Consider an object with s =12cm that produces an image with s ?=15cm. Note that

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.

SubmitMy AnswersGive Up

Part B

This question will be shown after you complete previous question(s).

Part C

This question will be shown after you complete previous question(s).

Part D

This question will be shown after you complete previous question(s).

Now consider a diverging lens with focal length f=?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?.

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

SubmitMy AnswersGive Up

Part F

This question will be shown after you complete previous question(s).

Part G

This question will be shown after you complete previous question(s).

A lens placed at the origin with its axis pointing along the x axis produces a real inverted image at x=?24cm that 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.

Submit

Explanation / Answer

A) Givn that S = 12 cm
and S' = 15 cm

(1/f) = (1/S)+(1/S')

f = (S*S')/(S+S') = (12*15)/(12+15) = 6.67 cm

B) no question
C) no question

D) no question
===========================

For diverging lens

F = -15 cm

image height hi = (5/9)*object hight ho

hi = (5/9)*ho

hi/ho = (5/9)

maginification m = -S'/S = hi/ho = 5/9

S' = - (5/9)*S

since S' is negative the formed image is virtual

F) no question

G) no question

H) image distance S' = 24 cm

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects