Answer to Question #191826 in Physics for Dominic

Question

A 4.8 cm tall image is located 3 cm in front of a lens whose focal length is - 6 cm

a) Solve for the distance in which the object is placed in front of the lens

b) Solve for the magnification

c) Solve for the height of the object

d) Is the image upright or inverted?

Answer 1

(a) We can find the object distance from the lens equationb ($d_i=-3\ cm$ because the image is in front of the lens and, therefore, is virtual):

$$\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f},$$ $$d_o=\dfrac{1}{\dfrac{1}{f}-\dfrac{1}{d_i}},$$ $$d_o=\dfrac{1}{\dfrac{1}{-6\ cm}-\dfrac{1}{-3\ cm}}=6\ cm.$$

(b) We can find the magnification of the lens from the magnification equation:

$$M=-\dfrac{d_i}{d_o},$$ $$M=-\dfrac{-3\ cm}{6\ cm}=0.5.$$

The positive magnification means that the image is upright.

(c) We can find the height of the object from the magnification equation:

$$M=\dfrac{h_i}{h_o},$$ $$h_o=\dfrac{h_i}{M}=\dfrac{4.8\ cm}{0.5}=9.6\ cm.$$

(d) The image is upright.