# An object is 6 cm in front of a concave mirror with a focal length of 4 cm. The magnification of...

## Question:

An object is 6 cm in front of a concave mirror with a focal length of 4 cm. The magnification of the object is _____.

## Mirror Equation:

Fundamentally, the mirror equation looks the same as the thin lens equation. The sign convention for images is still followed for mirrors, where positive values mean real images and negative values mean virtual images. As for the focal length, concave mirrors have positive focal lengths while convex mirrors have negative focal lengths.

## Answer and Explanation:

Given:

- {eq}\displaystyle \rm o = 6\ cm {/eq} is the object distance from the mirror

- {eq}\displaystyle \rm f = 4\ cm {/eq} is the focal length of the concave mirror

Here, we can use the mirror equation to first calculate the image distance:

{eq}\displaystyle \rm \frac{1}{f} = \frac{1}{o} + \frac{1}{i} {/eq}

We isolate the image distance:

{eq}\displaystyle \rm \frac{1}{i} = \frac{1}{f} - \frac{1}{o} {/eq}

We substitute:

{eq}\displaystyle \rm \frac{1}{i} = \frac{1}{4\ cm} - \frac{1}{6\ cm} {/eq}

We simplify:

{eq}\displaystyle \rm \frac{1}{i} = \frac{0.25}{cm} - \frac{0.16667}{cm} {/eq}

{eq}\displaystyle \rm \frac{1}{i} = \frac{0.08333}{cm} {/eq}

We take the reciprocal to get:

{eq}\displaystyle \rm i = 12\ cm {/eq}

The magnification of an image can be expressed in terms of its image and object distances as:

{eq}\displaystyle \rm m = -\frac{i}{o} {/eq}

We substitute:

{eq}\displaystyle \rm m = -\frac{12\ cm}{6\ cm} {/eq}

We will get:

{eq}\displaystyle \rm \boxed{\rm m = -2} {/eq}

Negative magnification values imply not only that the object is real, but that the image is also **inverted**.

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from Physics 101: Help and Review

Chapter 14 / Lesson 17