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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 person swimming \(0.80 \mathrm{~m}\) below the surface of the water in a swimming pool looks at the diving board that is directly overhead and sees the image of the board that is formed by refraction at the surface of the water. This image is a height of \(5.20 \mathrm{~m}\) above the swimmer. What is the actual height of the diving board above the surface of the water?

A frog can see an insect clearly at a distance of \(10 \mathrm{~cm}\). At that point the effective distance from the lens to the retina is \(8 \mathrm{~mm} .\) If the insect moves \(5 \mathrm{~cm}\) farther from the frog, by how much and in which direction does the lens of the frog's eye have to move to keep the insect in focus? (a) \(0.02 \mathrm{~cm}\), toward the retina; (b) \(0.02 \mathrm{~cm}\), away from the retina; (c) \(0.06 \mathrm{~cm}\), toward the retina; (d) \(0.06 \mathrm{~cm}\), away from the retina.

A converging lens with a focal length of \(9.00 \mathrm{~cm}\) forms an image of a \(4.00-\mathrm{mm}\) -tall real object that is to the left of the lens. The image is \(1.30 \mathrm{~cm}\) tall and erect. Where are the object and image located? Is the image real or virtual?

A transparent rod \(30.0 \mathrm{~cm}\) long is cut flat at one end and rounded to a hemispherical surface of radius \(10.0 \mathrm{~cm}\) at the other end. A small object is embedded within the rod along its axis and halfway between its ends, \(15.0 \mathrm{~cm}\) from the flat end and \(15.0 \mathrm{~cm}\) from the vertex of the curved end. When the rod is viewed from its flat end, the apparent depth of the object is \(8.20 \mathrm{~cm}\) from the flat end. What is its apparent depth when the rod is viewed from its curved end?

The radii of curvature of the surfaces of a thin converging meniscus lens are \(R_{1}=+12.0 \mathrm{~cm}\) and \(R_{2}=+28.0 \mathrm{~cm} .\) The index of refraction is \(1.60 .\) (a) Compute the position and size of the image of an object in the form of an arrow \(5.00 \mathrm{~mm}\) tall, perpendicular to the lens axis, \(45.0 \mathrm{~cm}\) to the left of the lens. (b) A second converging lens with the same focal length is placed \(3.15 \mathrm{~m}\) to the right of the first. Find the position and size of the final image. Is the final image erect or inverted with respect to the original object? (c) Repeat part (b) except with the second lens \(45.0 \mathrm{~cm}\) to the right of the first.

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