An interference pattern is produced by light of wavelength 580 nm from a distant source incident

An interference pattern is produced by light of wavelength...

An interference pattern is produced by light of wavelength 580 $\mathrm{nm}$ from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.530 $\mathrm{mm}$ . (a) If the slits are very narrow, what would be the angular positions of the first-order and second-order, two-slit, interference maxima? (b) Let the slits have width 0.320 $\mathrm{mm}$ , In terms of the intensity $I_{0}$ at the center of the central maximum, what is the intensity at each of the angular positions in part (a)?

Key Concepts

Double-Slit Interference

This concept involves the phenomenon where light waves from two coherent sources (the two slits) combine, producing regions of constructive and destructive interference. The condition for constructive interference (maxima) is given by d sin ? = m?, where d is the separation between the centers of the slits, ? is the wavelength of the light, ? is the angle of the maximum, and m is an integer representing the order of the maximum. This equation allows the calculation of angular positions where bright fringes occur in an interference pattern.

Single-Slit Diffraction

When the slits have a finite width, each slit produces its own diffraction pattern, characterized by a central maximum and diminishing side lobes. The overall intensity observed in a two-slit experiment is therefore modulated by this diffraction envelope. The intensity distribution from a single slit is typically described by the sinc-squared function [ (sin ? / ?)^2 ], where ? is proportional to the slit width and sin ?. This modulation affects the brightness of the interference maxima and must be considered when determining the intensity at various angular positions.

Transcript

00:01 In this question we have interference from two slits which are 0 .530 nanometers apart and the light incident on them has a wavelength of 580 nanometers.

00:19 We want to look at the interference pattern.

00:23 Now for the first case, the width of the slits is really small, which means the diffraction pattern will have almost, the diffraction pattern will have no change.

00:39 The intensity pattern will be constant and we only need to consider the interference pattern.

00:49 We know that for just interference, the intensity varies as i not cost square fee by 2, if fee is equal to 2 pi by lambda d .s.

01:03 Theta where theta is the angle that the point makes from the center.

01:19 For maximus we want p by 2 to be pi and 2 pi.

01:27 Cos pi will be minus 1 and cost 2 pi will be 1 and that will be the maximum value for cost square p by 2.

01:34 For this we want fee to be 2 pi which means 2 pi is equal to 2 pi lambda d sine theta and for the second case we want 4 pi is equal to 2 pi lambda d sine theta giving us sine theta is equal to lambda by d and sine theta is equal to 2 lambda by d now solving this equation is pretty easy uh here theta is equal to sine inverse of we know that the wavelength is 580 nanometers i'm sorry this is millimeter this is 0 .530 millimeters.

02:21 So that's 580 times 10 power minus 9.

02:24 And the distance d is 0 .530 into 10 bar minus 3.

02:36 This turns out to be 1 .094 into 10 bar minus 3 radiance.

02:46 And for the second case, this same angle turns out to be sine inverse of 2 times lambda by d, which turns out to be 2.

02:55 0 .188 into 10 power minus 3 radiance...

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