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Chapter 33: Problem 27

A physics student epoxies two converging lenses to the opposite ends of a \(2.0 \cdot 10^{1}-\mathrm{cm}\) -long tube. One lens has a focal length of \(f_{1}=6.0 \mathrm{~cm}\) and the other has a focal length of \(f_{2}=3.0 \mathrm{~cm}\). She wants to use this device as a microscope. Which end should she look through to obtain the highest magnification of an object?

Short Answer

Expert verified
Answer: The student should look through the end with the lens of focal length 3.0 cm to obtain the highest magnification.

Step by step solution

01

Write down the magnification formula for converging lenses

The magnification formula for a converging lens is given by: $$M = 1 + \frac{D}{f}$$
02

Calculate the magnification power for the first lens

Given the first lens has a focal length of \(f_1=6.0\,\mathrm{cm}\), we can now calculate the magnification power of the first lens: $$M_1 = 1 + \frac{D}{f_1}$$
03

Calculate the magnification power for the second lens

Given the second lens has a focal length of \(f_2=3.0\,\mathrm{cm}\), we can now calculate the magnification power of the second lens: $$M_2 = 1 + \frac{D}{f_2}$$
04

Compare the magnification powers of both lenses

Let's compare the magnification powers of both lenses as a ratio of \(M_1\) to \(M_2\). This will help us determine which lens provides the highest magnification. $$\frac{M_1}{M_2} = \frac{1 + \frac{D}{f_1}}{1 + \frac{D}{f_2}}$$ Substitute the values of \(f_1\) and \(f_2\): $$\frac{M_1}{M_2} = \frac{1 + \frac{D}{6.0}}{1 + \frac{D}{3.0}}$$ Since \(D>0\) and \(f_2<f_1\), this ratio will be greater than 1, which means \(M_1<M_2\). Therefore, the student should look through the end with the lens of focal length \(f_2=3.0\,\mathrm{cm}\) to obtain the highest magnification of an object.

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Most popular questions from this chapter

When a diver with \(20 / 20\) vision removes her mask underwater, her vision becomes blurry. Why is this the case? Does the diver become nearsighted (eye lens focuses in front of retina) or farsighted (eye lens focuses behind retina)? As the index of refraction of the medium approaches that of the lens, where does the object get imaged? Typically, the index of refraction for water is 1.33 , while the index of refraction for the lens in a human eye is \(1.40 .\)

A microscope has a \(2.0-\mathrm{cm}\) focal length eyepiece and a \(0.80-\mathrm{cm}\) objective lens. For a relaxed normal eye, calculate the position of the object if the distance between the lenses is \(16.2 \mathrm{~cm}\)

An object is \(6.0 \mathrm{~cm}\) from a thin lens along the axis of the lens. If the lens has a focal length of \(9.0 \mathrm{~cm},\) determine the image distance.

Jack has a near point of \(32 \mathrm{~cm}\) and uses a magnifier of 25 diopter. a) What is the magnification if the final image is at infinity? b) What is the magnification if the final image is at the near point?

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: \(f=g / 2 \omega^{2}\).
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