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Chapter 9: Problem 38

Derive a formula for the magnification of a pinhole magnifier. Check the formula as follows: Make two marks \(2 \mathrm{~cm}\) apart on one piece of paper; make two marks \(2 \mathrm{~mm}\) apart on another. Put the pinhole over one eye and nothing over the other. Both pieces of paper should be illuminated from behind (at least, that is easiest to look at). With both eyes open, look with one eye through the pinhole at the \(2 \mathrm{~mm}\) marks and with your other eye look at the \(2 \mathrm{~cm}\) marks. Bring the \(2 \mathrm{~mm}\) marks closer until you superpose the two sets of marks with your two eyes. Measure appropriate distances.

Short Answer

Expert verified
The magnification formula for a pinhole is \(M = \frac{d_1}{d_2}\). Use relative sizes and distances for verification.

Step by step solution

01

Understand the concept of magnification in a pinhole magnifier

A pinhole magnifier works on the principle that light passing through a small hole creates a clear image due to limited light scattering. The magnification is determined by comparing the size of the image formed by the pinhole to the actual size of the object being viewed.
02

Establish the condition for superposition

When using a pinhole magnifier, the marks seen through the pinhole appear to overlap (superpose) with marks seen by the other eye. This means both views appear the same size鈥攊ndicating the point of magnification equality.
03

Set up the geometric relationship

Consider the geometry of the situation. Let the distance from the pinhole to the eye be denoted as \(d_1\), and the distance from the pinhole to the object be \(d_2\). Given that the perceived size (image) and actual size are proportionate to these distances, the magnification \(M\) is calculated as \(M = \frac{d_1}{d_2}\).
04

Calculate the expected visual magnification

To superpose the marks correctly, the distance ratios should match. The formula becomes \(M = \frac{ ext{Size of the larger marks (seen directly)}}{ ext{Size of the smaller marks (seen through pinhole)}}\). If you look at 2 cm marks and 2 mm marks, \(M\) is determined by the relativity of these sizes and distances.
05

Measure and implement the check

Measure the distance from the eye to the pinhole (\(d_1\)) and from the pinhole to the object (\(d_2\)). By bringing the 2 mm marks closer until they appear superimposed with the 2 cm marks, experiment with these distances and verify with measured values locked in M = \(\frac{d_1}{d_2}\).

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Key Concepts

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

Magnification
Magnification is the process of enlarging the appearance of an object. In a pinhole magnifier, the size of the image is determined by the ratio of distances from the pinhole to the observer and the object. This ratio forms the basis of the magnification formula:\[ M = \frac{d_1}{d_2} \]where
  • \(M\) is the magnification,
  • \(d_1\) is the distance from the pinhole to your eye,
  • and \(d_2\) is the distance from the pinhole to the object.
To experiment with magnification, you engage both eyes: one looks through the pinhole at a set of smaller marks, and the other at a set of larger marks without any aid. The challenge is to adjust the distances such that the two sets of marks appear the same size, indicating magnification parity.
This is a clear demonstration of how distance influences perceived size in optics, providing deeper understanding of magnification's effects.
Optical Physics
Optical physics deals with the study of light's behavior and characteristics. In the context of a pinhole magnifier, optical physics explains how light passes through a tiny aperture to form images. This process relies on controlling light diffraction, the bending of light around the edges of the pinhole. Light behaves predictably by following straight-line paths. It forms sharp images when passing through a sufficiently small opening. A pinhole limits scattered light, emphasizing clarity over brightness. This highlights one of the magical principles of optical physics. Through such explorations, learners can appreciate how light's wave properties affect visualization in practical applications like pinhole magnification.
Experimental Setup
Setting up an experiment with a pinhole magnifier involves a few precise steps to ensure valid and clear observations. You start by making two sets of marks on paper鈥攐ne set with 2 cm spacing and the other with 2 mm spacing. These serve as your objects of comparison. Place a pinhole between one eye and the paper with smaller marks, ensuring it is well-lit from behind. With the other eye, look at the larger marks at a further distance. The task is to move the smaller marks closer until they appear to align in size with the larger marks as viewed normally. This setup is crucial to effectively explore and understand magnification, providing a hands-on approach to learning about real-life applications of geometric optics.
Geometric Optics
Geometric optics is a branch of optics that uses concepts like rays, images, and magnification to explain how light interacts with materials to form images. It assumes light travels in straight lines, which are essential for understanding pinhole magnifiers.When using a pinhole magnifier, geometric optics helps us calculate the magnification by considering the relationship between object distance and image size. The key formula \[ M = \frac{d_1}{d_2} \]comes from these geometric principles. This formula shows that the perceived size of the object is a direct function of the distances involved.Understanding these optics helps learners link physical movements in experiments to theoretical equations, enriching comprehension of magnifying effects through simple, yet profound, geometry in physics.

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Most popular questions from this chapter

Use your diffraction grating as follows to measure the wavelengths of the red and green passed by your filters. Put a line (or point) source right next to a wall or door. Make a mark on the wall about a foot to the side of the source. Look at the source through the grating, holding the filter over your grating (or put the filter over the source-but don't melt it!) Move closer and farther from the source until the color of interest appears to be superposed with your mark on the wall. Measure the appropriate distances and calculate \(\lambda\). Thus calibrate the wavelengths transmitted by your red, green, and purple filters. Memorize the results. (Then you can use your filters and the grating to find the wavelengths of other colors when you wish to, without repeating the geometric measurement of this experiment.)

