Chapter 10: Problem 1
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 \(\frac{1}{2}\) ?
Short Answer
Step by step solution
Key Concepts
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