Answer to Question #263665 in Physics for Rethabile

Question

Question 29 (4 points)

 

You place an object 22 cm in front of a concave lens with a focal length of -35 cm. Find the image distance and magnification that results from this arrangement.

Question 29 options:

A) 

-0.34

B) 

No option is correct

C) 

-0.64

D) 

-0.55

E) 

-0.45

Answer 1

(a) We can find the image distance from the thin lens equation:

$$\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f},$$ $$\dfrac{1}{d_i}=\dfrac{1}{f}-\dfrac{1}{d_o},$$ $$d_i=\dfrac{1}{\dfrac{1}{f}-\dfrac{1}{d_o}}=\dfrac{1}{\dfrac{1}{-35\ cm}-\dfrac{1}{22\ cm}}=-13.5\ cm.$$

The sign minus means that the image is virtual (it appears on the same side as the object).

(b) We can find the magnification of the lens as follows:

$$m=-\dfrac{d_i}{d_o}=-\dfrac{-13.5\ cm}{22\ cm}=0.61.$$

The sign plus means that the image is upright.

Therefore, no option is correct.