Chapter 38, Problem 4P

(a)

To determine

The width of aperture.

Answer to Problem 4P

The width of aperture is $51.8\text{μm}$.

Explanation of Solution

On looking at the figure P38.4, width of the rectangular patch is more than that of its height.

Write the equation for tangent of angular width of aperture.

    $\tan {\theta }_{\text{width}}=\frac{y_{\text{width}}}{L}$

Here, ${\theta }_{\text{width}}$ is the angular width of aperture, $y_{\text{width}}$ is half the height of rectangle, and $L$ is the separation between the aperture and wall.

Since ${\theta }_{\text{width}}$ is very small, the following approximation can be used.

    $\tan {\theta }_{\text{width}}\approx \sin {\theta }_{\text{width}}$

Write the relation between width of aperture and the wavelength of light used for first order diffraction pattern.

    $a_{\text{width}}\sin {\theta }_{\text{width}}=\left(1\right)\lambda$

Here, $a_{\text{width}}$ is the width of aperture and $\lambda$ is the wavelength of light used.

Conclusion:

Substitute $\frac{110\text{mm}}{2}$ for $y_{\text{width}}$ and $4.50\text{m}$ for $L$ in the equation for $\tan {\theta }_{\text{width}}$.

    $\begin{array}{c}\tan {\theta }_{\text{width}}=\frac{\left(\frac{110\text{mm}\frac{\text{1}\text{m}}{{\text{10}}^{\text{3}}\text{mm}}}{2}\right)}{4.50\text{m}}\\ =0.0122\end{array}$

Substitute $632.8\text{nm}$ for $\lambda$ and $0.0122$ for $\sin {\theta }_{\text{width}}$ in the equation for $\lambda$.

    $a_{\text{width}}\left(0.0122\right)=\left(1\right)\left(632.8\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)$

Rewrite the above expression in terms of $a_{\text{width}}$.

    $\begin{array}{c}{a}_{\text{width}}=\frac{632.8\times {10}^{-9}\text{m}}{0.0122}\\ =51,868\times {10}^{-9}\text{m}\left(\frac{1\text{μm}}{{10}^{-6}\text{m}}\right)\\ =51.8\text{μm}\end{array}$

Therefore, the width of aperture is $51.8\text{μm}$.

(b)

To determine

The height of aperture.

Answer to Problem 4P

The height of aperture is $949\text{μm}$.

Explanation of Solution

On looking at the figure P38.4, width of the rectangular patch is more than that of its height.

Write the equation for tangent of angular width of aperture.

    $\tan {\theta }_{\text{height}}=\frac{y_{\text{height}}}{L}$

Here, ${\theta }_{\text{height}}$ is the angular height of aperture, $y_{\text{height}}$ is the half the height of rectangle, and $L$ is the separation between the aperture and wall.

Since ${\theta }_{\text{width}}$ is very small, the following approximation can be used.

    $\tan {\theta }_{\text{height}}\approx \sin {\theta }_{\text{height}}$

Write the relation between width of aperture and the wavelength of light used for first order diffraction pattern.

    $a_{\text{height}}\sin {\theta }_{\text{height}}=\left(1\right)\lambda$

Here, $a_{\text{height}}$ is the height of aperture.

Conclusion:

Substitute $\frac{6.00\text{mm}}{2}$ for $y_{\text{width}}$ and $4.50\text{m}$ for $L$ in the equation for $\tan {\theta }_{\text{height}}$.

    $\begin{array}{c}\tan {\theta }_{\text{width}}=\frac{\left(\frac{6.000\text{mm}\left(\frac{\text{1}\text{m}}{{\text{10}}^{\text{3}}\text{mm}}\right)}{2}\right)}{4.50\text{m}}\\ =0.000667\end{array}$

Substitute $632.8\text{nm}$ for $\lambda$ and $0.000667$ for $\sin {\theta }_{\text{height}}$ in the equation for $\lambda$.

    $a_{\text{width}}\left(0.000667\right)=\left(1\right)\left(632.8\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)$

Rewrite the above expression in terms of $a_{\text{height}}$.

    $\begin{array}{c}{a}_{\text{width}}=\frac{632.8\times {10}^{-9}\text{m}}{0.000667}\\ =9.49\times {10}^{-4}\text{m}\left(\frac{1\text{μm}}{{10}^{-6}\text{m}}\right)\\ =949\text{μm}\end{array}$

Therefore, the height of aperture is $949\text{μm}$.

(c)

To determine

Check whether the horizontal or vertical dimension of central bright portion is greater.

Answer to Problem 4P

Horizontal dimension of central bright portion is longer than its vertical dimension.

Explanation of Solution

Draw the diagram showing the diffraction pattern on light passing through a circular aperture.

From the diagram, it can be seen that the central bright patch has an ellipse shape. It has greater length in horizontal direction than in vertical direction.

Therefore, the horizontal dimension of central bright portion is longer than its vertical dimension.

(d)

To determine

Check whether the horizontal or vertical dimension of aperture is greater.

Answer to Problem 4P

Vertical dimension of aperture is greater.

Explanation of Solution

Refer the diagram shown in part (c). From the diagram, it is understood that to obtain diffraction pattern with greater horizontal dimension its vertical length, the vertical length of aperture must be greater than that of horizontal length. If the horizontal dimension of aperture is greater, the vertical dimension of bright becomes greater than that of the horizontal dimension.

Therefore, the vertical dimension of aperture is greater.

(e)

To determine

Identify the relation between the two rectangles given in question with the help of a diagram.

Answer to Problem 4P

The distances between edges of rectangular aperture is inversely proportional to size of central maxima rectangle on the wall.

Explanation of Solution

Refer the figure 1shown in part (c). The size of aperture is inversely proportional to the size of diffraction pattern. Smaller the size of aperture, larger will be the size of diffraction pattern. It is found that the width of aperture is $51.8\text{μm}$ and the height is $\text{949}\text{μm}$. It means that the larger dimension of aperture is around $18.3$ times longer than its smaller dimension.

Therefore, the distances between edges of rectangular aperture is inversely proportional to size of central maxima rectangle on the wall.