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Chapter 10: Problem 3

Work out the theory of double-slit Fraunhofer diffraction when the two slits have different widths \(a_{1}=2 a_{2}\) and \(b\), the distance between slit centers, is \(4 a_{2}\).

Short Answer

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**Question**: Using the fundamental concepts of wave optics, calculate the intensity of the light at various points on a screen due to double-slit Fraunhofer diffraction of two slits with different widths, where \(a_1 = 2a_2\) and the distance between their centers, \(b\), is equal to \(4 a_2\). **Short Answer**: To calculate the intensity of the light at various points on the screen, follow these steps: 1. Calculate the path difference, \(\Delta\), using: \(\Delta = b \sin\theta - a_1 \sin\theta\) 2. Calculate the phase difference, \(\delta\), using: \(\delta = \frac{2\pi}{\lambda}(b \sin\theta - a_1 \sin\theta)\) 3. Evaluate the intensity of the light, \(I(\theta)\), using: \(I(\theta) \propto 2A_{0}^2\left(1 + \cos\left(\frac{2\pi}{\lambda}(b \sin\theta - a_1 \sin\theta)\right)\right)\) This will allow you to determine the intensity of the light at any point on the screen due to the double-slit Fraunhofer diffraction interference pattern with different slit widths \(a_1\) and \(a_2\).

Step by step solution

01

Identifying the variables

For this problem, we have two slits with different widths, \(a_1 = 2a_2\), and the distance between the centers of the two slits, \(b = 4a_2\). The light from these slits will interfere to produce a diffraction pattern on a screen located at a distance \(L\) from the slits. The variables we need to consider are: - \(a_1\) and \(a_2\): Widths of the two slits - \(b\): Distance between the centers of the two slits - \(L\): Distance between the slits and the screen - \(y\): Vertical distance of a point on the screen from the center
02

Calculating the path difference

We first need to find the path difference between the waves reaching a point on the screen from both slits. Let the light reaching this point have angles \(\theta_1\) and \(\theta_2\) from the slits. The path difference, \(\Delta\), can be calculated as: \(\Delta = b \sin\theta_2 - a_1 \sin\theta_1\) We have \(\sin\theta_1 \approx \sin\theta_2\) due to the small-angle approximation, as \(L\) is usually much larger than \(b\) and \(a_1\). Thus: \(\Delta = b \sin\theta - a_1 \sin\theta\)
03

Calculating the phase difference

In order to find the intensity of the light on the screen, we need to consider the phase difference between the waves originating from the two slits. The phase difference, \(\delta\), is given by: \(\delta = \frac{2\pi}{\lambda}\Delta\) Where \(\lambda\) is the wavelength of the light and \(\Delta\) is the path difference we calculated in Step 2: \(\delta = \frac{2\pi}{\lambda}(b \sin\theta - a_1 \sin\theta)\)
04

Evaluating the intensity of the light

The total amplitude, \(A\), at a point on the screen, is given by the sum of the amplitudes of the waves from both slits: \(A(\theta) = A_1 + A_2 = A_{0} + A_{0}\text{exp}(i\delta)\) Where \(A_{0}\) is the amplitude of each individual wave, and the phase difference, \(\delta\), is as calculated in Step 3. The intensity, \(I(\theta)\), is proportional to the square of the total amplitude: \(I(\theta) \propto |A(\theta)|^2 = (A_{0} + A_{0}\text{exp}(i\delta))(A_{0} + A_{0}\text{exp}(-i\delta))\) \(I(\theta) \propto 2A_{0}^2(1 + \cos\delta)\) Replacing \(\delta\) with the expression from Step 3, we get the intensity of the light at different points on the screen: \(I(\theta) \propto 2A_{0}^2\left(1 + \cos\left(\frac{2\pi}{\lambda}(b \sin\theta - a_1 \sin\theta)\right)\right)\) This equation gives us the intensity of the light at any point on the screen due to the double-slit Fraunhofer diffraction interference pattern with different slit widths \(a_1\) and \(a_2\).

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Most popular questions from this chapter

Review the elementary arguments for discussing single-slit Fraunhofer diffraction by pairing up portions of the wavefront in the aperture that are \(180^{\circ}\) out of phase to show that (a) a null is observed when the slit may be divided into two equal portions such that $$ \frac{\delta}{2}=\frac{a}{2} \sin \theta=\frac{\lambda}{2} $$ Prob. 10.4.1 (b) Similarly, nulls result when the aperture is divided up into \(2 n\) portions such that $$ \frac{a}{2 n} \sin \theta=\frac{\lambda}{2} $$ where \(n\) is an integer; \((c)\) the secondary maxima occur approximately when the aperture is divided up into \(2 n+1\) portions such that $$ \frac{a}{2 n+1} \sin \theta=\frac{\lambda}{2} $$ and that the intensity of these maxima, relative to the principal maximum, is approximately $$ \frac{1}{2}\left(\frac{1}{2 n+1}\right)^{2} . $$ How do you justify the factor of \(\frac{1}{2}\) ?

Show that when a circular aperture is illuminated by a plane wave whose amplitude varies with radius as \(T(\rho)\), from (9.4.19), the diffraction pattern is given by $$ \psi(u)=2 \pi \breve{C}^{\prime} \int_{0}^{a} T(\rho) J_{c}\left(\frac{u \rho}{a}\right) \rho d \rho, $$ which reduces to \((10.5 .4)\) when \(T(\rho)=1\).

The largest fully steerable radio telescope at the National Radio Astronomy Observatory (Green Bank, W.Va.) has a parabolic mirror \(140 \mathrm{ft}\) in diameter. Approximately what angular resolution does it have for the 1,420-M Hz \((21-\mathrm{cm})\) line radiated by interstellar atomic hydrogen? (In practice, the sensitivity of the detector at the parabola focus is not isotropic. A careful analysis would include a nonuniform aperture-illumination function analogous to Probs. \(10.4 .8\) and 10.4.9.) The mirror was constructed to conform to an ideal paraboloid within a tolerance of \(0.030 \mathrm{in}\). What minimum wavelength can be used with this instrument if the criterion is that the construction errors not exceed \(\lambda / 8\) ? What is the approximate resolution for the minimum wavelength? Answer: 21 minutes of arc; \(\lambda=6 \mathrm{~mm}, 0.6\) minute of arc.

Show that the fraction of the total energy in the circular-aperture diffraction pattern out to the radius specified by \(u_{\operatorname{msx}}\) is $$ \frac{1}{2} \int_{0}^{u_{\max }}\left[\frac{2 J_{1}(u)}{u}\right]^{2} u d u=1-J_{0}^{2}\left(u_{\max }\right)-J_{1}^{2}\left(u_{\max }\right) . $$ Hence, verify the final column of Table \(10.2\).

A hi-fi tweeter has a rectangular aperture 5 by \(12 \mathrm{~cm}\). Which way would you mount this so that the sound pattern is broad in the horizontal plane and narrow in the vertical? What is the approximate radiation pattern at \(10,000 \mathrm{~Hz}\) ?
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