Chapter 37, Problem 37.2OQ

Four trials of Young's double-slit experiment are conducted. (a) In the first trial, blue light passes through two fine slits 400 μm apart and forms an interference pattern on a screen 4 in away, (b) In a second trial, red light passes through the same slits and falls on the same screen. (c) A third trial is performed with red light and the same screen, but with slits 800 μm apart, (d) A final trial is performed with red light, slits 800 μm apart, and a screen 8 m away. (i) Rank the trials (a) through (d) from the largest to the smallest value of the angle between the central maximum and the first-order side maximum. In your ranking, note any cases of equality, (ii) Rank the same trials according to the distance between the central maximum and the First-order side maximum on the screen.

To determine

Ranking of the trials from largest to the smallest angular distance.

Answer to Problem 37.2OQ

The Ranking of the trials from largest to the smallest angular distance is $b>a>c=d$.

Explanation of Solution

Given info: For case (a) and case (b) the slit separation is $400\text{μm}$ and the screen is $4\text{m}$ away, for case (c) the slit separation is $800\text{μm}$ the screen is $4\text{m}$ away and for case (d) the slit separation is $800\text{μm}$ the screen is $8\text{m}$ away.

Write the expression for constructive interference for the bright fringes.

$\begin{array}{c}d\sin \theta =m\lambda \\ \sin \theta =\left(\frac{m\lambda }{d}\right)\end{array}$

Here,

$\sin \theta$ is the angular distance.

$m$ is the order number of fringes.

$d$ is the separation between the fringes.

$\lambda$ is the wavelength.

The value of $m$ is $1$.

Substitute $1$ for $m$ in above equation.

$\begin{array}{c}\sin \theta =\left(\frac{1\times \lambda }{d}\right)\\ \sin \theta =\frac{\lambda }{d}\end{array}$ (1)

Case (a);

The wavelength of the blue light is $450\text{nm}$.

Substitute $400\text{μm}$ for $d$ and $450\text{nm}$ for $\lambda$ in equation (1).

$\begin{array}{c}\sin \theta =\left(\frac{450\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{400\text{μm}\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)}\right)\\ =1.125\times {10}^{-3}\end{array}$

Thus, the value of angular distance is $1.125\times {10}^{-3}$.

Case (b);

The wavelength of the red light is $620\text{nm}$.

Substitute $400\text{μm}$ for $d$ and $620\text{nm}$ for $\lambda$ in equation (1).

$\begin{array}{c}\sin \theta =\left(\frac{620\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{400\text{μm}\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)}\right)\\ =1.55\times {10}^{-3}\end{array}$

Thus, the value of angular distance is $1.55\times {10}^{-3}$.

Case (c);

The wavelength of the red light is $620\text{nm}$.

Substitute $800\text{μm}$ for $d$ and $620\text{nm}$ for $\lambda$ in equation (1).

$\begin{array}{c}\sin \theta =\left(\frac{620\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{800\text{μm}\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)}\right)\\ =0.775\times {10}^{-3}\end{array}$

Thus, the value of angular distance is $0.775\times {10}^{-3}$.

Case (d);

The wavelength of the red light is $620\text{nm}$.

Substitute $800\text{μm}$ for $d$ and $620\text{nm}$ for $\lambda$ in equation (1).

$\begin{array}{c}\sin \theta =\left(\frac{620\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{800\text{μm}\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)}\right)\\ =0.775\times {10}^{-3}\end{array}$

Thus, the value of angular distance is $0.775\times {10}^{-3}$.

The ranking of the angular distances of all the cases is,

$\begin{array}{c}1.55\times {10}^{-3}>1.125\times {10}^{-3}>0.775\times {10}^{-3}=0.775\times {10}^{-3}\\ b>a>c=d\end{array}$

Conclusion:

Therefore, the ranking of the trials from largest to the smallest angular distance is $b>a>c=d$.

To determine

Ranking of the trials on the basis of their distance between the central maximum and the first order side maximum on the screen.

Answer to Problem 37.2OQ

The Ranking of the trials on the basis of their distance between the central maximum and the first order side maximum on the screen is $b=d>a>c$.

Explanation of Solution

Given info: For case (a) and case (b) the slit separation is $400\text{μm}$ and the screen is $4\text{m}$ away, for case (c) the slit separation is $800\text{μm}$ the screen is $4\text{m}$ away and for case (d) the slit separation is $800\text{μm}$ the screen is $8\text{m}$ away.

Write the expression for distance between the central maximum and the screen.

$y=\frac{L\lambda }{d}$ (2)

Here,

$y$ is the distance between the central maximum and the screen.

$d$ is the slit separation.

$\lambda$ is the wavelength.

$L$ is the distance of screen from slit.

Case (a);

The wavelength of the blue light is $450\text{nm}$ and the value of $L$ is $4\text{m}$.

Substitute $400\text{μm}$ for $d$ and $450\text{nm}$ for $\lambda$ in equation (2).

$\begin{array}{c}y=\frac{\left(4\text{m}\right)450\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{400\text{μm}\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)}\\ =4.5\times {10}^{-3}\text{m}\end{array}$

Thus, the value of $y$ is $4.5\times {10}^{-3}\text{m}$.

Case (b);

The wavelength of the red light is $620\text{nm}$ and the value of $L$ is $4\text{m}$.

Substitute $400\text{μm}$ for $d$ and $620\text{nm}$ for $\lambda$ in equation (2).

$\begin{array}{c}y=\frac{\left(4\text{m}\right)620\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{400\text{μm}\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)}\\ =6.2\times {10}^{-3}\text{m}\end{array}$

Thus, the value of $y$ is $6.2\times {10}^{-3}\text{m}$.

Case (c);

The wavelength of the red light is $620\text{nm}$ and the value of $L$ is $4\text{m}$.

Substitute $800\text{μm}$ for $d$ and $620\text{nm}$ for $\lambda$ in equation (2).

$\begin{array}{c}y=\left(4\text{m}\right)\frac{620\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{800\text{μm}\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)}\\ =3.1\times {10}^{-3}\text{m}\end{array}$

Thus, the value of $y$ is $3.1\times {10}^{-3}\text{m}$.

Case (d);

The wavelength of the red light is $620\text{nm}$ and the value of $L$ is $8\text{m}$.

Substitute $800\text{μm}$ for $d$ and $620\text{nm}$ for $\lambda$ in equation (2).

$\begin{array}{c}y=\left(8\text{m}\right)\frac{620\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{800\text{μm}\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)}\\ =6.2\times {10}^{-3}\text{m}\end{array}$

Thus, the value of $y$ is $6.2\times {10}^{-3}\text{m}$.

The ranking of the distance between the central maximum and the screen of all the cases is,

$\begin{array}{c}6.2\times {10}^{-3}\text{m}=6.2\times {10}^{-3}\text{m}>4.5\times {10}^{-3}\text{m}>3.1\times {10}^{-3}\text{m}\\ b=d>a>c\end{array}$

Conclusion:

Therefore, the ranking of the trials on the basis of their distance between the central maximum and the first order side maximum on the screen is $b=d>a>c$.