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Chapter 33: Problem 9

An object is placed on the left of a converging lens at a distance that is less than the focal length of the lens. The image produced will be a) real and inverted. c) virtual and inverted. b) virtual and erect. d) real and erect.

Short Answer

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Question: When an object is placed at a distance less than the focal length of a converging lens, the image formed will be: a) real and inverted b) virtual and erect c) virtual and inverted d) real and erect Answer: b) virtual and erect

Step by step solution

01

Review the lens formula and lensmaker's equation

The lens formula for a converging lens is given by (1/f) = (1/u) + (1/v), where f is the focal length, u is the object distance, and v is the image distance. The lensmaker's equation is given by 1/f = (n-1)((1/R1)-(1/R2)), where n is the refractive index, R1 and R2 are the radii of the lens surfaces.
02

Apply the lens formula with given information

Using the lens formula, we have to find v, the image distance, to determine the type of image formed. In this case, the object distance (u) is less than the focal length (f). Taking u < f, (1/f) = (1/u) + (1/v) Now, since u < f, the left side of the equation (1/f) will be smaller than the middle term (i.e., 1/u). Therefore, the term (1/v) must be negative to balance the equation. If 1/v is negative, this means that v must be negative. When the image distance (v) is negative, it indicates that the image is virtual.
03

Determine the orientation of the image

When a virtual image is formed using a converging lens, the image is always erect. This occurs because the light rays appear to diverge from the image point, but they never actually cross. The image is created by the extension of these diverging rays.
04

Choose the correct answer

Based on the analysis and calculations above, the image produced will be virtual and erect. The correct option is: b) virtual and erect.

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Most popular questions from this chapter

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.

Two refracting telescopes are used to look at craters on the Moon. The objective focal length of both telescopes is \(95.0 \mathrm{~cm}\) and the eyepiece focal length of both telescopes is \(3.80 \mathrm{~cm} .\) The telescopes are identical except for the diameter of the lenses. Telescope A has an objective diameter of \(10.0 \mathrm{~cm}\) while the lenses of telescope \(\mathrm{B}\) are scaled up by a factor of two, so that its objective diameter is \(20.0 \mathrm{~cm}\). a) What are the angular magnifications of telescopes \(A\) and \(B\) ? b) Do the images produced by the telescopes have the same brightness? If not, which is brighter and by how much?

What kind of lens is used in eyeglasses to correct the vision of someone who is a) nearsighted? b) farsighted?

LASIK surgery uses a laser to modify the a) curvature of the retina. b) index of refraction of the aqueous humor. c) curvature of the lens. d) curvature of the cornea.

What is the magnification of a telescope with \(f_{0}=1.00 \cdot 10^{2} \mathrm{~cm}\) and \(f_{e}=5.00 \mathrm{~cm} ?\)
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