Chapter 35, Problem 14PQ

(a)

To determine

The linear distance on the screen between the central maximum and the $n=2$ minimum.

Answer to Problem 14PQ

Therefore the linear distance on the screen between the central maximum and the $n=2$ minimum is $4.75\text{cm}$.

Explanation of Solution

Write the expression for the path difference for bright fringes in young’s double slit experiment.

$d\sin {\theta }_{\text{n}}=\left(n+\frac{1}{2}\right)\lambda$ (I)

Here, $\lambda$ is the wavelength of microwaves, $d$ is the separation between two slits and $n$ is the integer.

Calculate the angle made by a particular interference fringe on the screen from the central maximum.

$\tan {\theta }_{\text{n}}=\frac{y_n}{x}$ (II)

Here, $y_{\text{n}}$ is the distance of the particular fringe from the central maximum, ${\theta }_{\text{n}}$ is the angle made by the $n^{\text{th}}$  fringe and $x$ is the distance from slits to the screen.

For small angle approximation $\tan \theta \approx \sin \theta$.

Substitute $\left(n+\frac{1}{2}\right)\frac{\lambda }{d}$ for $\tan \theta$ in equation (II).

$\begin{array}{c}\frac{y_{\text{n}}}{x}=\left(n+\frac{1}{2}\right)\frac{\lambda }{d}\\ y_{\text{n}}=\left(n+\frac{1}{2}\right)\frac{\lambda x}{d}\end{array}$ (III)

Conclusion:

Substitute $50.0\text{μm}$ for d, $633\text{nm}$ for $\lambda$, $2$ for n and $1.50\text{m}$ for x in equation (III) to calculate $y_2$.

$\begin{array}{c}{y}_2=\frac{\left(2+\frac{1}{2}\right)\times \left(633\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\left(1.50\text{m}\right)}{50.0\text{μm}\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)}\\ =\frac{\left(\frac{5}{2}\right)\times \left(633\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\left(1.50\text{m}\right)}{50.0\text{μm}\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)}\\ =4.75\times {10}^{-2}\text{m}\left(\frac{{10}^2\text{cm}}{1\text{m}}\right)\\ =4.75\text{cm}\end{array}$

Therefore the linear distance on the screen between the central maximum and the $n=2$ minimum is $4.75\text{cm}$.

(b)

To determine

The linear distance on the screen between the central maximum and the $n=20$ minimum.

Answer to Problem 14PQ

The linear distance on the screen between the central maximum and the $n=20$ maximum is $40.2\text{cm}$.

Explanation of Solution

Write the expression for the path difference for dark fringes in young’s double slit experiment.

$\begin{array}{c}d\sin {\theta }_{\text{n}}=\left(n+\frac{1}{2}\right)\lambda \\ \sin {\theta }_{\text{n}}=\left(n+\frac{1}{2}\right)\frac{\lambda }{d}\\ {\theta }_{\text{n}}={\sin }^{-1}\left(\left(n+\frac{1}{2}\right)\frac{\lambda }{d}\right)\end{array}$ (IV)

Here, $\lambda$ is the wavelength of microwaves, $d$ is the separation between two slits and $n$ is the integer.

Consider the equation (II)

$\begin{array}{c}\tan {\theta }_{\text{n}}=\frac{y_{\text{n}}}{x}\\ y_{\text{n}}=x\tan {\theta }_{\text{n}}\end{array}$ (V)

Here, $y_{\text{n}}$ is the distance of the particular fringe from the central maximum, ${\theta }_{\text{n}}$ is the angle made by the $n^{\text{th}}$  fringe and $x$ is the distance from slits to the screen.

Conclusion:

Substitute $633\text{nm}$ for $\lambda$, $20$ for $n$ and $50.0\text{μm}$ for $d$ in equation (IV) to calculate ${\theta }_{20}$.

$\begin{array}{c}{\theta }_{20}={\sin }^{-1}\left(\frac{\left(20+\frac{1}{2}\right)\left(633\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{50.0\text{μm}\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)}\right)\\ ={\sin }^{-1}\left(\frac{\left(\frac{41}{2}\right)\left(633\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{50.0\text{μm}\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)}\right)\\ =15.0°\end{array}$

Substitute $1.50\text{m}$ for $x$, $20$ for $n$ and $15.0°$ for ${\theta }_{20}$ in equation (V) to calculate $y_{20}$.

$\begin{array}{c}{y}_{20}=\left(1.50\text{m}\right)\tan \left(15.0°\right)\\ =0.402\text{m}\left(\frac{{10}^2\text{cm}}{1\text{m}}\right)\\ =40.2\text{cm}\end{array}$

Therefore, the linear distance on the screen between the central maximum and the $n=20$ maximum is $40.2\text{cm}$.

(c)

To determine

Whether the minima are evenly spaced.

Answer to Problem 14PQ

No, the minima are not evenly spaced.

Explanation of Solution

No, the minima are not evenly spaced. The fringes with small angles have equal separation but when the integer n increases the separation between the subsequent minima increases. That is why the ${20}^{\text{th}}$ minima will have more angular separation than the $2^{\text{nd}}$ minima.