Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A camera lens has a focal length of 180.0 mm and an aperture diameter of 16.36 mm. (a) What is the \(f\)-number of the lens? (b) If the correct exposure of a certain scene is \(1\over 30\)s at \(f/\)11, what is the correct exposure at \(f/\)2.8?

Short Answer

Expert verified
(a) The f-number is 11. (b) The correct exposure at f/2.8 is approximately 0.5 seconds.

Step by step solution

01

Understand the f-Number Formula

The f-number, often denoted as \( f/\) or \( N \), is calculated using the formula \( N = \frac{f}{D} \), where \( f \) is the focal length of the lens, and \( D \) is the aperture diameter.
02

Calculate the f-Number

Using the given focal length \( f = 180.0 \) mm and aperture diameter \( D = 16.36 \) mm, we apply the formula: \[ N = \frac{180.0}{16.36} \approx 11.00. \] Thus, the f-number is approximately 11.
03

Understand Exposure Relation

The exposure time needed for a lens is related to the square of the f-number, given a constant amount of light, by the formula \( t_2 = t_1 \left( \frac{N_2}{N_1} \right)^2 \). This formula helps us find the new exposure time given a different f-number.
04

Calculate New Exposure Time

We're given \( t_1 = \frac{1}{30} \) seconds at \( f/11 \), and we need to find \( t_2 \) at \( f/2.8 \). Using the formula: \[ t_2 = \frac{1}{30} \left( \frac{11}{2.8} \right)^2 \approx \frac{1}{30} \times 15.49 \approx \frac{1}{2}. \] Thus, the correct exposure at \( f/2.8 \) is approximately \( 0.5 \) seconds.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Focal Length
In the world of camera optics, understanding focal length is essential. Focal length is the distance between the lens and the image sensor when the subject is in focus. It is usually measured in millimeters (mm). This length determines the field of view and magnification of the image.

A longer focal length results in a narrower field of view, making it easier to zoom in on distant subjects. Conversely, a shorter focal length offers a wider view, suitable for capturing vast landscapes. Using a lens with a focal length of 180.0 mm, for instance, means that it's likely geared toward zoomed-in shots or portraits where the background is less emphasized.
Aperture
Aperture refers to the size of the opening in the lens through which light enters. This opening can be adjusted, much like the pupil of an eye. A larger aperture allows more light to reach the camera sensor.

Aperture size is typically expressed in terms of the diameter of the aperture opening, measured in millimeters. In the given example, the lens has an aperture diameter of 16.36 mm. This indicates how wide the lens opening is, impacting the brightness of the captured image. A larger aperture not only increases light intake but also affects the depth of field, enabling beautifully blurred backgrounds, which is especially popular in portrait photography.
Exposure Time
Exposure time, sometimes known as shutter speed, is the duration for which the camera's sensor is exposed to light. It is crucial in determining the overall brightness of the photo. Shorter exposure times freeze motion, while longer times can capture motion blur.

In our exercise, the initial exposure time is \( \frac{1}{30} \) seconds. Adjustment of exposure time is necessary when changing the f-number of the lens.Using a mathematical relationship involving the square of the f-number, we can compute the new exposure time required when the aperture size is altered to maintain consistent illumination.
f-number
The f-number, also known as the f-stop, is a critical concept in optics that influences both exposure and depth of field. It is calculated by dividing the focal length by the aperture diameter, represented as \(N = \frac{f}{D}\).

A smaller f-number (e.g., f/2.8) means a larger aperture and more light entering the camera. This is ideal for shooting in low-light conditions. On the other hand, a larger f-number (e.g., f/11) denotes a smaller aperture, suitable for brightly lit scenes.

In practical terms, understanding the f-number helps photographers adjust exposure settings correctly, as demonstrated in our exercise's calculations. Switching from an f/11 to an f/2.8 aperture requires recalculating the exposure time to ensure the photographed scene remains properly exposed.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A converging meniscus lens (see Fig. 34.32a) with a refractive index of 1.52 has spherical surfaces whose radii are 7.00 cm and 4.00 cm. What is the position of the image if an object is placed 24.0 cm to the left of the lens? What is the magnification?

A screen is placed a distance \(d\) to the right of an object. A converging lens with focal length \(f\) is placed between the object and the screen. In terms of \(f\), what is the smallest value \(d\) can have for an image to be in focus on the screen?

Given that frogs are nearsighted in air, which statement is most likely to be true about their vision in water? (a) They are even more nearsighted; because water has a higher index of refraction than air, a frog's ability to focus light increases in water. (b) They are less nearsighted, because the cornea is less effective at refracting light in water than in air. (c) Their vision is no different, because only structures that are internal to the eye can affect the eye's ability to focus. (d) The images projected on the retina are no longer inverted, because the eye in water functions as a diverging lens rather than a converging lens.

A convex spherical mirror with a focal length of magnitude 24.0 cm is placed 20.0 cm to the left of a plane mirror. An object 0.250 cm tall is placed midway between the surface of the plane mirror and the vertex of the spherical mirror. The spherical mirror forms multiple images of the object. Where are the two images of the object formed by the spherical mirror that are closest to the spherical mirror, and how tall is each image?

In one form of cataract surgery the person's natural lens, which has become cloudy, is replaced by an artificial lens. The refracting properties of the replacement lens can be chosen so that the person's eye focuses on distant objects. But there is no accommodation, and glasses or contact lenses are needed for close vision. What is the power, in diopters, of the corrective contact lenses that will enable a person who has had such surgery to focus on the page of a book at a distance of 24 cm?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free