# A 2.00 \ cm high object is placed 3.00 \ cm in front of a concave mirror. If the image is 5.00...

## Question:

A {eq}2.00 \ cm {/eq} high object is placed {eq}3.00 \ cm {/eq} in front of a concave mirror. If the image is {eq}5.00 \ cm {/eq} high and virtual, what is the focal length of the mirror?

## Concave Mirror:

Focal length in concave mirror is positive since the center of curvature is on the same side of outgoing light. It is given by the formula:

{eq}\frac{1}{f} = \frac{1}{s} + \frac{1}{s'} {/eq}

Given:

y = 2.00 cm

y' = 5.00 cm

s = 3.00 cm

Solution:

The image is virtual therefore it is upright and positive. Using magnification formula, we can solve for the image distance.

Magnification is given by the formula:

{eq}m = \frac{y'}{y} = \frac{-s'}{s} \\ m = \frac{5.00 \ cm}{2.00 \ cm} \\ m = 2.5 {/eq}

Therefore, the image distance is:

{eq}m = \frac{-s'}{s} \\ 2.5 = \frac{-s'}{3.00 \ cm} \\ s' = -7.5 \ cm {/eq}

Since the image is said to be virtual and upright, it is formed behind the mirror with a negative magnitude of distance.

Now, for the focal length, we can use the formula {eq}\frac{1}{f} = \frac{1}{s} + \frac{1}{s'} {/eq}

{eq}\frac{1}{f} = \frac{1}{3.00 \ cm} + \frac{1}{-7.5 \ cm'} \\ f = 5 \ cm {/eq}