Chapter 37, Problem 18P

(a)

To determine

The phase difference between two waves arriving at $P$ when $\theta =0.500°$.

Answer to Problem 18P

The phase difference between two waves arriving at $P$ when $\theta =0.500°$ is $\underset{\_}{13.2\text{rad}}$.

Explanation of Solution

Write the expression for phase difference.

    $\varphi =\frac{2\pi }{\lambda }d\sin \theta$

Here, $\lambda$ is the wavelength, $d$ is the separation of slits and $\theta$ is the angle.

Conclusion:

Substitute $500\times {10}^{-9}\text{m}$ for $\lambda$, $0.120\times {10}^{-3}\text{m}$ for $d$ and $0.500°$ for $\theta$ in the above equation.

    $\begin{array}{c}\varphi =\frac{2\pi }{500\times {10}^{-9}\text{m}}\left(0.120\times {10}^{-3}\text{m}\right)\sin \left(0.500°\right)\\ =13.2\text{rad}\end{array}$

Therefore, the phase difference between two waves arriving at $P$ when $\theta =0.500°$ is $\underset{\_}{13.2\text{rad}}$.

(b)

To determine

The phase difference between two waves arriving at $P$ when $y=5.00\text{mm}$.

Answer to Problem 18P

The phase difference between two waves arriving at $P$ when $y=5.00\text{mm}$ is $\underset{\_}{6.28\text{rad}}$.

Explanation of Solution

Write the expression for phase difference.

    $\varphi =\frac{2\pi }{\lambda }d\left(\frac{y}{L}\right)$

Here, $\lambda$ is the wavelength, $d$ is the separation of slits and $\theta$ is the angle.

Conclusion:

Substitute $500\times {10}^{-9}\text{m}$ for $\lambda$, $0.120\times {10}^{-3}\text{m}$ for $d$, $5.00\times {10}^{-3}\text{m}$ for $y$ and $1.20\text{m}$ for $L$ in the above equation.

    $\begin{array}{c}\varphi =\frac{2\pi }{500\times {10}^{-9}\text{m}}\left(0.120\times {10}^{-3}\text{m}\right)\left(\frac{5.00\times {10}^{-3}\text{m}}{1.20\text{m}}\right)\\ =6.28\text{rad}\end{array}$

Therefore, the phase difference between two waves arriving at $P$ when $y=5.00\text{mm}$ is $\underset{\_}{6.28\text{rad}}$

(c)

To determine

The value of $\theta$ for which the phase difference is $0.333\text{rad}$.

Answer to Problem 18P

The value of $\theta$ for which the phase difference is $0.333\text{rad}$ is $\underset{\_}{1.27\times {10}^{-2}°}$.

Explanation of Solution

Write the expression for phase difference.

    $\varphi =\frac{2\pi }{\lambda }d\sin \theta$

Here, $\lambda$ is the wavelength, $d$ is the separation of slits and $\theta$ is the angle.

Conclusion:

Substitute $500\times {10}^{-9}\text{m}$ for $\lambda$, $0.120\times {10}^{-3}\text{m}$ for $d$ and $0.333\text{rad}$ for $\varphi$ in the above equation.

    $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(500\times {10}^{-9}\text{m}\right)\left(0.333\text{rad}\right)}{2\pi \left(0.120\times {10}^{-3}\text{m}\right)}\right)\\ =1.27\times {10}^{-2}°\end{array}$

Therefore, the value of $\theta$ for which the phase difference is $0.333\text{rad}$ is $\underset{\_}{1.27\times {10}^{-2}°}$.

(d)

To determine

The value of $\theta$ for which the path difference is $0.333\text{rad}$.

Answer to Problem 18P

The value of $\theta$ for which the path difference is $0.333\text{rad}$ is $\underset{\_}{5.97\times {10}^{-2}°}$.

Explanation of Solution

Write the expression for path difference.

    $d\sin \theta =\frac{\lambda }{4}$

Here, $\lambda$ is the wavelength, $d$ is the separation of slits and $\theta$ is the angle.

Conclusion:

Substitute $500\times {10}^{-9}\text{m}$ for $\lambda$ and $0.120\times {10}^{-3}\text{m}$ for $d$ in the above equation.

    $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(500\times {10}^{-9}\text{m}\right)}{4\left(0.120\times {10}^{-3}\text{m}\right)}\right)\\ =5.97\times {10}^{-2}°\end{array}$

Therefore, the value of $\theta$ for which the path difference is $0.333\text{rad}$ is $\underset{\_}{5.97\times {10}^{-2}°}$.