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Chapter 2: Problem 62

An object of height \(3.0 \mathrm{cm}\) is placed at \(5.0 \mathrm{cm}\) in front of a diverging lens of focal length \(20 \mathrm{cm}\) and observed from the other side. Where and how large is the image?

Short Answer

Expert verified
The image is located at a distance of \(4 \mathrm{cm}\) on the same side as the object, and its size is \(2.4 \mathrm{cm}\).

Step by step solution

01

Identify the lens type and values given

Since we are dealing with a diverging lens, its focal length will be negative. Hence, the focal length \(f = -20 cm\). The object's distance from the lens \(u = 5.0 cm\) and the object's height \(h_o = 3.0 cm\).
02

Use the lens formula to find the image distance

The lens formula is given by \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\), where \(f\) is the focal length, \(u\) is the object's distance from the lens, and \(v\) is the image's distance from the lens. First, we will substitute the given values into the lens formula: \[ \frac{1}{-20} = \frac{1}{5} + \frac{1}{v} \] Now, we need to solve for \(v\).
03

Solve for the image distance \(v\)

Subtract \(\frac{1}{5}\) from both sides of the equation: \[ \frac{1}{v} = \frac{1}{-20} - \frac{1}{5} \] Now, find a common denominator and combine fractions: \[ \frac{1}{v} = \frac{-1-4}{20} \] Simplify the right side of the equation: \[ \frac{1}{v} = \frac{-5}{20} \] Now, to find \(v\), take the reciprocal of both sides: \[ v = -4 \mathrm{cm} \] The negative sign indicates that the image is on the same side as the object.
04

Use the magnification formula to find the image size

The magnification formula is given by \(M = \frac{h_i}{h_o} = \frac{-v}{u}\), where \(M\) is the magnification, \(h_i\) is the image height, and \(h_o\) is the object height. First, substitute the known values into the magnification formula: \[ M = \frac{-(-4)}{5} = \frac{4}{5} \] Now, we need to solve for the image height, \(h_i\).
05

Solve for the image height \(h_i\)

Use the magnification formula to find the image height: \[ h_i = M\cdot h_o = \frac{4}{5} \cdot 3.0 \mathrm{cm} \] Multiply the fractions: \[ h_i = 2.4 \mathrm{cm} \]
06

Write the final answer

The image is located at a distance of 4 cm on the same side as the object, and its size is 2.4 cm.

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Most popular questions from this chapter

For normal distant vision, the eye has a power of 50.0 D. What was the previous far point of a patient who had laser vision correction that reduced the power of her eye by \(7.00 \mathrm{D},\) producing normal distant vision?

What is the far point of a person whose eyes have a relaxed power of \(50.5 \mathrm{D} ?\)

A large reflecting telescope has an objective mirror with a \(10.0-\mathrm{m}\) radius of curvature. What angular magnification does it produce when a \(3.00 \mathrm{m}\) -focal length eyepiece is used?

Two convex lenses of focal lengths \(20 \mathrm{cm}\) and 10 \(\mathrm{cm}\) are placed \(30 \mathrm{cm}\) apart, with the lens with the longer focal length on the right. An object of height \(2.0 \mathrm{cm}\) is placed midway between them and observed through each lens from the left and from the right. Describe what you will see, such as where the image(s) will appear, whether they will be upright or inverted and their magnifications.

An object is located in air \(30 \mathrm{cm}\) from the vertex of a convex surface made of glass with a radius of curvature 80 cm. Where does the image by refraction form and what is its magnification?
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