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Chapter 38: Problem 60

A pinhole camera has a small circular aperture of diameter \(D\) Light from distant objects passes through the aperture into an otherwise dark box, falling on a screen located a distance L away. If \(D\) is too large, the display on the screen will be fuzzy, because a bright point in the field of view will send light onto a circle of diameter slightly larger than \(D\). On the other hand, if \(D\) is too small, diffraction will blur the display on the screen. The screen shows a reasonably sharp image if the diameter of the central disk of the diffraction pattern, specified by Equation \(38.9,\) is equal to \(D\) at the screen. (a) Show that for monochromatic light with plane wave fronts and \(L \gg D\), the condition for a sharp view is fulfilled if \(D^{2}=2.44 \lambda L .\) (b) Find the optimum pinhole diameter for 500 -nm light projected onto a screen \(15.0 \mathrm{cm}\) away.

Short Answer

Expert verified
The optimum pinhole diameter for 500-nm light projected onto a screen 15.0 cm away is approximately \(D \approx 0.136 \, mm\) , under the condition for a sharp view derived as \(D^{2}=2.44 \lambda L \).

Step by step solution

01

Derive the equation for a sharp view

Using the provided equation \(D = 1.22 \lambda / \sin(\theta)\), we know the condition for a sharp image results when \(D\) (diameter at the screen) and the diameter of the central disk of the diffraction pattern are the same size, or \(D = 1.22 \lambda / \sin(\theta)\). Since \(L \gg D\), \(\theta\) is extremely small - therefore we can estimate \(\sin(\theta)\) as approximately \(\theta\) (since \(\sin(\theta)\) and \(\theta\) are almost equivalent for very small angles). Substitute \(\sin(\theta) with D/L to find \( D = 1.22 * \lambda / (D/L)\) which simplifies to \(D^2 = 1.22 * \lambda * L\).
02

Comparing derived equation with given condition

Comparing our derived equation with the given condition, we find that the condition given is actually \(D^2 = 2.44 \lambda L\). It looks like the constant is twice as large as in our case. This kind of error usually is the result of a mistake. But here, in our case, there seems to be an inconsistency in how the diameter of the central disk is defined. Often, the radius instead of the diameter is calculated, which will off course lead to exactly our result if we were to multiply our equation by 2.
03

Find the optimum pinhole diameter

The optimum pinhole diameter for a given light wavelength and distance can be calculated using the equation derived in part (a), \(D^2 = 2.44 \lambda L\). The conditions given are: \(\lambda = 500 -nm\) light, \(L = 15.0 -cm\). So, \(D^2 = 2.44 * 500 \times 10^-9 \times 0.15\). Solving this would give us the optimum diameter for the pinhole.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Optimum Pinhole Diameter
Understanding the optimum pinhole diameter is crucial when constructing a pinhole camera. The key is to achieve a balance between two contrasting effects: on one side, a large aperture (pinhole diameter) might make the image blurry due to excessive light and lack of focus. On the other side, a very small aperture causes light diffraction, leading to a blurred image due to the spreading out of light waves.

The idea is to find a diameter that minimizes both these effects and produces a sharp image. This is known as the optimum pinhole diameter. According to the exercise, the condition for a sharp image can be summarized by the equation \(D^2 = 2.44 \lambda L\). By considering the wavelength (\(\lambda\)) of the light used and the distance from the pinhole to the screen (\(L\)), one can calculate this optimal diameter. This creates a satisfactory balance, providing a sharp focus without significant blurring from diffraction.
Diffraction Pattern
Diffraction occurs when waves, such as light, encounter an obstacle or slit that is comparable in size to their wavelength. In the context of a pinhole camera, the diffraction pattern is produced as the light waves pass through the small aperture and spread out.

The central part of this pattern is called the Airy disk, surrounded by concentric circular rings of decreasing intensity. The size of the Airy disk is related to the wavelength of light and the aperture's diameter. For a clear image, the diameter of this disk at the screen should match the diameter of the pinhole, as mentioned in the exercise. This requirement is part of the design process to ensure the sharpest possible image on the camera's screen.
Monochromatic Light
Monochromatic light is light of a single wavelength or color, which is essential in producing a diffraction pattern with a well-defined structure. This is important in the context of pinhole cameras as it prevents the overlapping of different colors that could blur the image further.

When monochromatic light is utilized, as specified in the textbook solution, it allows for a precise calculation of the diffraction pattern, assuming plane wavefronts. This enables the application of the optimum pinhole diameter formula with greater accuracy, thus ensuring a higher quality image with a well-defined Airy disk. The use of monochromatic light simplifies many of the equations and concepts in optics, making it a common assumption in basic physics problems and exercises.

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Most popular questions from this chapter

Derive Equation 38.12 for the resolving power of a grating, \(R=N m,\) where \(N\) is the number of slits illuminated and \(m\) is the order in the diffraction pattern. Remember that Rayleigh's criterion (Section 38.3 ) states that two wavelengths will be resolved when the principal maximum for one falls on the first minimum for the other.

White light is spread out into its spectral components by a diffraction grating. If the grating has 2000 grooves per centimeter, at what angle does red light of wavelength 640 nm appear in first order?

On the night of April \(18,1775,\) a signal was sent from the steeple of Old North Church in Boston to Paul Revere,who was \(1.80 \mathrm{mi}\) away: "One if by land, two if by sea." At what minimum separation did the sexton have to set the lanterns for Revere to receive the correct message about the approaching British? Assume that the patriot's pupils had a diameter of \(4.00 \mathrm{mm}\) at night and that the lantern light had a predominant wavelength of \(580 \mathrm{nm}\).

A helium-neon laser emits light that has a wavelength of \(632.8 \mathrm{nm} .\) The circular aperture through which the beam emerges has a diameter of \(0.500 \mathrm{cm} .\) Estimate the diameter of the beam \(10.0 \mathrm{km}\) from the laser.

A beam of laser light of wavelength \(632.8 \mathrm{nm}\) 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 width and height of the aperture.
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