Chapter 37, Problem 37.18P

In Figure P36.10 (not to scale), let L = 1.20 m and d = 0.120 mm and assume the slit system is illuminated with monochromatic 500-nm light. Calculate the phase difference between the two wave fronts arriving at P when (a) θ = 0.500° and (b) y = 5.00 mm. (c) What is the value of θ for which the phase difference is 0.333 rad? (d) What is the value of θ for which the path difference is λ/4?

Figure P36.10

To determine

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

Answer to Problem 37.18P

The phase difference between the two waves fronts arriving at $P$ when $\theta =0.500°$ is $13.2\text{radian}$

Explanation of Solution

Given info:  The separation between the slits is $0.120\text{mm}$ and the distance between the slit and screen is $1.20\text{m}$ and wavelength of light is $500\text{nm}$.

The given diagram is shown below.

Figure 1

The formula to calculate the phase difference is,

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

Here,

$\lambda$ is the wavelength.

$d$ is the separation between the slits.

$\theta$ is the angle between the point $P$ and horizontal.

Substitute $0.120\text{mm}$ for $d$, $0.500°$ for $\theta$, $500\text{nm}$ for $\lambda$, in the above formula as,

$\begin{array}{c}\varphi =\frac{2\pi }{\lambda }d\sin \theta \\ =\frac{2\pi }{\left(500\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\left(.120\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\sin \left(0.500°\right)\\ =13.2\text{radian}\end{array}$

Conclusion:

Therefore, the phase difference between the two waves fronts arriving at $P$ when $\theta =0.500°$ is $13.2\text{radian}$

To determine

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

Answer to Problem 37.18P

The phase difference between the two waves fronts arriving at $P$ when $y=5.00\text{mm}$ is $6.28\text{radian}$.

Explanation of Solution

Given info:  The separation between the slits is $0.120\text{mm}$ and the distance between the slit and screen is $1.20\text{m}$ and wavelength of light is $6.28\text{radian}$.

The formula to calculate the phase difference is,

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

Here,

$\lambda$ is the wavelength.

$d$ is the separation between the slits.

$\theta$ is the angle between the point $P$ and horizontal.

From the right angle triangle $\sin \theta$ is

$\begin{array}{c}\sin \theta =\frac{\text{perpendicular}}{\text{hypotenuse}}\\ =\frac{y}{L}\end{array}$

Here,

$y$ is the distance from $O$ to $P$.

$L$ is the distance between slit and screen.

Substitute $\frac{y}{L}$ for  $\sin \theta$ in the above formula as,

$\begin{array}{l}\varphi =\frac{2\pi }{\lambda }d\sin \theta \\ =\frac{2\pi }{\lambda }d\left(\frac{y}{L}\right)\end{array}$

Substitute $0.120\text{mm}$ for $d$, $500\text{nm}$ for $\lambda$, $1.20\text{m}$ for $L$, $5.00\text{mm}$ for $y$ in the above formula as,

$\begin{array}{c}\varphi =\frac{2\pi }{\lambda }d\sin \theta \\ =\frac{2\pi }{\left(500\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\left(.120\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\left(\frac{5.00\text{mm}\times \frac{{10}^{-3}\text{m}}{\text{1}\text{mm}}}{1.20\text{m}}\right)\\ =6.28\text{radian}\end{array}$

Conclusion:

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

To determine

The value of $\theta$ for which the path difference is $\frac{\lambda }{4}$.

Answer to Problem 37.18P

The value of $\theta$ for which the path difference is $\frac{\lambda }{4}$ is $1.27\times {10}^{-2}\text{degree}$.

Explanation of Solution

Given info:  The separation between the slits is $0.120\text{mm}$ and the distance between the slit and screen is $1.20\text{m}$ and wavelength of light is $6.28\text{radian}$, phase difference is $0.33\text{radian}$

The formula to calculate the phase difference is,

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

Here,

$\lambda$ is the wavelength.

$d$ is the separation between the slits.

$\theta$ is the angle between the point $P$ and horizontal.

Rearrange the above formula to find $\theta$ is,

$\begin{array}{c}\varphi =\frac{2\pi }{\lambda }d\sin \theta \\ \sin \theta =\frac{\lambda \varphi }{2\pi d}\\ \theta ={\sin }^{-1}\left(\frac{\lambda \varphi }{2\pi d}\right)\end{array}$

Substitute $0.120\text{mm}$ for $d$, $500\text{nm}$ for $\lambda$, $0.33\text{radian}$ for $\varphi$ in the above formula as,

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\lambda \varphi }{2\pi d}\right)\\ ={\sin }^{-1}\left(\frac{500\text{nm}\times \frac{{10}^{-9}\text{mm}}{1\text{mm}}\left(0.33\text{radian}\right)}{2\pi \left(0.120\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}\right)\\ =1.27\times {10}^{-2}\text{degree}\end{array}$

Conclusion:

Therefore, the value of $\theta$ is $1.27\times {10}^{-2}\text{degree}$.

To determine

The value of $\theta$

Answer to Problem 37.18P

The value of $\theta$ is $5.97\times {10}^{-2}\text{degree}$.

Explanation of Solution

Given info:  The separation between the slits is $0.120\text{mm}$ and the distance between the slit and screen is $1.20\text{m}$ and wavelength of light is $6.28\text{radian}$, path difference is $\frac{\lambda }{4}$

The formula to calculate the phase difference is,

$\varphi =\frac{2\pi }{\lambda }d\sin \theta$ (1)

Here,

$\lambda$ is the wavelength.

$d$ is the separation between the slits.

$\theta$ is the angle between the point $P$ and horizontal.

The path difference is $\frac{\lambda }{4}$ as,

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

Substitute $\frac{\lambda }{4}$ for $d\sin \theta$ in the above formula as,

$\begin{array}{c}\varphi =\frac{2\pi }{\lambda }d\sin \theta \\ =\frac{2\pi }{\lambda }\left(\frac{\lambda }{4}\right)\\ =\frac{\pi }{2}\end{array}$

Rearrange the equation (1) to find $\theta$ as,

$\theta ={\sin }^{-1}\left(\frac{\lambda \varphi }{2\pi d}\right)$

Substitute $0.120\text{mm}$ for $d$, $500\text{nm}$ for $\lambda$, $\frac{\pi }{2}$ for $\varphi$ in the above formula as,

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

Conclusion:

Therefore, the value of $\theta$ is. $5.97\times {10}^{-2}\text{degree}$.