An interference pattern is produced by light of wave- length 580 nm from a distant source incide

An interference pattern is produced by light of wave- length...

An interference pattern is produced by light of wave- length 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

Single?Slit Diffraction

Even when light passes through a narrow slit, it diffracts, spreading out as it propagates. The intensity pattern from a single slit is described by an envelope function, often given by (sin?/?)², where ? is proportional to the slit width and sin?. This diffraction envelope determines the overall modulation of the intensity distribution observed in multi-slit experiments, especially when the slits have finite widths.

Combined Intensity Distribution

In experiments with finite slit widths, the observed pattern is a product of the two-slit interference pattern and the single-slit diffraction envelope. The bright interference fringes occur at angles predicted by the double-slit interference condition, but their intensities are modulated by the diffraction pattern of each slit. As a result, the net intensity at any given angle is calculated by multiplying the pure interference intensity with the corresponding diffraction factor, providing a complete description of the light distribution.

Double?Slit Interference

This concept involves the formation of interference patterns when coherent light passes through two narrow, closely spaced slits. The condition for constructive interference (bright fringes) is given by d sin? = m?, where d is the distance between the slits, ? is the wavelength, ? is the angle relative to the central axis, and m is the order number of the maximum. This relationship is fundamental in predicting the angular positions of the interference maxima.

Transcript

00:01 Okay, i wrote down the given information and i converted from nanometers and millimeters to meters.

00:08 We know that the sign of theta is lambda m over d.

00:17 So theta is the inverse sign of lambda m over d.

00:29 So the positions would be, put this in a calculator, i'm going to set m.

00:35 Equal 1, inverse sign of 580 times 10 to the negative 9th power times 1 over d, which is 0 .000 0 .00530.

00:56 So that's going to give me 0 .0019 degrees.

01:24 I just want to make sure that i did not type in anything incorrectly because that's not the answer that's given in the book.

01:47 And i found my mistake.

01:48 I had my calculator set to radians, not degrees.

01:54 And so when i set it to degrees, i got 0 .0627 degrees.

02:00 Okay, now i'm going to set m equal 2 to 2, and i get 0 .125 .5.

02:08 Degrees.

02:15 If the width of the slits is 0 .320 millimeters, which is going to be 0 .00320 meters, yes, what is the intensity of each of these? so i over i .0 .0 is going to be.

02:42 I over i sub zero is going to be.

02:46 The sign of pi a sine over lambda over pi a sine theta over pi a sine theta over lambda squared so let's put that into a calculator, the sign, and i need to make sure that i use radian mode now, of pi and a times sine, oops, a is 0 .00032 times sine, times sine, 0 .00032 times sign of theta.

03:54 But theta is times pi over 180 to convert it to radiance.

04:09 Okay.

04:11 And then i have to take that over lambda.

04:22 Lambda is 580 times 10 to the negative 9th power...

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