Question

The radius of curvature of a concave mirror used by a dentist is 30 cm. How far from

the teeth of a patient must the mirror be placed to give a virtual image which is

magnified five times?​

Answer 1

Given :

▪ Radius of curvature = 30cm

▪ Magnification = 5

▪ Type of mirror : concave

▪ Nature of image : virtual

To Find :

▪ Distance of object (u).

Concept :

→ X-coordinate of centre of curvature and focus of concave mirror are negative and those for convex mirror are positive. In case of mirrors sice light rays reflects back in X-direction, therefore -ve sign of v indicates real image and +ve sign of v indicates virtual image.

Calculation :

❇ Lateral Magnification :

$\implies\sf\:m=-\dfrac{v}{u}\\ \\ \implies\sf\:5=-\dfrac{(+v)}{(-u)}\\ \\ \implies\sf\:5=\dfrac{v}{u}\\ \\ \implies\underline{\underline{\bf{v=5u}}}$

❇ Mirror Formula :

$\implies\sf\:\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{2}{R}\\ \\ \implies\sf\:\dfrac{1}{5u}+\dfrac{1}{(-u)}=-\dfrac{2}{30}\\ \\ \implies\sf\:\dfrac{4}{5u}=-\dfrac{1}{15}\\ \\ \implies\sf\:\dfrac{4}{u}=-\dfrac{1}{3}\\ \\ \implies\underline{\boxed{\bf{\purple{u=-12cm}}}}\:\orange{\bigstar}$

Answer 2

Given :

• Radius of curvature (C) = 30 cm
• Magnification = 5

To Find :

• How far from the teeth of a patient must the mirror be placed to give a virtual image which is 5 times magnified?

Formula used :

• Mirror formula

$\</u></strong><strong><u>d</u></strong><strong><u>frac{1}{v} + \</u></strong><strong><u>d</u></strong><strong><u>frac{1}{u} = \</u></strong><strong><u>d</u></strong><strong><u>frac{1}{f}$

Here, v is image distance, u is object distance and f is the focal length.

Solution :

As, centre of curvature (C) = 30 cm

Then focal length (f) = C/2 = 15 cm.

In case of concave mirror, object distance & focal length is negative.

According to the question,

$Magnification (m) = \: - \dfrac{ v}{ \: u}$

Then,

$- (\dfrac{ v}{ - u}) = 5$

$\dfrac{v}{u} = 5$

$v = 5u$

Now upon substituting the values in the formula we get,

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

$\dfrac{1}{ 5u} + \dfrac{1}{ - u} = - \dfrac{1}{15}$

$\dfrac{4}{5u} = - \dfrac{1}{15}$

$5u = - 15 \times 4$

$u = \dfrac{ - 15 \times 4}{5} = - 12 \: cm$