Chapter 38, Problem 38.4P

A horizontal laser beam of wavelength 632.8 nm has a circular cross section 2.00 nun in diameter. A rectangular aperture is to lie placed in the center of the beam so that when the light falls perpendicularly on a wall 4.50 m away, the central maximum fills a rectangle 110 mm wide and 6.00 mm high. The dimensions are measured between the minima bracketing the central maximum. Find the required (a) width and (b) height of the aperture. (c) Is the longer dimension of the central bright patch in the diffraction pattern horizontal or vertical? (d) Is the longer dimension of the aperture horizontal or vertical? (e) Explain the relationship between these two rectangles, using a diagram.

To determine

The width of the aperture.

Answer to Problem 38.4P

The width of the aperture is $51.77\text{μm}$.

Explanation of Solution

Given info: The wavelength of the laser beam is $632.8\text{nm}$ for $\lambda$, the diameter of the beam is $2.00\text{mm}$, the distance to the wall is $4.50\text{m}$, the width of the rectangle is $110\text{mm}$ and the height of the rectangle is $6.00\text{mm}$.

Write the expression for the destructive interference.

$\sin \theta =m\frac{\lambda }{a}$ (1)

Here,

$m$ is the order.

$\lambda$ is the wavelength of the light.

$a$ is the width of the aperture.

Write the expression for the distance of the minimum from the central maximum.

$y=\tan \theta \times L$

Here,

$\theta$ is the angle with the horizontal.

$L$ is the distance of the screen from the slit.

The tangent is approximately equal to the sine if the angle is very small.

Substitute $\sin \theta$ for $\tan \theta$ in above equation.

$\begin{array}{c}y=\sin \theta \times L\\ \sin \theta \approx \frac{y}{L}\end{array}$ (2)

Write the expression for the width of the central maximum.

$\begin{array}{c}w=2y\\ y=\frac{w}{2}\end{array}$

Here,

$y$ is the distance of the first minimum from the central maximum.

Equate equation (1) and equation (2).

$\frac{y}{L}=m\frac{\lambda }{a}$ (3)

Substitute $\frac{w}{2}$ for $y$ in the above equation.

$\frac{\frac{w}{2}}{L}=m\frac{\lambda }{a}$

Substitute $4.50\text{m}$ for $L$, $632.8\text{nm}$ for $\lambda$, $1$ for $m$ and $110\text{mm}$ for $w$ in the above equation.

$\begin{array}{c}\frac{\frac{110\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}}{2}}{4.50\text{m}}=1\times \frac{632.8\text{nm}\times \frac{1\text{m}}{{10}^{{}^9}\text{nm}}}{a}\\ a\approx 51.77\times {10}^{-6}\text{m}\times \frac{1\text{μm}}{{10}^{-6}\text{m}}\\ =51.77\text{μm}\end{array}$

Conclusion:

Therefore, the width of the aperture is $51.77\text{μm}$.

To determine

The height of the aperture.

Answer to Problem 38.4P

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

Explanation of Solution

Given info: The wavelength of the laser beam is $632.8\text{nm}$ for $\lambda$, the diameter of the beam is $2.00\text{mm}$, the distance to the wall is $4.50\text{m}$, the width of the rectangle is $110\text{mm}$ and the height of the rectangle is $6.00\text{mm}$.

Write the expression for the height of the central maximum.

$\begin{array}{c}h=2y\\ y=\frac{h}{2}\end{array}$

Here,

$h$ is the height of the rectangle.

$y$ is the distance of the first minimum from the central maximum.

Substitute $\frac{h}{2}$ for $y$ in the equation (3).

$\frac{\frac{h}{2}}{L}=m\frac{\lambda }{a}$

Substitute $4.50\text{m}$ for $L$, $632.8\text{nm}$ for $\lambda$, $1$ for $m$ and $6.00\text{mm}$ for $h$ in the above equation.

$\begin{array}{c}\frac{\frac{\left(6.00\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}{2}}{4.50\text{m}}=1\times \frac{632.8\text{nm}\times \frac{1\text{m}}{{10}^{{}^9}\text{nm}}}{a}\\ a\approx 949\times {10}^{-6}\text{m}\times \frac{1\text{μm}}{{10}^{-6}\text{m}}\\ =949\text{μm}\end{array}$

Conclusion:

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

To determine

Whether the longer dimension of the central bright patch is horizontal or vertical.

Answer to Problem 38.4P

The longer dimension of the central bright patch is horizontal.

Explanation of Solution

Given info: The wavelength of the laser beam is $632.8\text{nm}$ for $\lambda$, the diameter of the beam is $2.00\text{mm}$, the distance to the wall is $4.50\text{m}$, the width of the rectangle is $110\text{mm}$ and the height of the rectangle is $6.00\text{mm}$.

From the given information, the width of the rectangle in the central bright patch is $110\text{mm}$ and the height of the rectangle in the central bright patch is $6.00\text{mm}$. The value of the height of the rectangle in the central bright patch is smaller than the width of the rectangle in the central bright patch. So, the longer dimension of the central bright patch is $110\text{mm}$. The width lies horizontally and the height is vertical. Thus, the longer dimension of the aperture is horizontal.

Conclusion:

Therefore, the longer dimension of the central bright patch is horizontal.

To determine

Whether the longer dimension of the aperture is horizontal or vertical.

Answer to Problem 38.4P

The longer dimension of the aperture is vertical.

Explanation of Solution

Given info: The wavelength of the laser beam is $632.8\text{nm}$ for $\lambda$, the diameter of the beam is $2.00\text{mm}$, the distance to the wall is $4.50\text{m}$, the width of the rectangle is $110\text{mm}$ and the height of the rectangle is $6.00\text{mm}$.

From part (a), the width of the aperture is $51.77\text{μm}$ and from part (b), the height of the aperture is $949\text{μm}$. The value of the height is larger than the width of the aperture. So, the longer dimension of the aperture is $949\text{μm}$. The width lies horizontally and the height is vertical. Thus, the longer dimension of the aperture is vertical.

Conclusion:

Therefore, the longer dimension of the aperture is vertical

To determine

The relationship between the two rectangles.

Answer to Problem 38.4P

The longer dimension is $18.33$ times the smaller dimension of the aperture.

Explanation of Solution

Given info: The wavelength of the laser beam is $632.8\text{nm}$ for $\lambda$, the diameter of the beam is $2.00\text{mm}$, the distance to the wall is $4.50\text{m}$, the width of the rectangle is $110\text{mm}$ and the height of the rectangle is $6.00\text{mm}$.

From part (a), the width of the aperture is $51.77\text{μm}$ and from part (b), the height of the aperture is $949\text{μm}$. From the given information, the width of the rectangle in the central bright patch is $110\text{mm}$ and the height of the rectangle in the central bright patch is $6.00\text{mm}$.

The smaller distance between aperture edges causes a wider diffraction angle.

Write the expression for the ratio of larger dimension to the smaller dimension of the aperture.

$x=\frac{\text{longer}\text{dimension}}{\text{smaller}\text{dimension}}$

Substitute $949\text{μm}$ for $\text{longer}\text{dimension}$ and $51.77\text{μm}$ for $\text{smaller}\text{dimension}$ in the above equation.

$\begin{array}{c}x=\frac{949\text{μm}}{51.77\text{μm}}\\ \approx 18.33\end{array}$

Thus, the longer dimension is $18.33$ times the smaller dimension of the aperture.

Conclusion:

Therefore, the longer dimension is $18.33$ times the smaller dimension of the aperture.