Question

A diverging lens has a focal length of magnitude 22.8 cm. (a) Locate the images for each of the following object distances. 45.6 cm distance 22.8 cm distance 11.4 cm distance

Answer 1

(a) -15.2 cm

We can solve the problem by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f = -22.8 cm is the focal length of the lens (negative because it is a diverging lens)

p = 45.6 cm is the distance of the object from the lens

q is the distance of the image from the lens

Solving the equation for q, we find the position of the image:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-22.8 cm}-\frac{1}{45.6 cm}=-0.066 cm^{-1}$

$q=\frac{1}{-0.066 cm^{-1}}=-15.2 cm$

and the negative sign means that the image is virtual.

(b) -11.4 cm

In this case, the distance of the object from the lens is

p = 22.8 cm

Substituting into the lens equation, we can find the new image distance, q:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-22.8 cm}-\frac{1}{22.8 cm}=-\frac{2}{22.8 cm}$

$q=\frac{-22.8 cm}{2}=-11.4 cm$

and the negative sign means that the image is virtual.

(c) -7.6 cm

In this case, the distance of the object from the lens is

p = 11.4 cm

Substituting into the lens equation, we can find the new image distance, q:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-22.8 cm}-\frac{1}{11.4 cm}=-0.132 cm^{-1}$

$q=\frac{1}{0.132 cm^{-1}}=-7.6 cm$

and again, the negative sign means that the image is virtual.