## Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both the source a

Question

Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both the source and screen are far enough from the slit for Fraunhofer diffraction to apply.
A) If the first diffraction minima are at ±90.0∘, so the central maximum completely fills the screen, what is the width of the slit?
B) For the width of the slit as calculated in part (a), what is the ratio of the intensity at θ = 45.0∘ to the intensity at θ = 0?

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3 years 2021-08-08T23:16:42+00:00 1 Answers 186 views 0

## Answers ( )

A) d = 580 nm

B) I/I_o = 0.81

Explanation:

A) To find the width of the slit, we will use the formula;

sin θ = mλ/d

Where;

d = distance between the slits

θ = angle relative to the incident direction

m is an integer which is the order of interference.

We are given,wavelength;λ = 580 nm; θ = 90° and that it’s a first diffraction minima, thus m = 1

So;

sin θ = mλ/d is now;

sin 90 = 1(λ/d)

1 = (λ/d)

d = λ

Thus, d = 580 nm

B) To get the ratio of the intensity at θ = 45.0∘ to the intensity at θ = 0, we will use the formula;

I/I_o = ((sin θ)/θ)²

Thus,at 45°,we have;

I/I_o = ((sin 45)/(π/4))²

I/I_o = (0.7071/0.7854)²

I/I_o = 0.81