Physics: How wide is the central diffraction peak on a screen 2.30 m behind a 0.0368-mm-wide slit illuminated by 558-nm light

I see this is quite late of an answer, but i as well could use help on this ...

How wide is the central diffraction peak on a

shaydog6353

31.01.2022

The value $y = 0.0349 \ m$

Explanation:

From the question we are told that

The distance of the screen is $D = 2.30 \ m$

The width of the slit is $d = 0.0368 \ nm = 0.0368 *10^{-3} \ m$

The wavelength is $\lambda = 558 \ nm = 558 *10^{-9} \ m$

The width of the central diffraction peak is mathematically represented as

k = 2 * y

Where y is the distance from the center to the high peak which is mathematically represented as

$y = \frac{\lambda * D }{d }$

substituting values

$y = \frac{ 558 *10^{-8} * 2.30 }{0.0368 *10^{-3} }$

$y = 0.0349 \ m$

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