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Chapter 25: Problem 2

A camera has a lens with an aperture diameter of \(8.00 \mathrm{~mm}\). It is used to photograph a pet dog. What aperture diameter would correspond to an increase in the intensity of the dog's image on the film by a factor of \(2 ?\)

Short Answer

Expert verified
The aperture diameter should increase to approximately 11.32 mm.

Step by step solution

01

Understanding Light Intensity and Aperture

The intensity of an image captured by a camera is proportional to the area of the aperture of the camera lens. The aperture area is calculated as the area of a circle with radius equal to half the diameter. Intensity is proportional to \( \pi \left( \frac{D}{2} \right)^2 \) where \(D\) is the diameter of the aperture.
02

Calculating the Original Aperture Area

First, we calculate the area of the original aperture with a diameter of 8.00 mm: \[ A_1 = \pi \left( \frac{8.00}{2} \right)^2 = \pi (4)^2 = 16\pi \mathrm{~mm^2} \]
03

Determine Required Aperture Area for Increased Intensity

The problem states that we need to increase the intensity by a factor of 2. This means the area of the aperture must also increase by a factor of 2. Therefore, we want the new area \(A_2 = 2 \times A_1 = 2 \times 16\pi = 32\pi \mathrm{~mm^2} \).
04

Solving for the New Aperture Diameter

Set up the equation for the new area in terms of the new diameter \(D_2\): \[ \pi \left( \frac{D_2}{2} \right)^2 = 32\pi \mathrm{~mm^2} \] Simplify and solve for \(D_2\): \[ \left( \frac{D_2}{2} \right)^2 = 32 \] \[ \left( \frac{D_2}{2} \right) = \sqrt{32} \] \[ D_2 = 2 \times \sqrt{32} \] Since \( \sqrt{32} = 4\sqrt{2} \approx 5.66 \), \[ D_2 \approx 11.32 \mathrm{~mm} \]
05

Conclusion

The new aperture diameter, corresponding to the increased intensity of the image, is approximately 11.32 mm.

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Key Concepts

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

Aperture and Image Intensity
In photography, the aperture refers to the opening of a camera lens through which light passes to enter the camera. It works similarly to the pupil of an eye, expanding and contracting to allow more or less light in.
The size of the aperture is crucial because it directly affects the image's brightness, called image intensity. This is key when photographing in different lighting conditions.
The aperture is expressed in f-stops, like f/2.8, f/4, etc., which are ratios of the lens's focal length to the aperture diameter. A smaller f-stop number corresponds to a larger aperture, allowing more light into the camera sensor, thereby increasing image intensity.
**Aperture Impact on Image Intensity**
- Larger aperture (smaller f-stop) results in higher light intensity, creating brighter images. - Smaller aperture (larger f-stop) reduces light intensity, which can make an image appear darker.
Understanding the relationship between aperture size and image intensity helps photographers adapt to varying environments, whether capturing a sunny landscape or a dimly lit indoor scene.
Camera Lens Diameter
The diameter of a camera lens aperture plays a pivotal role in determining how much light can pass through.
The larger the diameter, the greater the area through which light can enter.
This is mathematically defined as the area of a circle, which is calculated using the formula: \[ \text{Area} = \pi \left( \frac{D}{2} \right)^2 \]where \(D\) is the aperture diameter.
**Why Diameter Matters**
- A larger aperture diameter means more light enters, which increases image brightness.- It also influences the depth of field鈥攖he area in sharp focus in an image.Photographers often adjust the lens diameter consciously to play with lighting and background focus. A larger diameter is useful in low-light conditions, allowing adequate exposure without needing a flash or increasing ISO settings, which can lead to noise in the image.
Light Intensity and Area
Light intensity in photography is essentially the amount of light that hits the film or image sensor when taking a photo.
It's a vital component because it determines the exposure and overall brightness of an image. It's important to understand that light intensity is related to the aperture area.
The principle here is simple: - The intensity is proportional to the area of the aperture, which is calculated as: \[ \text{Intensity} \propto \pi \left( \frac{D}{2} \right)^2 \]This illustrates how even a small change in diameter leads to a significant change in image brightness.
**Key Points on Light Intensity**
- Doubling the aperture area doubles the image intensity, allowing more light to create a brighter image in darker settings.- Controlling light intensity through the aperture gives photographers significant creative control, like achieving sharp subjects with blurred backgrounds or capturing vivid details in shadowed areas.This knowledge empowers photographers to optimize their camera settings based on situational lighting and desired photographic effects.

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Most popular questions from this chapter

A camera is focused on an object that is \(1.2 \mathrm{~m}\) from the lens. The distance between the CCD image sensor and the lens is \(5 \mathrm{~cm}\). What is the focal length of the lens?

A camera lens has a focal length of \(200 \mathrm{~mm}\). How far from the lens should the subject for the photo be if the lens is \(20.4 \mathrm{~cm}\) from the film?

Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or image) is the same as the distance from the lens to that object (or image). A certain person can see distant objects well, but his near point is \(45.0 \mathrm{~cm}\) from his eyes instead of the usual \(25.0 \mathrm{~cm}\). (a) Is this person nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct his vision? (c) If the correcting lenses will be contact lenses, what focal-length lens is needed and what is its power in diopters?

A thin lens with a focal length of \(6.00 \mathrm{~cm}\) is used as a simple magnifier. (a) What angular magnification is obtainable with the lens if the object is at the focal point? (b) When an object is examined through the lens, how close can it be brought to the lens? Assume that the image viewed by the eye is at infinity and that the lens is very close to the eye.

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?
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