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 fr…

Question

asked May 9, 2021 136k views

1 Answer

Brian Walker · Answer 1 · 2021-05-15T08:39:15+0000

Answer:

The value is
$a = 0.000137 \ m$

Step-by-step 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

$y = (m * D * \lambda )/(a)$

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

So

$a = ( m * \lambda * D)/(y)$

=>
a = (1 * 560 *10^(-9 ) * 1.30 )/( 5.30 *10^(-3))

=>
a = (1 8 560 *10^(-9 ) * 1.30 )/( 5.30 *10^(-3))

=>
$a = 0.000137 \ m$