A dentist's mirror has a radius of curvature of 3 cm. How far must it be placed from a small dental cavity to give a virtual image of the cavity that is magnified five times?

The dentist's mirror is a concave mirror.

Radius of curvature, $R$ = $-$3 cm

Magnification, $m$ = 5

To find: Distance of the object $(u)$.

Solution:

We know that, 

$f=\frac {R}{2}$, where, $f$ = focal length, and $R$ = radius of curvature.

Putting the vlaue of $R$, we get-

$f=\frac {-3}{2}$

$f=-1.5cm$

So, the focal length is 1.5cm.

From the magnification formula, we know that-

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

Substituting the given values in the magnification formula we get-

$5=-\frac{v}{u}$

$v=-5u$

Now, from the mirror formula, we know that-

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

Substituting the given values in the mirror formula we get-

$\frac{1}{(-1.5)}=\frac{1}{(-5u)}+\frac{1}{u}$

$\frac{1}{-1.5}=-\frac{1}{5u}+\frac{1}{u}$

$\frac{1}{-1.5}=\frac{-1+5}{5u}$

$\frac{1}{-1.5}=\frac{4}{5u}$

$5u=4\times {(-1.5)}$

$u=\frac{-6}{5}$

$u=-1.2cm$

Thus, the distance of the object, $u$ is 1.2 cm from the mirror, and the negative sign implies that the object is placed in front of the mirror (on the left).

Hence, the dentist should place the object at a distance of 1.2 cm from the mirror.

Tutorialspoint

Updated on: 10-Oct-2022

1K+ Views

Kickstart Your Career

Get certified by completing the course

Get Started