Chapter 27, Problem 7P

In Figure P27.7 (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?

To determine

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

Answer to Problem 7P

Phase difference is $13.2\text{rad}$.

Explanation of Solution

Write the relation between phase difference and path difference.

    $\varphi =\frac{2\pi }{\lambda }\delta$        (I)

Here, $\varphi$ is the phase difference, $\lambda$ is the wavelength and $\delta$ is the path difference.

Write the expression for $\delta$.

  $\delta \approx d\sin \theta$        (II)

Here, $d$ is the separation between slits and $\theta$ is the angle of diffraction.

Write the expression for $\delta$ in terms of separation between slits and distance from central maxima to the fringe.

    $\delta \approx d\frac{y}{L}$        (III)

Here, $d$ is the separation between slits, $y$ is the distance from central maxima to the fringe and $L$ is the distance from slit to screen.

Rewrite the equation (I) by substituting equation (II).

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

Conclusion:

Substitute $500\text{nm}$ for $\lambda$, $0.120\text{mm}$ for $d$ and $0.500°$ for $\theta$ in equation (IV).

    $\begin{array}{c}\varphi =\frac{2\pi }{500\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\left(0.120\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)\left(\text{sin}0.500°\right)\\ =13.2\text{rad}\end{array}$

Thus, the phase difference is $13.2\text{rad}$.

To determine

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

Answer to Problem 7P

The phase difference is $6.28\text{rad}$.

Explanation of Solution

Rewrite equation (I) by substituting equation (III).

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

Conclusion:

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

    $\begin{array}{c}\varphi =\frac{2\pi }{500\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\left(0.120\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)\left(\frac{5.00\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}{1.20\text{m}}\right)\\ =6.28\text{rad}\end{array}$

Thus, the phase difference is $6.28\text{rad}$.

To determine

The phase difference between waves at point P when $\theta =0.333\text{rad}$.

Answer to Problem 7P

The phase difference is $0.0127°$.

Explanation of Solution

Rewrite equation (IV) in terms of $\theta$.

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

Conclusion:

Substitute $500\text{nm}$ for $\lambda$, $0.120\text{mm}$ for $d$ and $0.333\text{rad}$ for $\theta$ in the above equation.

    $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(500\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)\left(0.333\text{rad}\right)}{2\pi \left(0.120\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)}\right)\\ =0.0127°\end{array}$

Thus, the phase difference is $0.0127°$.

To determine

The value of $\theta$ corresponding to $\frac{\lambda }{4}$.

Answer to Problem 7P

The value of $\theta$ is $0.0597°$.

Explanation of Solution

Write the relation between path difference and wavelength.

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

Rewrite the above expression in terms of $\theta$.

    $\theta ={\sin }^{-1}\left(\frac{\lambda }{4d}\right)$

Conclusion:

Substitute $500\text{nm}$ for $\lambda$ and $0.120\text{mm}$ for $d$ in the above equation to find $\theta$.

    $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{500\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{4\left(0.120\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)}\right)\\ =0.0597°\end{array}$

Thus, the value of $\theta$ is $0.0597°$.