Question

An object of 6 cm height is 16 cm in front of a converging lens with focal length of magnitude 4 cm. What is the dimension of the image(include the unit and if it is the case the sign)?

Answer 1

Answer:

$- 2$ cm

Explanation:

$h_{o}$ = height of the object = 6 cm

$d_{o}$ = distance of the object from the lens = 16 cm

$f$ = focal length of the lens = 4 cm

$d_{i}$ = image distance

using the lens equation

$\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}$

$\frac{1}{16} + \frac{1}{d_{i}} = \frac{1}{4}$

$d_{i} = \frac{16}{3}$ cm

$h_{i}$ = height of the image

Using the equation

$\frac{h_{i}}{h_{o}} = \frac{- d_{i}}{d_{o}}$

$\frac{h_{i}}{6} = \frac{- \frac{16}{3}}{16}$

$h_{i} =-2$ cm