Question

A makeup/shaving mirror provides 5times magnified virtual image. A person positions their face 15 cm away from the mirror. Determine a. The type of image formed. b. The location of the person's face in the mirror. c. the focal length of the mirror (explicitly state the algebraic sign in your answer) d. the radius of curvature of the mirror e. The type of mirror.

Answer 1

Given:

A makeup/shaving mirror provides 5times magnified virtual image. The face is 15 cm away from mirror.

To find:

• Type of image
• Location of image
• Focal length of mirror
• Radius of curvature of mirror
• Type of mirror

Calculation:

• First of all, shaving mirrors are always CONCAVE MIRRORS.

• Since , the face is kept quite close to mirror (within focus), image is virtual, erect and magnified (i.e. type of image).

$\rm \: \dfrac{ h_{i}}{ h_{o} } = - \dfrac{v}{u}$

$\rm \implies\: \dfrac{ 5 \times h_{o}}{ h_{o} } = - \dfrac{v}{u}$

$\rm \implies\: 5 = - \dfrac{v}{( - 15)}$

$\rm \implies\: v = 75 \: cm$

So, image is formed 75 cm behind mirror.

• Let focal length be f :

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

$\rm \implies\: \dfrac{1}{f} = \dfrac{1}{75} + \dfrac{1}{( - 15)}$

$\rm \implies\: \dfrac{1}{f} = \dfrac{ - 4}{75}$

$\rm \implies\: f = - 18.75 \: cm$

So, focal length is -18.75 cm.

$\rm \implies\: R = 2f = - 37.5 \: cm$

So, radius of curvature is -37.5 cm.