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Chapter 33: Problem 77

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.

Short Answer

Expert verified
Answer: The image distance is -18.0 cm.

Step by step solution

01

Write down the known values

We are given the object distance \(d_o = 6.0 ~cm\) and the focal length \(f = 9.0 ~cm\). We want to find the image distance \(d_i\).
02

Apply the lens formula

Using the lens formula, we will plug in the known values and solve for \(d_i\): $$ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} $$ $$ \frac{1}{9.0} = \frac{1}{6.0} + \frac{1}{d_i} $$
03

Solve for the image distance

We need to find the value of \(\frac{1}{d_i}\) by isolating it: $$ \frac{1}{d_i} = \frac{1}{9.0} - \frac{1}{6.0} $$ Now, find the common denominator for the two fractions on the right side of the equation, which is 18: $$ \frac{1}{d_i} = \frac{2}{18} - \frac{3}{18} $$ Combine the two fractions: $$ \frac{1}{d_i} = -\frac{1}{18} $$ Finally, take the reciprocal of both sides to get the image distance: $$ d_i = -18 \mathrm{~cm} $$ Therefore, the image distance is \(-18.0 \mathrm{~cm}\). The negative sign indicates that the image is on the same side of the lens as the object, meaning it is a virtual image.

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Most popular questions from this chapter

A plastic cylinder of length \(3.0 \cdot 10^{1} \mathrm{~cm}\) has its ends ground to convex (from the rod outward) spherical surfaces, each having radius of curvature \(1.0 \cdot 10^{1} \mathrm{~cm}\). A small object is placed \(1.0 \cdot 10^{1} \mathrm{~cm}\) from the left end. How far will the image of the object lie from the right end, if the index of refraction of the plastic is \(1.5 ?\)

Two identical thin convex lenses, each of focal length \(f\), are separated by a distance \(d=2.5 f\). An object is placed in front of the first lens at a distance \(d_{\mathrm{a}, 1}=2 f .\) a) Calculate the position of the final image of an object through the system of lenses. b) Calculate the total transverse magnification of the system. c) Draw the ray diagram for this system and show the final image. d) Describe the final image (real or virtual, erect or inverted, larger or smaller) in relation to the initial object.

A classmate claims that by using a \(40.0-\mathrm{cm}\) focal length mirror, he can project onto a screen a \(10.0-\mathrm{cm}\) tall bird locat ed 100 . \(\mathrm{m}\) away. He claims that the image will be no less than \(1.00 \mathrm{~cm}\) tall and inverted. Will he make good on his claim?

A converging lens will be used as a magnifying glass. In order for this to work, the object must be placed at a distance a) \(d_{\mathrm{o}}>f\). c) \(d_{\mathrm{o}}

Suppose the near point of your eye is \(2.0 \cdot 10^{1} \mathrm{~cm}\) and the far point is infinity. If you put on -0.20 diopter spec tacles, what will be the range over which you will be able to see objects distinctly?
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