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Chapter 2: Problem 87

How far should you hold a 2.1 cm-focal length magnifying glass from an object to obtain a magnification of \(10 \times\) ? Assume you place your eye \(5.0 \mathrm{cm}\) from the magnifying glass.

Short Answer

Expert verified
The object should be placed \(5.25\ cm\) from the magnifying glass to obtain a magnification of \(10 \times\).

Step by step solution

01

List the provided information

Here is the information given in the problem: - Focal length (f) of the magnifying glass = 2.1 cm - Distance between magnifying glass and eye (v) = 5.0 cm - Magnification (M) = 10 times
02

Use the lens formula to find the image distance (v)

The lens formula is given by: \[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\] Where: - f is the focal length of the lens - u is the distance between the object and the lens (what we want to find) - v is the distance between the image and the lens We are given the focal length and distance from the eye, so we can substitute and rewrite the equation to get: \[\frac{1}{u} = \frac{1}{f} - \frac{1}{v}\]
03

Use the magnification formula to find the required image distance (v)

The magnification formula is given by: \[ M = \frac{v}{u}\] We know both the magnification (M) and the distance between the magnifying glass and eye (v), so we can rewrite the equation to get: \[u = \frac{v}{M}\]
04

Combine and solve the equations

Now we have the equations: \[u = \frac{v}{M}\] \[\frac{1}{u} = \frac{1}{f} - \frac{1}{v}\] Let's substitute the first equation into the second one: \[\frac{1}{\frac{v}{M}} = \frac{1}{f} - \frac{1}{v}\] Now, we can input the given values (M = 10, f = 2.1 cm, v = 5.0 cm) and solve for u: \[\frac{1}{\frac{5.0}{10}} = \frac{1}{2.1} - \frac{1}{5.0}\] Solving the equation: \[2 = \frac{1}{2.1} - \frac{1}{5.0}\] \[2 = \frac{5 - 2.1}{10.5}\] \[10.5 = 2(5 - 2.1)\] \[u = \frac{10.5}{2}\]
05

Calculate the required distance (u)

Finally, we can calculate the distance u: \[u = \frac{10.5}{2}\] \[u = 5.25\ cm\] The object should be placed 5.25 cm from the magnifying glass to obtain a magnification of 10 times.

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Most popular questions from this chapter

Derive the formula for the apparent depth of a fish in a fish tank using Snell's law.

A very myopic man has a far point of \(20.0 \mathrm{cm}\). What power contact lens (when on the eye) will correct his distant vision?

Unless otherwise stated, the lens-to-retina distance is 2.00 Calculate the power of the eye when viewing an object \(3.00 \mathrm{m}\) away.

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