/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 A pinhole camera doesn't have a ... [FREE SOLUTION] | 91影视

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Chapter 25: Problem 56

A pinhole camera doesn't have a lens; a small circular hole lets light into the camera, which then exposes the film. For the sharpest image, light from a distant point source makes as small a spot on the film as possible. What is the optimum size of the hole for a camera in which the film is $16.0 \mathrm{cm}$ from the pinhole? A hole smaller than the optimum makes a larger spot since it diffracts the light more. A larger hole also makes a larger spot because the spot cannot be smaller than the hole itself (think in terms of geometrical optics). Let the wavelength be \(560 \mathrm{nm}\).

Short Answer

Expert verified
Answer: The optimal size of the hole for this pinhole camera is about 21.3 渭m.

Step by step solution

01

Find the angular spread due to diffraction

We will use the formula for the angular width of the central maximum of the diffraction pattern produced by a circular aperture: $$\theta \approx 1.22 \frac{\lambda}{D}$$ where \(\theta\) is the angular width, \(\lambda\) is the wavelength of the light, and \(D\) is the diameter of the pinhole.
02

Calculate the width of the light spot on the film due to diffraction

The width \(w\) of the light spot on the film can be calculated using the distance \(d\) from the pinhole to the film and the angle \(\theta\): $$w \approx d \cdot \theta = d \cdot 1.22 \frac{\lambda}{D}$$ Since we want to minimize the value of \(w\), we will take the derivative of the expression with respect to \(D\) and set it to zero: $$\frac{dw}{dD} = - d \cdot 1.22 \frac{\lambda}{D^2} = 0$$
03

Find the optimal pinhole diameter

To minimize \(w\), the expression we just derived should be positive. This implies that \(D\) should be large enough to eliminate the effects of diffraction. However, we still need to satisfy the condition from geometrical optics that the spot size cannot be smaller than the hole size. For that, we will set the optimal diameter \(D_{opt}\) equal to the size of the smallest possible light spot on the film and, hence, equal to the width \(w\). So we have: $$D_{opt} = d \cdot 1.22 \frac{\lambda}{D_{opt}}$$
04

Calculate the optimal diameter

Rearranging the equation from Step 3, we find: $$D_{opt}^2 = d \cdot 1.22 \lambda$$ Now we can plug in the given values of \(d = 16.0 \mathrm{cm}\) and \(\lambda = 560 \mathrm{nm}\) to get: $$D_{opt}^2 = 16.0 \mathrm{cm} \cdot 1.22 \cdot 560 \mathrm{nm}$$ Calculating \(D_{opt}\): $$D_{opt} = \sqrt{16.0 \mathrm{cm} \cdot 1.22 \cdot 560 \mathrm{nm}} \approx 2.13 \times 10^{-3} \mathrm{cm}$$
05

Convert to appropriate units

The optimal pinhole diameter is approximately \(2.13 \times 10^{-3} \mathrm{cm}\), which can also be written as \(21.3 \, \mu\mathrm{m}\). So, the optimal size of the hole for this pinhole camera is about \(21.3 \, \mu\mathrm{m}\).

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Most popular questions from this chapter

A steep cliff west of Lydia's home reflects a \(1020-\mathrm{kHz}\) radio signal from a station that is \(74 \mathrm{km}\) due east of her home. If there is destructive interference, what is the minimum distance of the cliff from her home? Assume there is a \(180^{\circ}\) phase shift when the wave reflects from the cliff.
Suppose a transparent vessel \(30.0 \mathrm{cm}\) long is placed in one arm of a Michelson interferometer, as in Example 25.2. The vessel initially contains air at \(0^{\circ} \mathrm{C}\) and 1.00 atm. With light of vacuum wavelength \(633 \mathrm{nm}\), the mirrors are arranged so that a bright spot appears at the center of the screen. As air is slowly pumped out of the vessel, one of the mirrors is gradually moved to keep the center region of the screen bright. The distance the mirror moves is measured to determine the value of the index of refraction of air, \(n .\) Assume that, outside of the vessel, the light travels through vacuum. Calculate the distance that the mirror would be moved as the container is emptied of air.
A mica sheet \(1.00 \mu \mathrm{m}\) thick is suspended in air. In reflected light, there are gaps in the visible spectrum at \(450,525,\) and $630 \mathrm{nm} .$ Calculate the index of refraction of the mica sheet.
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