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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 μm.

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 lens \((n=1.52)\) is coated with a magnesium fluoride film \((n=1.38) .\) (a) If the coating is to cause destructive interference in reflected light for \(\lambda=560 \mathrm{nm}\) (the peak of the solar spectrum), what should its minimum thickness be? (b) At what two wavelengths closest to 560 nm does the coating cause constructive interference in reflected light? (c) Is any visible light reflected? Explain.
A grating has 5000.0 slits/cm. How many orders of violet light of wavelength \(412 \mathrm{nm}\) can be observed with this grating? (tutorial: grating).
If you shine a laser (wavelength \(0.60 \mu \mathrm{m}\) ) with a small aperture at the Moon, diffraction makes the beam spread out and the spot on the Moon is large. Making the aperture smaller only makes the spot on the Moon larger. On the other hand, shining a wide searchlight at the Moon can't make a tiny spot- the spot on the Moon is at least as wide as the searchlight. What is the radius of the smallest possible spot you can make on the Moon by shining a light from Earth? Assume the light is perfectly parallel before passing through a circular aperture.
Sketch a sinusoidal wave with an amplitude of \(2 \mathrm{cm}\) and a wavelength of \(6 \mathrm{cm} .\) This wave represents the electric field portion of a visible EM wave traveling to the right with intensity \(I_{0}\). (a) Sketch an identical wave beneath the first. What is the amplitude (in centimeters) of the sum of these waves? (b) What is the intensity of the new wave? (c) Sketch two more coherent waves beneath the others, one of amplitude \(3 \mathrm{cm}\) and one of amplitude \(1 \mathrm{cm},\) so all four are in phase. What is the amplitude of the four waves added together? (d) What intensity results from adding the four waves?
When a double slit is illuminated with light of wavelength \(510 \mathrm{nm},\) the interference maxima on a screen \(2.4 \mathrm{m}\) away gradually decrease in intensity on either side of the 2.40 -cm-wide central maximum and reach a minimum in a spot where the fifth-order maximum is expected. (a) What is the width of the slits? (b) How far apart are the slits?
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