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Chapter 33: Problem 4

An object is moved from a distance of \(30 \mathrm{~cm}\) to a distance of \(10 \mathrm{~cm}\) in front of a converging lens of focal length \(20 \mathrm{~cm}\). What happens to the image? a) Image goes from real and upright to real and inverted. b) Image goes from virtual and upright to real and inverted. c) Image goes from virtual and inverted to real and upright. d) Image goes from real and inverted to virtual and upright. e) None of the above.

Short Answer

Expert verified
Based on the given lens problem, the correct answer is: d) The image goes from real and inverted to virtual and upright.

Step by step solution

01

1. Use the Lens Formula

The lens formula relates the object distance (u), the image distance (v) and the focal length (f) as follows: $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$ Remember, for a converging lens, the focal length is positive.
02

2. Calculate Image Distance for the Initial Position (u = 30 cm)

For the initial position, the object is placed at a distance of 30 cm from the lens. Let's plug u = 30 cm and f = 20 cm into the lens formula and solve for v. $$\frac{1}{v} - \frac{1}{30} = \frac{1}{20}$$ Solving for v, we get v = 60 cm.
03

3. Calculate Image Distance for the Final Position (u = 10 cm)

Now, for the final position, the object is placed at a distance of 10 cm from the lens. Let's plug u = 10 cm and f = 20 cm into the lens formula and solve for v. $$\frac{1}{v} - \frac{1}{10} = \frac{1}{20}$$ Solving for v, we get v = -20 cm.
04

4. Determine the Nature of the Image for Both Positions

For the initial position (u=30 cm), the image distance (v) is positive (v=60 cm), meaning the image is real and on the opposite side of the object. Also, since the magnification (M) for a lens is given by: $$ M = -\frac{v}{u} $$ In this case, M = -2, which is negative. Thus, the initial image is real and inverted. For the final position (u=10 cm), the image distance (v) is negative (v=-20 cm), meaning the image is virtual and on the same side as the object. The magnification for this position is M = 2, which is positive. Thus, the final image is virtual and upright.
05

5. Choose the Correct Answer

From our analysis, we find the image goes from real and inverted to virtual and upright. Therefore, the correct answer is: d) Image goes from real and inverted to virtual and upright.

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Most popular questions from this chapter

Mirrors for astronomical instruments are invariably first-surface mirrors: The reflective coating is applied on the surface exposed to the incoming light. Household mirrors, on the other hand, are second-surface mirrors: The coating is applied to the back of the glass or plastic material of the mirror. (You can tell the difference by bringing the tip of an object close to the surface of the mirror. Object and image will nearly touch with a first-surface mirror; a gap will remain between them with a second-surface mirror.) Explain the reasons for these design differences.

How large does a \(5.0-\mathrm{mm}\) insect appear when viewed with a system of two identical lenses of focal length \(5.0 \mathrm{~cm}\) separated by a distance \(12 \mathrm{~cm}\) if the insect is \(10.0 \mathrm{~cm}\) from the first lens? Is the image real or virtual? Inverted or upright?

You have found in the lab an old microscope, which has lost its eyepiece. It still has its objective lens, and markings indicate that its focal length is \(7.00 \mathrm{~mm}\). You can put in a new eyepiece, which goes in \(20.0 \mathrm{~cm}\) from the objective. You need a magnification of about 200. Assume you want the comfortable viewing distance for the final image to be \(25.0 \mathrm{~cm}\). You find in a drawer eyepieces marked \(2.00-, 4.00-,\) and \(8.00-\mathrm{cm}\) focal length. Which is your best choice?

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 camera has a lens with a focal length of \(60 . \mathrm{mm} .\) Suppose you replace the normal lens with a zoom lens whose focal length can be varied from \(35 . \mathrm{mm}\) to \(250 . \mathrm{mm}\) and use the camera to photograph an object at infinity. Compared to a 60.-mm lens, what magnification of the image would be achieved using the \(240 .-\mathrm{mm}\) focal length?
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