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Intensity Pattern of N Slits. (a) Consider an
Chapter 36, Problem 74CP(choose chapter or problem)
Intensity Pattern of N Slits. (a) Consider an arrangement of N slits with a distance d between adjacent slits. The slits emit coherently and in phase at wavelength \(\lambda\). Show that at a time t, the electric field at a distant point P is
\(\begin{aligned} E_{P}(t)=& E_{0} \cos (k R-\omega t)+E_{0} \cos (k R-\omega t+\phi) \\ &+E_{0} \cos (k R-\omega t+2 \phi)+\cdots \\ &+E_{0} \cos (k R-\omega t+(N-1) \phi) \end{aligned}\)
where \(E_{0}\) is the amplitude at P of the electric field due to an individual slit, \(\phi=(2 \pi d \sin \theta) / \lambda, \theta\) is the angle of the rays reaching P (as measured from the perpendicular bisector of the slit arrangement), and R is the distance from P to the most distant slit. In this problem, assume that R is much larger than d. (b) To carry out the sum in part (a), it is convenient to use the complex-number relationship
\(e^{i z}=\cos z+i \sin z\)
where \(i=\sqrt{-1}\). In this expression, cos z is the real part of the complex number \(e^{i z}\), and sin z is its imaginary part. Show that the electric field \(E_{P}(t)\) is equal to the real part of the complex quantity
\(\sum_{n=0}^{N-1} E_{0} e^{i(k R-\omega t+n \phi)}\)
(c) Using the properties of the exponential function that \(e^{A} e^{B}=e^{(A+B)}\) and \(\left(e^{A}\right)^{n}=e^{n A}\), show that the sum in part (b) can be written as
\(\begin{array}{l} E_{0}\left(\frac{e^{i N \phi}-1}{e^{i \phi}-1}\right) e^{i(k R-\omega t)} \\ =E_{0}\left(\frac{e^{i N \phi / 2}-e^{-i N \phi / 2}}{e^{i \phi / 2}-e^{-i \phi / 2}}\right) e^{i[k R-\omega t+(N-1) \phi / 2]} \end{array}\)
Then, using the relationship \(e^{i z}=\cos z+i \sin z\), show that the (real) electric field at point P is
\(E_{P}(t)=\left[E_{0} \frac{\sin (N \phi / 2)}{\sin (\phi / 2)}\right] \cos [k R-\omega t+(N-1) \phi / 2]\)
The quantity in the first square brackets in this expression is the amplitude of the electric field at P. (d) Use the result for the electric field amplitude in part (c) to show that the intensity at an angle \(\theta\) is
\(I=I_{0}\left[\frac{\sin (N \phi / 2)}{\sin (\phi / 2)}\right]^{2}\)
where \(I_{0}\) is the maximum intensity for an individual slit. (e) Check the result in part (d) for the case N = 2. It will help to recall that sin 2A = 2 sin A cos A. Explain why your result differs from Eq. (35.10), the expression for the intensity in two-source interference, by a factor of 4. (Hint: Is \(I_{0}\) defined in the same way in both expressions?)
Questions & Answers
QUESTION:
Intensity Pattern of N Slits. (a) Consider an arrangement of N slits with a distance d between adjacent slits. The slits emit coherently and in phase at wavelength \(\lambda\). Show that at a time t, the electric field at a distant point P is
\(\begin{aligned} E_{P}(t)=& E_{0} \cos (k R-\omega t)+E_{0} \cos (k R-\omega t+\phi) \\ &+E_{0} \cos (k R-\omega t+2 \phi)+\cdots \\ &+E_{0} \cos (k R-\omega t+(N-1) \phi) \end{aligned}\)
where \(E_{0}\) is the amplitude at P of the electric field due to an individual slit, \(\phi=(2 \pi d \sin \theta) / \lambda, \theta\) is the angle of the rays reaching P (as measured from the perpendicular bisector of the slit arrangement), and R is the distance from P to the most distant slit. In this problem, assume that R is much larger than d. (b) To carry out the sum in part (a), it is convenient to use the complex-number relationship
\(e^{i z}=\cos z+i \sin z\)
where \(i=\sqrt{-1}\). In this expression, cos z is the real part of the complex number \(e^{i z}\), and sin z is its imaginary part. Show that the electric field \(E_{P}(t)\) is equal to the real part of the complex quantity
\(\sum_{n=0}^{N-1} E_{0} e^{i(k R-\omega t+n \phi)}\)
(c) Using the properties of the exponential function that \(e^{A} e^{B}=e^{(A+B)}\) and \(\left(e^{A}\right)^{n}=e^{n A}\), show that the sum in part (b) can be written as
\(\begin{array}{l} E_{0}\left(\frac{e^{i N \phi}-1}{e^{i \phi}-1}\right) e^{i(k R-\omega t)} \\ =E_{0}\left(\frac{e^{i N \phi / 2}-e^{-i N \phi / 2}}{e^{i \phi / 2}-e^{-i \phi / 2}}\right) e^{i[k R-\omega t+(N-1) \phi / 2]} \end{array}\)
Then, using the relationship \(e^{i z}=\cos z+i \sin z\), show that the (real) electric field at point P is
\(E_{P}(t)=\left[E_{0} \frac{\sin (N \phi / 2)}{\sin (\phi / 2)}\right] \cos [k R-\omega t+(N-1) \phi / 2]\)
The quantity in the first square brackets in this expression is the amplitude of the electric field at P. (d) Use the result for the electric field amplitude in part (c) to show that the intensity at an angle \(\theta\) is
\(I=I_{0}\left[\frac{\sin (N \phi / 2)}{\sin (\phi / 2)}\right]^{2}\)
where \(I_{0}\) is the maximum intensity for an individual slit. (e) Check the result in part (d) for the case N = 2. It will help to recall that sin 2A = 2 sin A cos A. Explain why your result differs from Eq. (35.10), the expression for the intensity in two-source interference, by a factor of 4. (Hint: Is \(I_{0}\) defined in the same way in both expressions?)
ANSWER:Solution 74CP