Physics, 08.12.2020 16:30 lakisa138

A Fraunhofer diffraction pattern is produced on a screen 1.30 m from a single slit. If a light source of 560 nm is used and the distance from the center of the central bright fringe to the first dark fringe is 5.30 * 10^-3 m, what is the slit width?

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Answer from: lezapancakes13

The value is $a = 0.000137 \ m$

Explanation:

From the question we are told that

The distance from the screen is D = 1.30 m

The wavelength of the light source is $\lambda = 560 \ nm = 560 *10^{-9} \ m$

The path difference is mathematically represented as $y = 5.30 *10^{-3} \ m$

Generally the path difference is mathematically represented as

$y = \frac{m * D * \lambda }{a}$

Here m is 1 given that we are considering the first dark fringe

$a = \frac{ m * \lambda * D}{y}$

=> $a = \frac{1 * 560 *10^{-9 } * 1.30 }{ 5.30 *10^{-3}}$

=> $a = \frac{1 8 560 *10^{-9 } * 1.30 }{ 5.30 *10^{-3}}$

=> $a = 0.000137 \ m$

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A Fraunhofer diffraction pattern is produced on a screen 1.30 m from a single slit. If a light sourc...