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At what distance a dentist’s mirror, having a radius of curvature of $3 cm,$ should be placed from a small dental cavity to give a five times enlarged virtual image of the cavity?

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2.5 cm

3 cm

1.2 cm

1.9 cm

answer is C.

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Detailed Solution

The dentist’s mirror should be placed at

1.2 cm

, from a small dental cavity to give a five times enlarged virtual image of the cavity.
Given,

f = - 1.5 cm

m = + 5 (as the image is virtual and enlarged, sign will be positive)

Substituting the given values in magnification formula we get,

\Rightarrow

m = - \frac{v}{u}

\Rightarrow

5 = - \frac{v}{u}

\Rightarrow

v = - 5 u

Now, using the mirror formula we get

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

\Rightarrow

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

\Rightarrow

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

\Rightarrow

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

\Rightarrow

u = - 1.2 cm

So, the mirror should be placed at

1.2 cm

from the dental cavity to get the magnification of

5 .

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