the first order image formed on the screen due to the violet and red wavelength is $4.33°$.

To determine

The angular separation between the second order image formed on the screen due to the violet and red wavelength if the spectrum of white light is obtained with a grating rule with $2500\text{ lines/cm}$.

Answer to Problem 39SP

Solution:

$8.95°$

Explanation of Solution

Given data:

The wavelength of the red light ${\lambda }_r$ is $700\text{ nm}$.

The wavelength of the violet light ${\lambda }_u$ is $400\text{ nm}$.

The number of lines in the diffraction grating is $2500{\text{ cm}}^{-1}$.

The order of the spectrum formed on the screen is $2$.

Formula used:

The expression for diffraction grating is written as:

$a\sin {\theta }_m=m\lambda$

Here, $a$ is the width of the slit, $\lambda$ is the wavelength of the light, and ${\theta }_m$ is the angle formed by the $m^{th}$ order of image.

The value of $m$ varies as $\left(1,2,3,4,\mathrm{...}\right)$.

Explanation:

Rewrite the expression for diffraction grating for the violet light.

${\theta }_1={\sin }^{-1}\left(\frac{m{\lambda }_u}{a}\right)$

Here, ${\lambda }_u$ is the wavelength of the violet light and ${\theta }_1$ is the angle formed by the first image due to the violet light.

Substitute $2$ for $m$, $\frac{1}{2500{\text{ cm}}^{-1}}$ for $a$, and $400\text{ nm}$ for ${\lambda }_u$

$\begin{array}{c}{\theta }_1={\sin }^{-1}\left(\frac{\left(2\right)\left(400\text{ nm}\right)}{\frac{1}{2500{\text{ cm}}^{-1}}}\right)\\ ={\sin }^{-1}\left(\frac{\left(2\right)\left(400\text{ nm}\left(\frac{{10}^{-9}\text{m}}{1\text{ nm}}\right)\right)}{\frac{1}{2500{\text{ cm}}^{-1}}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)}\right)\\ =11.54°\end{array}$

Rewrite the expression for diffraction grating for the red light.

${\theta }_2={\sin }^{-1}\left(\frac{m{\lambda }_r}{a}\right)$

Here, ${\lambda }_r$ is the wavelength of the red light and ${\theta }_2$ is the angle formed by the first image due to the red light.

Substitute $2$ for $m$, $\frac{1}{2500{\text{ cm}}^{-1}}$ for $a$, and $700\text{ nm}$ for ${\lambda }_r$

$\begin{array}{c}{\theta }_2={\sin }^{-1}\left(\frac{\left(2\right)\left(700\text{}\text{ nm}\right)}{\frac{1}{2500{\text{ cm}}^{-1}}}\right)\\ ={\sin }^{-1}\left(\frac{\left(2\right)\left(700\text{}\text{ nm}\left(\frac{{10}^{-9}\text{ m}}{1\text{ nm}}\right)\right)}{\frac{1}{2500{\text{ cm}}^{-1}}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)}\right)\\ =20.49°\end{array}$

The angular separation between the second order image due to the violet and red lights is written as:

$\Delta \theta ={\theta }_2-{\theta }_1$

Here, $\Delta \theta$ is the angular separation.

Substitute $20.49°$ for ${\theta }_2$ and $11.54°$ for ${\theta }_1$

$\begin{array}{c}\Delta \theta =\left(20.49°\right)-\left(11.54°\right)\\ =8.95°\end{array}$

Conclusion:

Therefore, the angular separation between the second order image formed on the screen due to the violet and red wavelength is $8.95°$.

To determine

Whether the yellow light (${\lambda }_y=600\text{ nm}$) in the third order overlap with the violet light (${\lambda }_u=400\text{ nm}$) in fourth orderor not.

Answer to Problem 39SP

Solution:

$\text{No}$

Explanation of Solution

Given data:

The wavelength of the yellow light ${\lambda }_y$ is $600\text{ nm}$.

The order of spectrum on the screen is third order.

The wavelength of the violet light ${\lambda }_u$ is $400\text{ nm}$.

The order of spectrum on the screen is fourth order.

The number of lines in the diffraction grating is $2500{\text{ cm}}^{-1}$.

Formula used:

The expression for diffraction grating is written as:

$a\sin {\theta }_m=m\lambda$

Here, $a$ is the width of the slit, $\lambda$ is the wavelength of the light, and ${\theta }_m$ is the angle formed by the $m^{th}$ order of image.

The value of $m$ varies as $\left(1,2,3,4,\mathrm{...}\right)$.

Explanation:

Rewrite the expression for diffraction grating for the yellow light.

${\theta }_1={\sin }^{-1}\left(\frac{m{\lambda }_y}{a}\right)$

Here, ${\lambda }_y$ is the wavelength of the yellow light and ${\theta }_1$ is the angle formed by the first image due to the yellow light.

Substitute $3$ for $m$, $\frac{1}{2500{\text{ cm}}^{-1}}$ for $a$, and $600\text{ nm}$ for ${\lambda }_r$

$\begin{array}{c}{\theta }_2={\sin }^{-1}\left(\frac{\left(3\right)\left(600\text{}\text{ nm}\right)}{\frac{1}{2500{\text{ cm}}^{-1}}}\right)\\ ={\sin }^{-1}\left(\frac{\left(3\right)\left(600\text{}\text{ nm}\left(\frac{{10}^{-9}\text{ m}}{1\text{ nm}}\right)\right)}{\frac{1}{2500{\text{ cm}}^{-1}}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)}\right)\\ ={\sin }^{-1}\left(\frac{9}{20}\right)\\ =26.74°\end{array}$

Rewrite the expression for diffraction grating for the violet light.

${\theta }_2={\sin }^{-1}\left(\frac{m{\lambda }_u}{a}\right)$

Here, ${\lambda }_u$ is the wavelength of the violet light and ${\theta }_2$ is the angle formed by the first image due to the violet light.

Substitute $4$ for $m$, $\frac{1}{2500{\text{ cm}}^{-1}}$ for $a$, and $400\text{ nm}$ for ${\lambda }_u$

$\begin{array}{c}{\theta }_1={\sin }^{-1}\left(\frac{\left(4\right)\left(400\text{ nm}\right)}{\frac{1}{2500{\text{ cm}}^{-1}}}\right)\\ ={\sin }^{-1}\left(\frac{\left(4\right)\left(400\text{ nm}\left(\frac{{10}^{-9}\text{m}}{1\text{ nm}}\right)\right)}{\frac{1}{2500{\text{ cm}}^{-1}}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)}\right)\\ ={\sin }^{-1}\left(\frac{2}{5}\right)\\ =23.6°\end{array}$

The angular separation between the third order image due to the violet and yellow lights is written as:

$\Delta \theta ={\theta }_2-{\theta }_1$

Here, $\Delta \theta$ is the angular separation.

Substitute $23.6°$ for ${\theta }_1$ and $26.74°$ for ${\theta }_2$

$\begin{array}{c}\Delta \theta =\left(26.74°\right)-\left(23.6°\right)\\ =3.14°\end{array}$

Conclusion:

Therefore, the angular separation between the thirdorder image formed on the screen due to the yellow wavelength and the fourth order image formed on the screen due to the violet wavelength is $3.14°$. Hence, they are not overlapped to each other.