Question

A lens with f= 20.0 cm creates a

virtual image at -37.5 cm (in front of
the lens). The object is 4.44 cm
tall. How tall is the image?
(Mind your minus signs.)
(Unit = cm)

Answer 1

12.8

Step-by-step explanation:

got the answer from chegg. Correct for acellus.

Answer 2

h ’= 12,768 cm

Step-by-step explanation:

For this exercise let's use the constructor equation

1 / f = 1 / p + 1 / q

where f is the focal length, p the distance to the object and q the distance to the image

the magnification equation is

m = h '/ h = -q / p

let's find the distance to the object

1 / p = 1 / f- 1 / q

1 / p = 1/20 - 1 / (- 37.5)

1 / p = 0.076666

p = 13.04 cm

now let's use the magnification equation

h ’= - q / p h

let's calculate

h ’= - (-37.5) / 13.04 4.44

h ’= 12,768 cm