Question

An object of length 10cm is placed in front of a convex

mirror of focal length 15cm. At what distance the object

should be placed so that the image is formed at a distance

of 20cm? Also find the nature and size of the image.​

Answer 1

Answer:

Hey Mate!!

Given,

height of object , h = + 10 cm

distance of image, v = + 20 cm

focal length, f = + 15 cm

distance of object, u = ?

height of image, h' = ?

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\\\=> \frac{1}{20}-\frac{1}{u}=\frac{1}{15} \\\\=> \frac{1}{20}-\frac{1}{15}=\frac{1}{u} \\\\=> \frac{3 - 4}{60}=\frac{1}{u}\\\\=> \frac{-1}{60}=\frac{1}{u}\\\\=> u = -60 \: cm$

Hence, the object should be placed at a distance of 60 cm in front of the mirror.

$\frac{h'}{h}=-\frac{v}{u} \\\\=> \frac{h'}{10}=-(\frac{20}{-60}) \\\\=> h' = \frac{20}{60} \times 10\\\\=> h' = \frac{20}{6}\\\\=> h' = +3.34 \: cm$

Therefore, height of image is 3.34 cm. It is a diminished and virtual image, as indicated by the plus sign.