A baby-food jar full of air and immersed in water is a diverging lens. Use a fish tank with glass sides, or use an ordinary pan with a mirror to change a vertically downward flashlight beam into a horizontal beam. Put a little milk in the water so you can see the beam. A good pencil-sized beam is obtained from a flashlight covered by an opaque piece of cardboard with an off-center hole. (The flashlight bulb is usually irregular at the tip. Also, you don't want the direct light from the bulb, which falls off as the inverse square of distance, but the parallel beam from the parabolic reflector.) You can study lenses of air and mineral oil and glass using a suspension of milk in water to see the beam.

Two thin lenses in series. Given two thin lenses of power \(f_{1}^{-1}\) and \(f_{2}^{-1}\) arranged in series along a common axis, with separation \(s\) between the two lenses. Take both lenses to be positive lenses. (The results will hold in general, with suitable interpretations of signs.) Consider a ray parallel to the axis, at distance \(h\) from the axis, and incident on the first lens. Say the ray is incident from the left, and the lenses are in order 1,2 from the left. The first lens deflects the ray toward the axis. Assume the ray hits the second lens before crossing the axis. Find the focal point \(F\) where the ray crosses the axis after leaving the second lens. Show the location of \(F\) is independent of \(h\) (for small-angle approximations). Now define the location \(P\) (which stands for "prineipal plane") as follows: Extrapolate the incident ray forward (to the right\\} and the emergent ray (the one that goes through \(F\) ) backward until they cross. They cross at the principal plane \(P\). Let \(x\) be the distance of \(F\) to the right of the second lens. Let \(y\) be the distance of \(P\) to the left of the second lens. Then \(x+y\) is the distance of the focal plane \(F\) to the right of the principal plane \(P\). This distance is called the focal length \(f\) of the combination of the two lenses, considered as though it were a single thin lens located at the principal plane \(P\). Find \(x, y\), and \(f\) in terms of \(f_{1}, f_{2}\), and \(s\). Once you have found \(f\) and \(P\) for rays going from left to right, do the same for rays traveling from right to left. Are the focal lengths the same? Are the principal planes at the same place? Ans. For rays incident from the left, $$ \begin{aligned} f^{-1} &=f_{1}-1+f_{2}-1-s f_{1}-1 f_{2}-1_{;} \\ x &=\left(1-s f_{1}-1\right) f ; \\ y &=s f_{1}^{-1} f . \end{aligned} $$

When you put your face under water and try looking without a face mask, everything looks blurred, because the change of index of refraction in going from water to eye is not very great. As a simplification, assume there is no change in index. Also assume your eye lens has very little effect, as if all the focusing were done at the first air-to-eye interface. (This is a crude approximation. Actually, you can see underwater to same extent.) Assume the focal length of that first surface is \(3 \mathrm{~cm}\), and that a parallel beam of light in air is brought to a focus at the retina. When you look underwater, you lose that focusing action. Design glasses that can be worn underwater so as to enable you to see clearly. Use glass with index of refraction 1.5. Show that if the focal length when used underwater is \(3 \mathrm{~cm}\), then the focal length when used in air is about \(1 \mathrm{~cm} .\) If one of these glass lenses is used as an ordinary magnifying glass what is its magnification? Suppose you use an ordinary glass marble for the lens. You want to form an image (of a parallel beam in water) \(3 \mathrm{~cm}\) behind the rear surface of the marble. What should be the diameter of the marble?

Pour some table salt on a wet knife or spoon (one that you don't mind ruining). Set the knife in the flame of a gas stove. Look at the yellow flame through your diffraction grating (this is easiest at night in a darkened room). Notice that the first-order (and higher-order) images of the yellow sodium flame are as sharp and clear as the zeroth-order "direct" image. That is because the yellow light is a "spectral line" having narrow bandwidth. (Actually the yellow light from sodium is a "doublet" of two lines with wavelengths 5890 and \(5896 \AA\).) Now look at a candle. In zeroth order, it does not look terribly different from the sodium flame; they are both yellow. But in the first-order diffraction image, the candle is very much spread out in color, whereas the sodium remains sharp. The "yellow" of the candle, which is due to hot particles of carbon, has a wavelength spectrum extending over (and beyond) the entire visible range. Here are other convenient sources of sharp spectral lines; look at them through your grating: Mercury vapor: Fluorescent lamps, mercury-vapor street lights, sunlamps. (A sunlamp is convenient in that it screws directly into an ordinary 110 -volt AC socket. It is probably the cheapest source of mercury-vapor spectral lines; the cost is about \(\$ 10 .\).) Neon: Many advertising signs. Neon has a profusion of lines; you see "many signs." A cheap broad monochromatic source is a G.E. bulb NE-34 which screws directly into a 110 -volt AC socket (the cost is about \(\$ 1.60)\). Others are a "circuit continuity tester," which plugs into any wall receptacle and which costs about \(\$ 1\) (at a hardware store), and a neon "night light." Strontium: Strontium chloride salt (available at a chemical supply house for about 25 cents \(/ \mathrm{oz}\) ); dissolve a little in a few drops of water and put it in the gas flame on your ruined spoon. The wavelength of the red line is a famous length standard. Copper: Copper sulfate; availability and technique as for strontium chloride. It gives a beautiful green color. Hydrocarbon: Look at your gas flame in the first-order spectrum. There are a sharp, clear blue image and a sharp, clear green image. The "blue" color of the flame is therefore due to one or more almost monochromatic spectral lines.
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