A horizontal laser beam of wavelength 632.8 nm has a...

A horizontal laser beam of wavelength 632.8 nm has a circular cross section $2.00 \mathrm{mm}$ in diameter. A rectangular aperture is to be placed in the center of the beam so that when the light falls perpendicularly on a wall $4.50 \mathrm{m}$ away, the central maximum fills a rectangle $110 \mathrm{mm}$ wide and $6.00 \mathrm{mm}$ high. The dimensions are measured between the minima bracketing the central maximum. Find the required (a) width and (b) height of the aperture. (c) Is the longer dimension of the central bright patch in the diffraction pattern horizontal or vertical? (d) Is the longer dimension of the aperture horizontal or vertical? (e) Explain the relationship between these two rectangles, using a diagram.

Key Concepts

Diffraction

Diffraction is the phenomenon whereby waves bend around obstacles or spread after passing through an aperture. It is a key principle in wave optics and explains how light (or other waves) produces interference patterns when its wavefront is disrupted by openings or edges. This concept is central to understanding the behavior of waves in various media and is used in numerous optical instruments and experiments.

Fraunhofer Diffraction

Fraunhofer diffraction refers to the far-field diffraction pattern observed when both the source and the screen (or observation plane) are at large distances from the diffracting aperture. In this regime, the wavefronts can be approximated as parallel, simplifying the analysis of the diffraction pattern. This form of diffraction is typically used to study patterns generated by apertures since it permits the use of Fourier analysis to predict the intensity distribution.

Diffraction by a Rectangular Aperture

When a wavefront passes through a rectangular aperture, the resulting diffraction pattern is a product of independent diffraction effects along the two perpendicular dimensions. This means that the intensity distribution in the far field can be analyzed separately for the horizontal and vertical dimensions. The pattern typically shows a central maximum and a series of minima and secondary maxima, with the positions of these features being dependent on the dimensions of the aperture.

Single?Slit Diffraction Minima Condition

The condition for minima in a single?slit (or single dimension of a rectangular aperture) diffraction pattern is given by a sin? = m?, where a is the width of the slit in that dimension, m is a nonzero integer, ? is the wavelength, and ? is the angle measured from the central axis. This relationship is crucial for calculating the angular positions of the dark fringes that bracket the central maximum in the diffraction pattern.

Inverse Relationship between Aperture Size and Diffraction Spread

An important concept in wave optics is the inverse relationship between the size of an aperture and the angular spread of the diffraction pattern. A larger aperture in a given dimension results in a narrower diffraction spread, while a smaller aperture produces a wider spread. This principle is essential when designing optical systems, as it links the physical dimensions of an aperture to the scale of the diffraction features observed in the far field.

Transcript

00:01 So in this problem, we have a horizontal laser beam of wavelength 632 .8 nanometer with a circular cross -section of 2mm in diameter.

00:14 We want to put a rectangular slit so that when the screen is at 4 .5 meter away from the slit, the central maximum is of dimension 110 millimeter to 6 millimeter.

00:28 Now in part a, in part a, we want to find the width of the aperture.

00:40 Since the rectangle formed in the wall is wider than it is taller, the slit has to be taller than it is wider.

00:49 So for the horizontal spreading of central maximum, we use 10 theta, theta, 1w, because that's for the distance.

01:04 You know this is w is the width of the central maximum okay so 10 theta 1 w is going to be equal to y1 w over l which is y is the distance of the first dark fringe from the center so this is going to be half of the w right w by 2 over l and this is going to be 110 by 2 which is 55 so 55 times 10 to the negative 3 over the length was 4 .5 meter.

01:48 And this is going to give us 1 .22 times 10 to the negative 2.

02:00 Now, note that i'm using y1w equals w by 2, because the distance of the first bright fringe from the center is half the weight of the central bright fringe.

02:10 Again the force for the first dark fringe sign theta 1 sine theta 1 w has to be lambda over a w we're just doing the calculation for the width part only just be careful about that here so this is actually 10 theta 1 w because we can use the small angle approximation here and that's gonna mean that lambda over a w equals 1 .22 times 10 to the negative 2 so this actually means weight of the slate equals lambda which is 632 .8 nanometer over 1 .22 times 10 to the to the negative 2 and this is going to give us 5 .18 times 10 to negative 6 10 to negative 7 meter or you can also say 51 .8 micrometer now in part b part b similarly for part b for height we see from part a that it's just going to be h over 2 over l equals lambda over ah you can see that here instead of w now it's going to be changed to h you can see from here now y1 h over l right so that's going to be h over 2 here so i'm just going to plug in the value here a h equals twice lambda l over h which is going to going to be 2 times 632 .8 times 10 to the negative 9 times 4 .5.

04:51 That is the distance between the slit and the screen over 6 times 10 to the negative 3, which is the height of the central bright finch.

05:03 And this is going to give us 4 .49 times 10 to the negative 4 meter...

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