91Ó°ÊÓ

Pinhole Camera. 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 \(\mathrm{cm}\) square and 20.0 \(\mathrm{cm}\) deep, with the hole in the middle of one of the 25 \(\mathrm{cm} \times 25 \mathrm{cm}\) faces. If this camera is used to photograph a fierce chicken that is 18 \(\mathrm{cm}\) high and 1.5 \(\mathrm{m}\) in front of the camera, how large is the image of this bird on the film? What is the magnification of this camera?

Step by step solution

01

Drawing the Ray Diagram

Begin by drawing a rectangle to represent the box of the pinhole camera. Place a small dot on one side near the center to represent the pinhole. Position an object (like a fierce chicken) opposite the pinhole. Draw rays from the top and bottom of the object through the pinhole projecting onto the opposite side, where the film is. These rays will form an inverted image of the object inside the box.

02

Determine Image Size

Use similar triangles to compute the size of the image. The ratio of the object height to the distance from the pinhole is equal to the ratio of the image height to the depth of the camera box. Therefore, \( \frac{h_o}{d_o} = \frac{h_i}{d_i} \), where \( h_o = 18 \, \mathrm{cm} \), \( d_o = 150 \, \mathrm{cm} \), and \( d_i = 20 \, \mathrm{cm} \). Rearrange to solve for \( h_i \), the image height.

03

Calculate Image Height

Substitute the known values into the equation: \( \frac{18}{150} = \frac{h_i}{20} \). Solving this gives \( h_i = \frac{18 \times 20}{150} = 2.4 \, \mathrm{cm} \). The image of the chicken on the film is 2.4 cm tall.

04

Determine Magnification

Magnification (\(M\)) is calculated by the formula \(M = \frac{h_i}{h_o}\). Using the values \( h_i = 2.4 \mathrm{cm} \) and \( h_o = 18 \mathrm{cm} \), we find \( M = \frac{2.4}{18} = 0.133 \). Hence, the magnification is 0.133.

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Key Concepts

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

Ray Diagram

A ray diagram for a pinhole camera is a simple yet effective way to visualize how images are formed without a lens. Imagine a rectangular box representing the camera body, with a pinhole on one face. This pinhole acts as the camera's "eye." Outside this pinhole, place an object, such as a fierce chicken, standing upright. Through this tiny hole, light rays travel in straight lines from the object to the opposite side where the film senses the image.

• Draw rays from the topmost point of the chicken, passing through the pinhole and hitting the back of the camera box. These continue straight without bending as there's no lens to refract them.
• Similarly, draw rays from the bottom of the chicken through the pinhole.

The image formed on the film will be upside down. This occurs because the light rays crossing through the pinhole invert the image diagonally across the box's rear face.

Image Formation

In a pinhole camera, the formation of the image is solely based on the pinhole's light-passing ability. This tiny hole allows images to be created through a process called "light inversion."

• The light emanates from every point on the object and enters the hole, projecting directly onto the film located on the opposite side.
• Each ray travels in a direct, linear path, producing an inverted image on the film due to the crossing points of these paths.

The pinhole camera does not have a lens to assist in bending the light and thus relies entirely on the principle that light travels straight. The image's clarity depends on the pinhole size; smaller holes give sharper images, but they also let less light in, making the image dimmer.

Magnification

Magnification in a pinhole camera is an important concept as it determines how much larger or smaller the image will appear compared to the actual object. Calculating it involves a simple formula: \[ M = \frac{h_i}{h_o} \]where \( h_i \) is the height of the image and \( h_o \) is the height of the object.
For the chicken example, using the values \( h_i = 2.4 \mathrm{cm} \) and \( h_o = 18 \mathrm{cm} \), the magnification becomes \( M = 0.133 \).

• If \( M \) is less than 1, the image is smaller than the object.
• If \( M \) is more than 1, the image is larger than the object.
• An \( M \) of 1 means the image and the object are of identical size.

In this example, the image of the chicken is smaller than the actual chicken due to the camera box's constraints.

Similar Triangles

The concept of similar triangles is pivotal for understanding how image size is computed in a pinhole camera. When light rays pass through the pinhole, they form similar triangles with the object and the image.

• The triangle formed by the chicken and the light rays to the pinhole corresponds proportionally to the triangle formed by the image and those rays on the film side.
• These triangles have the same shape, meaning their angles are identical, but their sizes differ due to the distances from the pinhole.

This similarity allows us to use the triangle proportion: \[ \frac{h_o}{d_o} = \frac{h_i}{d_i} \]where \( h_o \) and \( h_i \) are the heights of the object and the image, respectively, and \( d_o \) and \( d_i \) are the distances from the pinhole to the object and image. This key formula is why the image of the chicken formed in the pinhole camera box can be precisely calculated.