A camera has a lens with a focal length of \(60 . \mathrm{mm} .\) Suppose you replace the normal lens with a zoom lens whose focal length can be varied from \(35 . \mathrm{mm}\) to \(250 . \mathrm{mm}\) and use the camera to photograph an object at infinity. Compared to a 60.-mm lens, what magnification of the image would be achieved using the \(240 .-\mathrm{mm}\) focal length?

Short Answer

Expert verified
Answer: The magnification of the image when using a focal length of 240 mm compared to a normal 60 mm lens is 4 times.

Step by step solution

01

Identify the given information

In this exercise, we know the focal length of the normal lens is 60 mm and the focal length of the zoom lens when comparing magnification is 240 mm.
02

Calculate the magnification factor

The magnification factor is given by the ratio of the focal lengths of the zoom lens and the normal lens. Magnification factor = \(\frac{Focal\: length\: of\: zoom\: lens}{Focal\: length\: of\: normal\: lens}\)
03

Substitute the known values

Now, let's substitute the focal lengths we know into the equation: Magnification factor = \(\frac{240\: mm}{60\: mm}\)
04

Simplify the result

We can simplify the fraction to get the magnification factor: Magnification factor = \(4\) Compared to the 60-mm lens, a 240-mm focal length would achieve a 4 times magnification of the image.

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
Focal length is a fundamental concept in the field of optics and it has vast implications in photography and vision science. To put it simply, the focal length of a lens is the distance from the lens to the point where the light rays converge to form a sharp image. This point is known as the focal point. When an object is at infinity, as mentioned in the exercise, the light rays arriving are parallel and the focal point is on the image sensor of the camera.

Focal length determines the lens's angle of view, which is how much of the scene will be captured, and also influences the magnification of the image. Shorter focal lengths provide a wider angle of view and a smaller magnification, while longer focal lengths offer a narrower angle of view and a larger magnification. Hence, a zoom lens, which has a variable focal length, allows for a range of magnification options and angles of view. For example, adjusting the zoom lens from 35 mm to 240 mm changes both the magnification and the angle of view, allowing for close-up details or wider scenes as needed.
Zoom Lens
A zoom lens is a type of camera lens that offers the photographer the flexibility to change the focal length within a specified range, essentially allowing for 'zooming' in or out on a subject without changing their position. This adjustability comes in handy across various photographic scenarios, offering versatility where multiple prime lenses (lenses with a fixed focal length) would have been required. When discussing a zoom lens, it is useful to talk about the zoom range, which describes the limits between which the focal length can be adjusted, such as from 35 mm to 250 mm in the given example.

With a zoom lens, the photographer can make objects appear closer or further away, and can also control the composition and how the subject fits within the frame. This flexibility makes zoom lenses particularly popular for sports, wildlife photography, and any situation where the subject distance from the camera is highly variable.
Magnification Factor
The magnification factor, or simply magnification, of an optical system like a camera lens, is the ratio of the image size to the actual size of the object being photographed. It tells you how much larger or smaller the image is compared to the object itself. In the context of the exercise, the magnification factor is determined by comparing the focal length of the zoom lens to that of a standard lens.

In practical terms, when we talk about a '4 times magnification', it means that the image projected by the lens onto the camera's sensor is four times larger than the size of the object as seen with a lens of standard (or reference) focal length. We can calculate this factor by dividing the focal length of the zoom lens by the focal length of the normal lens, as shown in the solution steps. It is important to note that this form of magnification does not take into account the final magnification seen when viewing the photo, which can be affected by factors such as print size and viewing distance.

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

You have found in the lab an old microscope, which has lost its eyepiece. It still has its objective lens, and markings indicate that its focal length is \(7.00 \mathrm{~mm}\). You can put in a new eyepiece, which goes in \(20.0 \mathrm{~cm}\) from the objective. You need a magnification of about 200. Assume you want the comfortable viewing distance for the final image to be \(25.0 \mathrm{~cm}\). You find in a drawer eyepieces marked \(2.00-, 4.00-,\) and \(8.00-\mathrm{cm}\) focal length. Which is your best choice?

Some reflecting telescope mirrors utilize a rotating tub of mercury to produce a large parabolic surface. If the tub is rotating on its axis with an angular frequency \(\omega,\) show that the focal length of the resulting mirror is: \(f=g / 2 \omega^{2}\).

A classmate claims that by using a \(40.0-\mathrm{cm}\) focal length mirror, he can project onto a screen a \(10.0-\mathrm{cm}\) tall bird locat ed 100 . \(\mathrm{m}\) away. He claims that the image will be no less than \(1.00 \mathrm{~cm}\) tall and inverted. Will he make good on his claim?

How large does a \(5.0-\mathrm{mm}\) insect appear when viewed with a system of two identical lenses of focal length \(5.0 \mathrm{~cm}\) separated by a distance \(12 \mathrm{~cm}\) if the insect is \(10.0 \mathrm{~cm}\) from the first lens? Is the image real or virtual? Inverted or upright?

Suppose the near point of your eye is \(2.0 \cdot 10^{1} \mathrm{~cm}\) and the far point is infinity. If you put on -0.20 diopter spec tacles, what will be the range over which you will be able to see objects distinctly?

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