Chapter 33: Q. 75 (page 930)

The pinhole camera of FIGURE images distant objects by allowing only a narrow bundle of light rays to pass through the hole and strike the film. If light consisted of particles, you could make the image sharper and sharper (at the expense of getting dimmer and dimmer) by making the aperture smaller and smaller. In practice, diffraction of light by the circular aperture limits the maximum sharpness that can be obtained. Consider two distant points of light, such as two distant streetlights. Each will produce a circular diffraction pattern on the film. The two images can just barely be resolved if the central maximum of one image falls on the first dark fringe of the other image. (This is called Rayleigh’s criterion, and we will explore its implication for optical instruments in Chapter $35$ .)
a. Optimum sharpness of one image occurs when the diameter of the central maximum equals the diameter of the pinhole. What is the optimum hole size for a pinhole camera in which the film is $20 c m$ behind the hole? Assume localid="1649089848422" $λ = 550 nm$ an average value for visible light.
b. For this hole size, what is the angle a (in degrees) between two distant sources that can barely be resolved?
c. What is the distance between two street lights localid="1649089839579" $1 k m$ away that can barely be resolved?

Short Answer

Expert verified

(a) Optimum hole size for a pinhole camera $D = 0.52 mm$

(b) Angle between two distant source $α = {0.074}^{\circ}$

Step by step solution

Find optimum hole size (Part a)

The size of the central maximum of a coping up patter is, within the small-angle approach,

$w = \frac{2.44 λ L}{D}$

We could use $w = D$ so we are contemplating an optimum sharpness circumstance in which the radius of the related topic equals the aperture. Therefore

localid="1649092283758" $\begin{matrix} D = \frac{2.44 λ L}{D} \end{matrix}$

$D^{2} = 2.44 λ L$

$D = \sqrt{2.44 λ L} = \sqrt{2.44 \times (550 \times 10^{- 9} m) \times (0.2 m)}$

$D = 0.52 mm$

Find angle between two distant sources (Part b)

For the image pairs to be hardly resolved, the central maxima of one photo must fall on the first minimum of the other image, and in this case, the angle ( $α$ ) will be similar to the angle $θ_{1}$ of one of the image pairs, which is given by the following.

localid="1649092582124" $\begin{aligned} α = θ_{1} = \frac{1.22 λ}{D} \end{aligned}$

$\begin{array}{r} = \frac{1.22 \times (550 \times 10^{- 9} m)}{0.52 \times 10^{- 3} m} \end{array}$

$\begin{array}{r} 1.29 \times 10^{- 3} rad \end{array}$

$= {0.074}^{\circ}$

Find distance betwwen two street lights (Part c)

We could use arc length equation, which has the form, as we are working with small angle ( $α$ )

$s = r θ$

The duration of the $(s)$ is indeed the distance between the two lights, and the radius $(r)$ is the length here between lights and the aperture, which is $1000 m$ in our example. Hence

$s = (1000 m) \times 1.29 \times 10^{- 3} = 1.29 m$