Light of wavelength 500nm shines on a pair of slits. This produces an interference pattern on a screen 2.00m behind the slits. Measuring from the center of the pattern, we find …

Question

asked Feb 3, 2021 50.2k views

1 Answer

Tensia · Answer 1 · 2021-02-08T19:30:04+0000

Answer:

The value is
$d = 0.000125 \ m$

Step-by-step explanation:

From the question we are told that

The wavelength is
$\lambda = 500 \ nm = 500 *10^(-9 ) \ m$

The distance is of the screen is
$D = 2.0 \ m$

The width of the fifth dark spot is
$y = 4.00 \ cm = 0.04 \ m$

Generally the width of a fringe is mathematically represented as

$y = ( n * \lambda * D )/(d )$

=>
0.04 = ( 5 * 500 *10^(-9) * 2 )/(d)

=>
d = ( 5 * 500 *10^(-9) * 2 )/(0.04)

=>
$d = 0.000125 \ m$