The pinhole camera of FIGURE CP22.75 images distant objects...

The pinhole camera of FIGURE CP22.75 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 $24 .$ ) 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 $20 \mathrm{cm}$ behind the hole? Assume $\lambda=550 \mathrm{nm},$ an average value for visible light. b. For this hole size, what is the angle $\alpha$ (in degrees) between two distant sources that can barely be resolved? c. What is the distance between two street lights $1 \mathrm{km}$ away that can barely be resolved? (FIGURE CAN'T COPY)

The pinhole camera of FIGURE CP22.75 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 $24 .$ )
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 $20 \mathrm{cm}$ behind the hole? Assume $\lambda=550 \mathrm{nm},$ an average value for visible light.
b. For this hole size, what is the angle $\alpha$ (in degrees) between two distant sources that can barely be resolved?
c. What is the distance between two street lights $1 \mathrm{km}$ away that can barely be resolved? (FIGURE CAN'T COPY)

Key Concepts

Diffraction

Diffraction is the phenomenon in which waves encounter an obstacle or aperture and bend around it, producing a spread-out pattern of light rather than a perfectly sharp shadow. For circular apertures, the resulting pattern is characterized by a central bright region (the Airy disk) surrounded by progressively dimmer rings. This effect limits the sharpness of images in optical systems such as pinhole cameras, particularly as the aperture size becomes comparable to the wavelength of light.

Airy Disk

The Airy disk refers to the central bright spot in the diffraction pattern produced by a circular aperture. Its diameter is determined by the wavelength of light and the size of the aperture. In imaging systems, the size of the Airy disk sets a fundamental limit on resolution because two point sources can only be resolved if their Airy disks are sufficiently separated, a concept that is central to understanding the resolving power of optical instruments.

Rayleigh Criterion

Rayleigh's criterion provides a quantitative measure for the resolution limit of an optical system. It states that two point sources are just resolvable when the central maximum of the diffraction pattern of one overlaps with the first dark fringe of the other. This criterion is used to define the minimal angular separation at which two sources can be distinguished, making it a key concept in the design and analysis of imaging systems including pinhole cameras.

Pinhole Camera Optimization

Pinhole camera optimization involves balancing the trade-off between geometric sharpness and diffraction blur. A smaller aperture increases geometric sharpness by reducing the spread of light rays but simultaneously increases diffraction effects which can blur the image. Optimizing the aperture size balances these two competing effects to achieve the highest practical resolution.

Angular Resolution and Small Angle Approximation

Angular resolution is the ability of an optical system to distinguish between two light sources separated by a small angle. When the angles involved are small, the small angle approximation (in which the sine of the angle is approximately equal to the angle in radians) is used to simplify calculations of linear separations at a given distance. This concept is essential when translating angular resolution limits, such as those provided by the Rayleigh criterion, into physical distances between objects.

Transcript

00:01 For this problem on the topic of wave optics, we are told how a pinhole camera operates, and we are told that the optimum sharpness of an image occurs when the diameter of the central maximum equals the diameter of the pinhole.

00:13 We want to know the optimal hole size for a pinhole camera in which the firm is 20 centimeters behind the hole.

00:19 We are given the wavelength, which is an average value for visible light, and we want to find the angle alpha between two distance sources that can be barely resolved.

00:30 And finally, we want to find the distance between two street lights, a given distance away, that can barely be resolved.

00:38 Now, diffraction patterns from two objects can barely be resolved if the central maximum of one image falls on the first dark fringe of the other image.

00:48 So using the equation that we know for the width and the aperture diameter, we know that the width is equal to this aperture diameter d.

00:59 And this is given by 2 .44 times the wavelength lambda times l over d.

01:12 Now we are given l to be 20 centimeters.

01:17 So we can rearrange this equation and we can solve for the aperture diameter.

01:24 And we get d to be the square root of 2 .44.

01:34 Times lambda times l and this is the square root we can now substitute our values in here that's 2 .44 times the wavelength given to be 550 times 10 to the minus 7 meters times 0 .2 meters which is l or 20 centimeters so calculating we get the aperture diameter or the optimum size for the hole camera where the film is 20 centimeters behind the hole to be 0 .52 millimeters.

02:19 And that's our first answer.

02:23 Now we move on to part b and in part b we are told that or we are asked rather what the angle is alpha between two distance sources that can barely be resolved.

02:37 Now to find the angle between two distance sources that can be resolved, we can use the equation alpha is equal to 1 .22 times the wavelength lambda over the aperture diameter d...

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