# A mirror produces an image that is located 32.0 cm behind the mirror when the object is located...

## Question:

A mirror produces an image that is located 32.0 cm behind the mirror when the object is located 8.00 cm in front of the mirror. What is the focal length of the mirror?

## Spherical Mirror Formula:

First we have to understand the spherical mirror formula to solve this problem:

The spherical mirror formula is given by: {eq}\displaystyle \frac{1}{f} = \frac{1}{v} + \frac{1}{u} {/eq} where:

- {eq}u {/eq} is the object distance from the mirror.

- {eq}v {/eq} is the image distance from the mirror.

- {eq}f {/eq} is the focal length of the mirror. It is also defined as: {eq}\displaystyle f = \frac{R}{2} {/eq} where {eq}R {/eq} is radius of sphere.

## Answer and Explanation:

Given:

- The image distance from the spherical mirror is: {eq}v = 32 \ \rm cm {/eq}.

- The object distance from the spherical mirror is: {eq}u = - 8 \ \rm cm {/eq}. Here the object distance is negative because the object is placed left of side of the image.

We will compute the object distance from the mirror.

Apply the spherical mirror formula:

$$\begin{align*} \displaystyle \frac{1}{f} &= \frac{1}{v} + \frac{1}{u} &\text{(Where } f = 34 \ \rm cm \text{ and } v = 13 \text{)}\\ \frac{1}{f} &= \frac{1}{32} + \frac{1}{- 8} &\text{(Plugging in all given values)}\\ \frac{1}{f} &= \frac{1}{32} - \frac{1}{8} \\ \frac{1}{f} &= \frac{1 - 4}{32} \\ \frac{1}{f} &= \frac{- 3}{32} \\ - 3 \ f &= 32 &\text{(Doing cross multiplication)}\\ f &= - \frac{32}{3} &\text{(Dividing both sides by 3)}\\ f \ &\boxed{= - 10.667 \ \rm cm } &\text{(Negative sign is showing, the mirror is concave mirror)}\\ \end{align*} $$

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from Physics 101: Help and Review

Chapter 14 / Lesson 17