Question

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, the object distance will be positive (in multiple optics setups, you will encounter "objects" that are actually images, but that is not a possibility in this problem). A positive image distance means that the image is formed on the side of the lens from which the light emerges.

Part A

Find the focal length of the lens that produces the image described in the problem introduction using the thin lens equation.

Express your answer in centimeters, as a fraction or to three significant figures.

f =

6.67

cm

Part B

Considering the sign of f, is the lens converging or diverging?

	converging
	diverging

Part C

What is the magnification m of the lens?

Express your answer as a fraction or to three significant figures.

m =

-1.25

Part D

Think about the sign of s′ and the sign of y′, which you can find from the magnification equation, knowing that a physical object is always considered upright. Which of the following describes the nature and orientation of the image?

	real and upright
	real and inverted
	virtual and upright
	virtual and inverted

Now consider a diverging lens with focal lengthf=−15cm, producing an upright image that is 5/9 as tall as the object.

Part E

Is the image real or virtual? Think about the magnification and how it relates to the sign of s′.

	real
	virtual

Part F

What is the object distance? You will need to use the magnification equation to find a relationship between sands′. Then substitute into the thin lens equation to solve for s.

Express your answer in centimeters, as a fraction or to three significant figures.

s =

12.0

cm

Part G

What is the image distance?

Express your answer in centimeters, as a fraction or to three significant figures.

s′ =	24	cm

A lens placed at the origin with its axis pointing along the x axis produces a real inverted image atx=−24cmthat is twice as tall as the object.

Part H

What is the image distance?

Express your answer in centimeters, as a fraction or to three significant figures.

s′ =

24.0

cm

Part I

What is the x coordinate of the object? Keep in mind that a real image and a real object should be on opposite sides of the lens.

Express your answer in centimeters, as a fraction or to three significant figures.

Answer 1

(A) The lens formula is. \(\frac{1}{f}=\frac{1}{s}+\frac{1}{s^{\prime}}\)

\(=\frac{1}{12}+\frac{1}{15}\)

\(f=\frac{(12)(15)}{12+15}\)

\(=6.67 \mathrm{~cm}\)

(B) The positive value of focal length implies that lens is converging.

(C) Magnification, \(m=-\frac{s^{\prime}}{s}\)

\(=-\frac{(15)}{(12)}\)

\(=-1.25\)

(D) The image is real and inverted.

(E)

\(\frac{5}{9}=-\frac{s^{\prime}}{s}\)

The image is virtual.

\((\mathrm{F})\)

\(\frac{1}{-15}=\frac{1}{s}+\frac{1}{\left(-\frac{5 s}{9}\right)}\)

\(s=12.0 \mathrm{~cm}\)

(G) The image distance is, \(s^{\prime}=-\frac{5(12)}{9}\)

\(=-6.67 \mathrm{~cm}\)

(H)

\(s^{\prime}=24.0 \mathrm{~cm}\)

(I) \(2=-\frac{s^{\prime}}{s}\)

\(s=\frac{-(-24)}{2}\)

\(=12.0 \mathrm{~cm}\)