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Chapter 34: Problem 49

When a camera is focused, the lens is moved away from or toward the digital image sensor. If you take a picture of your friend, who is standing \(3.90 \mathrm{~m}\) from the lens, using a camera with a lens with an \(85 \mathrm{~mm}\) focal length, how far from the sensor is the lens? Will the whole image of your friend, who is \(175 \mathrm{~cm}\) tall, fit on a sensor that is \(24 \mathrm{~mm} \times 36 \mathrm{~mm} ?\)

Short Answer

Expert verified
The distance of the lens from the sensor and whether the whole image of your friend will fit on the sensor can be determined after applying the appropriate variables into the respective formulas and executing the calculations. If the calculated image height is less than the height of the sensor, the image will fit completely.

Step by step solution

01

Calculate Image Distance

Rearrange the lens formula to solve for image distance (\(v\)): \(v = 1 / (1/f - 1/u)\). Substitute known values into the equation: \(v = 1 / (1/(-85mm) - 1/(-3.9m))\). Ensure that all quantities are in the same units (here, millimeters) before calculating.
02

Calculate Height of the Image

The height of the image can be calculated using the magnification formula, \(m = -v/u\). The negative sign indicates that the image is inverted. So, the image height, \(h'\), can be found by: \(h' = m \times h = -v/u \times h\). Substitute the known values to find \(h'\).
03

Check if the Image Fits on the Sensor

Once we have the height of the image, we can check if it fits on the image sensor by comparing it to the dimensions of the sensor. The height of the image should be less than or equal to the height of the sensor, to fit completely.

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Key Concepts

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

Lens Formula
The lens formula is a fundamental concept in optics and is used to relate the object distance, image distance, and the focal length of a lens. It is expressed as: \[ \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \]Where:- \( u \) is the object distance (the distance between the object and the lens).- \( v \) is the image distance (the distance between the image and the lens).- \( f \) is the focal length of the lens.To solve most exercises involving lenses, the lens formula allows us to calculate any one of the three parameters if the other two are known. For example, in the case of the camera optics exercise, we rearrange the formula to find the image distance given the focal length and the object distance. Remember, it's crucial to keep all measurements in the same unit when using the formula, typically converting everything to meters or millimeters as needed.
Image Distance
Image distance is the distance from the lens to the image that is formed. This concept is key when determining whether an image will focus correctly on a sensor. Using the lens formula, we can isolate the image distance \( v \) with the equation:\[ v = \frac{1}{\left(\frac{1}{f} - \frac{1}{u}\right)} \]In our exercise, the image distance tells us how far the lens needs to be from the sensor to focus the image correctly. Calculating \( v \) involves substituting the given focal length and object distance into the equation while ensuring they are in the same units.For a real-world application, as in a camera, the lens adjusts its position from the sensor to achieve this focused state.
Focal Length
The focal length \( f \) of a lens is an inherent property describing how strongly the lens converges or diverges light. It is measured from the lens to the focal point, where parallel rays of light meet after passing through the lens.In the camera scenario, the focal length of 85 mm determines the lens's capability to focus light on the sensor from objects at various distances. A shorter focal length implies a wider field of view, while a longer focal length provides a magnified view but a narrower field of view.Understanding focal length is crucial when setting up a camera to ensure that you capture the desired frame and focus on distant or near objects with clarity.
Magnification Formula
The magnification formula relates the height of the image to the height of the object, as well as the image and object distances:\[ m = \frac{h'}{h} = -\frac{v}{u} \]- \( m \) is the magnification.- \( h' \) is the height of the image.- \( h \) is the height of the object.- The negative sign indicates the image is inverted.In solving the problem, this formula helps determine whether the image of the friend will fit on the sensor. By calculating the image height using:\[ h' = m \times h \]It helps us assess if the image dimensions are suitable for the sensor's size, considering both the height and width constraints.

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Most popular questions from this chapter

A \(1.20-\mathrm{cm}\) -tall object is \(50.0 \mathrm{~cm}\) to the left of a converging lens of focal length \(40.0 \mathrm{~cm}\). A second converging lens, this one having a focal length of \(60.0 \mathrm{~cm},\) is located \(300.0 \mathrm{~cm}\) to the right of the first lens along the same optic axis. (a) Find the location and height of the image (call it \(I_{1}\) ) formed by the lens with a focal length of \(40.0 \mathrm{~cm}\). (b) \(I_{1}\) is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses.

You want to view through a magnifier an insect that is \(2.00 \mathrm{~mm}\) long. If the insect is to be at the focal point of the magnifier, what focal length will give the image of the insect an angular size of 0.032 radian?

It is your first day at work as a summer intern at an optics company. Your supervisor hands you a diverging lens and asks you to measure its focal length. You know that with a converging lens, you can measure the focal length by placing an object a distance \(s\) to the left of the lens, far enough from the lens for the image to be real, and viewing the image on a screen that is to the right of the lens. By adjusting the position of the screen until the image is in sharp focus, you can determine the image distance \(s^{\prime}\) and then use Eq. (34.16) to calculate the focal length \(f\) of the lens. But this procedure won't work with a diverging lens - by itself, a diverging lens produces only virtual images, which can't be projected onto a screen. Therefore, to determine the focal length of a diverging lens, you do the following: First you take a converging lens and measure that, for an object \(20.0 \mathrm{~cm}\) to the left of the lens, the image is \(29.7 \mathrm{~cm}\) to the right of the lens. You then place a diverging lens \(20.0 \mathrm{~cm}\) to the right of the converging lens and measure the final image to be \(42.8 \mathrm{~cm}\) to the right of the converging lens. Suspecting some inaccuracy in measurement, you repeat the lenscombination measurement with the same object distance for the converging lens but with the diverging lens \(25.0 \mathrm{~cm}\) to the right of the converging lens. You measure the final image to be \(31.6 \mathrm{~cm}\) to the right of the converging lens. (a) Use both lens-combination measurements to calculate the focal length of the diverging lens. Take as your best experimental value for the focal length the average of the two values. (b) Which position of the diverging lens, \(20.0 \mathrm{~cm}\) to the right or \(25.0 \mathrm{~cm}\) to the right of the converging lens, gives the tallest image?

A compound microscope has an objective lens with focal length \(14.0 \mathrm{~mm}\) and an eyepiece with focal length \(20.0 \mathrm{~mm}\). The final image is at infinity. The object to be viewed is placed \(2.0 \mathrm{~mm}\) beyond the focal point of the objective lens. (a) What is the distance between the two lenses? (b) Without making the approximation \(s_{1} \approx f_{1},\) use \(M=m_{1} M_{2}\) with \(m_{1}=-s_{1}^{\prime} / s_{1}\) to find the overall angular magnification of the microscope. (c) What is the percentage difference between your result and the result obtained if the approximation \(s_{1} \approx f_{1}\) is used to find \(M ?\)

A pencil that is \(9.0 \mathrm{~cm}\) long is held perpendicular to the surface of a plane mirror with the tip of the pencil lead \(12.0 \mathrm{~cm}\) from the mirror surface and the end of the eraser \(21.0 \mathrm{~cm}\) from the mirror surface. What is the length of the image of the pencil that is formed by the mirror? Which end of the image is closer to the mirror surface: the tip of the lead or the end of the eraser?
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