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Chapter 25: Problem 1

\(\cdot\) The focal length of an \(f / 4\) camera lens is 300 \(\mathrm{mm}\) . (a) What is the aperture diameter of the lens? (b) If the correct exposure of a certain scene is \(\frac{1}{250}\) s at \(f / 4,\) what is the correct exposure at \(f / 8 ?\)

Short Answer

Expert verified
(a) 75 mm; (b) \(\frac{1}{62.5}\) s.

Step by step solution

01

Understanding the Given Information

We start with an \(f/4\) camera lens which has a focal length of 300 mm. The \(f\)-number describes the aperture size in relation to the focal length. The correct exposure at \(f/4\) is given as \(\frac{1}{250}\) s, and we need to calculate the exposure time for \(f/8\).
02

Calculating the Aperture Diameter

The formula for the aperture diameter \(D\) in relation to the \(f\)-number \(N\) and the focal length \(f\) is \(D = \frac{f}{N}\). For \(f/4\), the aperture diameter is \(D = \frac{300\,\text{mm}}{4} = 75\,\text{mm}\).
03

Understand \(f/\) Number Changes

When the \(f\)-number changes from \(f/4\) to \(f/8\), the amount of light reaching the sensor decreases. Since \(f/8\) is two \(f\)-stops smaller than \(f/4\), the light is reduced by a factor of 4.
04

Calculating the Correct Exposure Time for \(f/8\)

To compensate for the reduced light at \(f/8\), we need to increase the exposure time by a factor of 4. The original exposure time at \(f/4\) is \(\frac{1}{250}\) s, so the new exposure time at \(f/8\) is \(4 \times \frac{1}{250} = \frac{1}{62.5}\) s.

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Key Concepts

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

Aperture Diameter
The term "aperture diameter" refers to the opening through which light enters a camera lens. This opening plays a crucial role in determining the amount of light that reaches the camera's sensor. A larger aperture diameter lets in more light, which can be particularly useful in low-light conditions. The aperture diameter can be calculated using the formula:
\[ D = \frac{f}{N} \]where:
  • \(D\) is the aperture diameter.
  • \(f\) is the focal length of the lens.
  • \(N\) is the \(f\)-number.
In our exercise, with a focal length of 300 mm and an \(f/4\) setting, the aperture diameter is 75 mm. This size allows ample light to penetrate the lens, making it suitable for a variety of lighting situations.
Exposure Time
Exposure time, also known as shutter speed, is the duration for which the camera's shutter is open to allow light to hit the sensor. It plays a vital role in photography, affecting the brightness and clarity of the photo. In simpler terms, a longer exposure time means more light is captured, resulting in a brighter image.In the problem provided, the correct exposure time at \(f/4\) is given as \(\frac{1}{250}\) seconds. When changing to \(f/8\), we need to adjust the exposure time to account for the reduced light. Since \(f/8\) lets in less light than \(f/4\), we need to increase the exposure time by a factor of 4 (as two \(f\)-stops reduce the light). Therefore, the adjusted exposure time for \(f/8\) is \(\frac{1}{62.5}\) seconds, which helps maintain the necessary light for proper image exposure.
f-stop
The f-stop, or \(f\)-number, is a measure of the lens's aperture size relative to its focal length. It controls the depth of field in a photo and influences how much light enters through the lens. The \(f\)-stop scale is such that:
  • Lower \(f\)-numbers (e.g., \(f/2\), \(f/4\)) have larger aperture openings and allow more light.
  • Higher \(f\)-numbers (e.g., \(f/8\), \(f/16\)) reduce the aperture size and allow less light.
Switching from \(f/4\) to \(f/8\) in the exercise reduces the light by a factor of 4. This is because each increase by one \(f\)-stop (e.g., from \(f/4\) to \(f/5.6\), then \(f/5.6\) to \(f/8\)) halves the amount of light reaching the sensor.

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Most popular questions from this chapter

\(\cdot\) Camera \(A\) has a lens with an aperture diameter of 8.00 \(\mathrm{mm}\) . It photographs an object, using the correct exposure time of \(\frac{1}{30}\) s. What exposure time should be used with camera \(\mathrm{B}\) in photographing the same object with the same film if camera \(B\) has a lens with an aperture diameter of 23.1 \(\mathrm{mm}\) ?

\(\bullet\) Physician, heal thyself! (a) Experimentally determine the near and far points for both of your own eyes. Are these points the same for both eyes? (All you need is a tape measure or ruler and a cooperative friend.) (b) Design correcting lenses, as needed, for your closeup and distant vision in one of your eyes. If you prefer contact lenses, design that type of lens. Otherwise design lenses for ordinary glasses, assuming that they will be 2.0 \(\mathrm{cm}\) from your eye. Specify the power (in diopters) of each correcting lens.

\(\bullet\) It's all done with mirrors. A photographer standing 0.750 \(\mathrm{m}\) in front of a plane mirror is taking a photograph of her image in the mirror, using a digital camera having a lens with a focal length of 19.5 \(\mathrm{mm}\) (a) How far is the lens from the light sensors of the camera? (b) If the camera is 8.0 \(\mathrm{cm}\) high, how high is its image on the sensors?

A telescope is constructed from two lenses with focal lengths of 95.0 \(\mathrm{cm}\) and \(15.0 \mathrm{cm},\) the 95.0 -cm lens being used as the objective. Both the object being viewed and the final image are at infinity. (a) Find the angular magnification of the telescope. (b) Find the height of the image formed by the objective of a building 60.0 \(\mathrm{m}\) tall and 3.00 \(\mathrm{km}\) away. (c) What is the angular size of the final image as viewed by an eye very close to the eyepiece?

\(\cdot\) A person can see clearly up close, but cannot focus on objects beyond 75.0 \(\mathrm{cm}\) . She opts for contact lenses to correct her vision. (a) Is she nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed? (b) What type vision? (c) What focal-length contact lens is needed, and what is its power in diopters?
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