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A pinhole camera is just a rectangular box with a tiny hole in one face. The film is on the face opposite this hole, and that is where the image is formed. The camera forms an image \(without\) a lens. (a) Make a clear ray diagram to show how a pinhole camera can form an image on the film without using a lens. (\(Hint\): Put an object outside the hole, and then draw rays passing through the hole to the opposite side of the box.) (b) A certain pinhole camera is a box that is 25 cm square and 20.0 cm deep, with the hole in the middle of one of the 25 cm \(\times\) 25 cm faces. If this camera is used to photograph a fierce chicken that is 18 cm high and 1.5 m in front of the camera, how large is the image of this bird on the film? What is the lateral magnification of this camera?

Short Answer

Expert verified
The image of the chicken on the film is 2.4 cm tall with a magnification of approximately 0.133.

Step by step solution

01

Understanding Pinhole Camera Functionality

In a pinhole camera, light rays from an object pass through a small hole and project an inverted image on the opposite side of the box. The key is that light travels in straight lines, so only rays traveling directly through the pinhole reach the film, creating the image without using a lens.
02

Drawing the Ray Diagram

Draw a box to represent the camera. Place an object in front of the hole. From the top and bottom of the object, draw two lines towards the pinhole. These lines should continue straight from the pinhole to the back side of the box. Where they intersect the back side is where the top and bottom of the image will form, thus creating an inverted image.
03

Calculating the Image Height

The dimensions of the box are 25 cm by 25 cm by 20 cm depth. The object height is 18 cm, and it is placed 1.5 m or 150 cm away from the pinhole. Using similar triangles, the ratio of the image height (\( h' \)) to the object height (\( h = 18 \text{ cm} \)) is equal to the ratio of the image distance (\( d' = 20 \text{ cm} \)) to the object distance (\( d = 150 \text{ cm} \)). Thus, \( \frac{h'}{18} = \frac{20}{150} \).
04

Solving for Image Height

From the equation \( \frac{h'}{18} = \frac{20}{150} \), cross-multiply to find \( h' \). This gives \( h' = \frac{18 \times 20}{150} \). Calculating this, we find \( h' = 2.4 \text{ cm} \).
05

Calculating Lateral Magnification

Lateral magnification (M) is defined as the ratio of the image height to the object height. Thus, \( M = \frac{h'}{h} = \frac{2.4}{18} \). Simplifying the fraction, we get \( M \approx 0.133 \).

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Key Concepts

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

Ray Diagram
In the world of optics, a ray diagram is a crucial tool to understand how light behaves in various conditions. This is particularly important when explaining how a pinhole camera functions.
In a pinhole camera, light travels in straight lines through a tiny hole to form an image on the opposite side of the camera box. Here's a simple way to visualize it:
  • Imagine a rectangular box, which serves as the pinhole camera.
  • Place an object, like a chicken, in front of the hole.
  • Draw straight lines representing light rays passing from the top and bottom of your object through the pinhole to the other side of the box.
These light rays intersect the back of the box, forming an inverted image. The inverted nature results from the crossing of light rays passing through the small hole. Only rays traveling directly through the pinhole will create an image, allowing us to capture clear, albeit upside-down, pictures without any lenses.
Understanding ray diagrams is key to recognizing how simply light can project complex images.
Lateral Magnification
Lateral magnification measures how much larger or smaller the image is compared to the object itself in a pinhole camera. It is calculated as the ratio of the height of the image (\(h'\)) to the height of the object (\(h\)).
Using the formula:
  • \( M = \frac{h'}{h} \)
We can substitute the values for our fierce chicken example:
  • \( h = 18 \text{ cm} \)
  • \( h' = 2.4 \text{ cm} \)
Thus, the lateral magnification is:
  • \( M = \frac{2.4}{18} \approx 0.133 \)
This means the image is approximately 13.3% the size of the actual chicken. In a pinhole camera, the magnification is often less than 1, indicating that the image will be smaller than the original object.
Similar Triangles
Viewing the imaging process through similar triangles allows us to calculate the size of the image formed in the pinhole camera relative to the original object. These triangles are formed by:
  • The object and its shadow on the film.
  • The path of light from the top and bottom of the object intersecting through the pinhole.
When we use similar triangles, we establish the relationship between different parts of the image and the object. The formula we use is:
  • \( \frac{h'}{h} = \frac{d'}{d} \)
Where \(d = 150\) cm is the distance from the object to the pinhole, and \(d' = 20\) cm is the distance from the pinhole to the film.
Using this relationship, we set up the proportion:
  • \( \frac{h'}{18} = \frac{20}{150} \)
Solving for \(h'\), we find that the image height, \(h'\), is 2.4 cm. This simplification thanks to similar triangles makes complex calculations straightforward and intuitive.

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Most popular questions from this chapter

A spherical, concave shaving mirror has a radius of curvature of 32.0 cm. (a) What is the magnification of a person's face when it is 12.0 cm to the left of the vertex of the mirror? (b) Where is the image? Is the image real or virtual? (c) Draw a principal-ray diagram showing the formation of the image.

An insect 3.75 mm tall is placed 22.5 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude 13.0 cm, and the index of refraction of the lens material is 1.70. (a) Calculate the location and size of the image this lens forms of the insect. Is it real or virtual? Erect or inverted? (b) Repeat part (a) if the lens is reversed.

A person with a near point of \(85 \mathrm{~cm},\) but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare. (a) If the lenses of the old pair have a power of \(+2.25\) diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest \(2.0 \mathrm{~cm}\) in front of his eye? (b) What would his near point be if his old glasses were contact lenses instead?

A person swimming 0.80 m below the surface of the water in a swimming pool looks at the diving board that is directly overhead and sees the image of the board that is formed by refraction at the surface of the water. This image is a height of 5.20 m above the swimmer. What is the actual height of the diving board above the surface of the water?

The eyepiece of a refracting telescope (see Fig. 34.53) has a focal length of 9.00 cm. The distance between objective and eyepiece is 1.20 m, and the final image is at infinity. What is the angular magnification of the telescope?

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