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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 36 / Problem 58P

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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The intensity of light in the Fraunhofer diffraction

Chapter 36, Problem 58P

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QUESTION:

The intensity of light in the Fraunhofer diffraction pattern of a single slit is

$I=I_{0}\left(\frac{\sin \gamma}{\gamma}\right)^{2}$

where

$\gamma=\frac{\pi a s i n \theta}{\lambda}$

(a) Show that the equation for the values of $\gamma$ at which $I$ is a maximum is tan $\gamma=\gamma$. (b) Determine the three smallest positive values of $\gamma$ that are solutions of this equation. (Hint: You can use a trial-and-error procedure. Guess a value of $\gamma$ and adjust your guess to bring tan $\gamma$ closer to $\gamma$. A graphical solution of the equation is very helpful in locating the solutions approximately, to get good initial guesses.)

Equation Transcription:

$I={I}_{0}(\frac{sin\ \gamma{}}{\gamma{}}{)}^{2}$

$\gamma{}=\frac{\pi{}asin\theta{}}{\lambda{}}$

$\gamma{}$

$\gamma{}=\gamma{}$

$\gamma{}$

Text Transcription:

I=I_0(sin gamma over gamma)^2

gamma=pi a sin theta over lambda

gamma

gamma=gamma

Gamma

Questions & Answers

QUESTION:

The intensity of light in the Fraunhofer diffraction pattern of a single slit is

$I=I_{0}\left(\frac{\sin \gamma}{\gamma}\right)^{2}$

where

$\gamma=\frac{\pi a s i n \theta}{\lambda}$

Equation Transcription:

$I={I}_{0}(\frac{sin\ \gamma{}}{\gamma{}}{)}^{2}$

$\gamma{}=\frac{\pi{}asin\theta{}}{\lambda{}}$

$\gamma{}$

$\gamma{}=\gamma{}$

$\gamma{}$

Text Transcription:

I=I_0(sin gamma over gamma)^2

gamma=pi a sin theta over lambda

gamma

gamma=gamma

Gamma

ANSWER:

Solution 58P

The intensity of light in the Fraunhofer diffraction pattern of a single slit is,

$I=I{\ }_{0}(\frac{sin\ \gamma{}}{\gamma{}}{)}^{2}$ …..(1)

For $I$ to be maximum, the differentiation of $I$ with respect to $\gamma{}$ , $\frac{dI}{d\gamma{}}$ is equal to zero.

(a) Differentiating equation (1) wrt $\gamma{}$ ,

$\frac{dI}{d\gamma{}}=I{\ }_{0}\frac{d}{d\gamma{}}\frac{si{n}^{2}\ \gamma{}}{\gamma{}{\ }^{2}}$

$\frac{dI}{d\gamma{}}=I{\ }_{0}\frac{\gamma{}{\ }^{2}\frac{d}{d\gamma{}}(si{n}^{2}\ \gamma{})-si{n}^{2}\ \gamma{}\frac{d}{d\gamma{}}{\gamma{}}^{2}}{{\gamma{}}^{4}}$

$\frac{dI}{d\gamma{}}=\frac{{\gamma{}}^{2}\times{}2sin\ \gamma{}.cos\ \gamma{}-si{n}^{2}\ \gamma{}.2\gamma{}}{{\gamma{}}^{4}\ }$

When $\frac{dI}{d\gamma{}}=0$

${\gamma{}}^{2}\times{}2sin\ \gamma{}.cos\ \gamma{}=si{n}^{2}\ \gamma{}.2\gamma{}$

${\gamma{}}^{2}\times{}2sin\ \gamma{}.cos\ \gamma{}=2\gamma{}\times{}sin\ \gamma{}\times{}sin\ \gamma{}$

$\gamma{}=tan\ \gamma{}$

$tan\ \gamma{}=\gamma{}$

Therefore, $I$ is maximum when $tan\ \gamma{}=\gamma{}$ .

(b) The values of the function $tan\ \gamma{}=\gamma{}$ can be found out online using Wolframalpha. In the calculation bar here, if we type the function the first three non-zero positive values of $\gamma{}$ obtained are 4.49, 7.72 and 10.90.

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