Question

A dentist uses a mirror in front of a decayed tooth at a distance 4cm from the tooth to get a 4 times magnified image in the mirror.Use mirror formula to find focal length and nature of the mirror used.

Answer 1

nature of mirror = Virtual and erect.

Answer 2

» Given

A dentist uses a mirror in front of a decayed tooth at a distance of 4 cm [means u = 4cm]

from tooth to get a four times magnified image [means h' = 4h]

• Now

u = distance between object and mirror.

v = distance between image and mirror.

f = focal length

So...

$\dfrac{1}{f}$ = $\dfrac{1}{v}$ + $\dfrac{1}{u}$ ....(1)

We also given that

u = - 4 cm [as object is always puts in left side so it's sign is -ve]

h' = 4h

$\dfrac{h'}{h}$ = 4 cm

Now

m = $\dfrac{-v}{u}$

4 = $\dfrac{-v}{-4}$

[As m = $\dfrac{h'}{h}$ = $\dfrac{-v}{u}$

v = + 16 cm

Put value of v and u in eq. (1)

$\dfrac{1}{f}$ = $\dfrac{1}{16}$ + $\dfrac{1}{-4}$

$\dfrac{1}{f}$ = $\dfrac{1}{16}$ - $\dfrac{1}{4}$

$\dfrac{1}{f}$ = $\dfrac{1\:-\:4}{16}$

$\dfrac{1}{f}$ = $\dfrac{-3}{16}$

f = $\dfrac{-16}{3}$

f = - 5.33 cm

As f = - 5.33 cm

v = + 16 cm

u = - 4 cm

So, it is clear from above that it is a Concave mirror . And also known as Converging mirror.

As object is placed between f and P. So, image is formed behind the mirror.

So, it's clear that nature of image is erect and virtual . Image formed is enlarged .