Question

A lens of focal length 0.1 m is placed 0.08 m away from a bug. Where does the image of the bug appear to be (i.e., what is the image distance)? How much is the bug magnified by the lens? Is the image of the bug upright or inverted? Is the image real or virtual?

Answer 1

Answer:

Image distance of the bug is - 0.4 m

Magnification is 5 times

Image is real and inverted

Solution:

As per the question:

The focal length of the lens, f = 0.1 m

Distance of the bud from the lens, v = 0.08 m

Now,

Using the lens maker formula:

$\frac{1}{f} = \frac{1}{v} + \frac{1}{v'}$

where

v' = distance of the bug's image from the lens

Now,

$\frac{1}{v'} = \frac{1}{f} - \frac{1}{v}$

$\frac{1}{v'} = \frac{1}{0.1} - \frac{1}{0.08}$

$\frac{1}{v'} = - \frac{1}{2.5}$

v' = - 0.4 m

Now,

Magnification of the bug's image is given by:

m = $\frac{-v'}{v}$

m = $\frac{-(- 0.4)}{0.08} = 5$

Since, negative sign shows that inverted image is formed and positive value of magnification shows that real image is formed.