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Chapter 17: Problem 41

You want to photograph a circular diffraction pattern whose central maximum has a diameter of \(1.0 \mathrm{cm} .\) You have a heliumneon laser \((\lambda=633 \mathrm{nm})\) and a 0.12 -mm-diameter pinhole. How far behind the pinhole should you place the viewing screen?

Short Answer

Expert verified
The viewing screen should be placed approximately 7.4 m behind the pinhole.

Step by step solution

01

Identify Given Values

The given values in this problem are the wavelength of the light from the laser (位) which is \(633 nm = 633 * 10^{-9} m\), the diameter of the diffraction pattern's central maximum (d) which is \(1.0 cm = 0.01 m\), the diameter of the pinhole (D) which is \(0.12 mm = 0.12 * 10^{-3} m\).
02

Understand the Formula for Diffraction Patterns

The equation that describes the diameter of the central maximum of a diffraction pattern is given as: \(d = 2.44 * 位 * \frac{L}{D}\) where L is the distance to the screen.
03

Rearrange the Formula

We need to solve for \(L\), the distance from the pinhole to the screen. Hence, rearrange the equation: \(L = \frac{d * D}{2.44 * 位}\).
04

Substitute the Known Values

Substituting the given values into the formula, we get: \(L = \frac{(0.01 m)*(0.12 * 10^{-3} m)}{2.44 * (633 * 10^{-9})}\).
05

Calculate Distance

Calculate the value of \(L\) which is the distance from the pinhole to the screen. This results in \(L = 7.4 m\).

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Most popular questions from this chapter

A double-slit experiment is performed with light of wavelength \(600 \mathrm{nm} .\) The bright interference fringes are spaced \(1.8 \mathrm{mm}\) apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of \(400 \mathrm{nm} ?\)

Two narrow slits \(50 \mu \mathrm{m}\) apart are illuminated with light of wavelength \(500 \mathrm{nm} .\) The light shines on a screen \(1.2 \mathrm{m}\) distant. What is the angle of the \(m=2\) bright fringe? How far is this fringe from the center of the pattern?

One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a small patch of light on the far wall. Having recently studied optics in your physics class, you're not too surprised to see that the patch of light seems to be a circular diffraction pattern. It appears that the central maximum is about \(3 \mathrm{cm}\) across, and you estimate that the distance from the window shade to the wall is about \(3 \mathrm{m} .\) Knowing that the average wavelength of sunlight is about \(500 \mathrm{nm},\) estimate the diameter of the pinhole.

Light illuminating a pair of slits contains two wavelengths, \(500 \mathrm{nm}\) and an unknown wavelength. The 10 th bright fringe of the unknown wavelength overlaps the 9 th bright fringe of the \(500 \mathrm{nm}\) light. What is the unknown wavelength?

Glass catfish, tropical fish popular with hobbyists, have no pigment, and matching of the index of refraction of their tis- sues leaves them largely transparent-you can see through them. When a beam of white light illuminates these fish, diffraction of light from evenly spaced striations in muscle fibers produces a rainbow pattern. As the muscles contract, the striations get closer together, and this affects the diffraction pattern. This phenomenon has been used to study muscle contraction in living fish as they swim. Investigators set up a chamber with moving water where the fish swam steadily to stay stationary. The investigators then passed a beam of laser light of wavelength \(632 \mathrm{nm}\) through the fish's muscle tissue as it swam. Muscle action produced periodic changes in the distances between striations, which ranged from 1.87 to \(1.94 \mu \mathrm{m} .\) Investigators measured the change of position of a bright spot corresponding to \(m=1\) on a screen. If the screen was \(30 \mathrm{cm}\) behind the fish, what was the distance spanned by the diffraction spot as it moved back and forth? The screen was in the tank with the fish, so that the entire path of the laser was in water and tissue with an index of refraction close to that of water. The properties of the diffraction pattern were thus determined by the wavelength in water.
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