# A concave spherical mirror has a radius of curvature of 18 cm. Locate the images for object...

## Question:

A concave spherical mirror has a radius of curvature of 18 cm. Locate the images for object distances as given below. In each case, state whether the image is real or virtual and upright or inverted, and find the magnification. (If an answer does not exist, enter DNE. If an answer is infinity, enter INFINITY.)

(a) p = 9 cm image distance cm image orientation magnification

(b) p = 18 cm image distance cm image orientation magnification (c) p = 31 cm image distance cm image orientation magnification

## Concave Mirror

Concave mirror is a spherical mirror that is bulging inward.The focal length of a concave mirror is positive. The focal length is related to the radius of curvature by {eq}\displaystyle f = \frac{R}{2} {/eq}.

## Answer and Explanation:

Let's begin by getting the focal length of the mirror.

{eq}\displaystyle f = \frac{R}{2} \\ \displaystyle f = \frac{18\,cm}{2} \\ \displaystyle f = 9\,cm {/eq}

**(a)**

We can get the image distance using the mirror equation.

{eq}\displaystyle \frac{1}{f} = \frac{1}{p} + \frac{1}{d_i} \\ \displaystyle \frac{1}{(9\,cm)} = \frac{1}{(9\,cm)} + \frac{1}{d_i} \\ \displaystyle \frac{1}{9\,cm} = \frac{1}{9\,cm} + \frac{1}{d_i} \\ \displaystyle \frac{1}{9\,cm} - \frac{1}{9\,cm} = \frac{1}{d_i} \\ \displaystyle \frac{1-1}{9\,cm} = \frac{1}{d_i} \\ \displaystyle \frac{0}{9\,cm} = \frac{1}{d_i} \\ \displaystyle d_i = \frac{9\,cm}{0} \\ \displaystyle d_i = 9\,cm \cdot \frac{1\,cm}{0} \\ \displaystyle d_i = 9\,cm\cdot \infty \\ \displaystyle d_i = \infty {/eq}

The image is at INFINITY.

**(b)**

{eq}\displaystyle \frac{1}{f} = \frac{1}{p} + \frac{1}{d_i} \\ \displaystyle \frac{1}{9\,cm} = \frac{1}{18\,cm} + \frac{1}{d_i} \\ \displaystyle \frac{1}{9\,cm} - \frac{1}{18\,cm} = \frac{1}{d_i} \\ \displaystyle \left (\frac{1}{9\,cm} \cdot \frac{2}{2} \right ) - \frac{1}{18\,cm} = \frac{1}{d_i} \\ \displaystyle \left (\frac{2}{18\,cm} \right ) - \frac{1}{18\,cm} = \frac{1}{d_i} \\ \displaystyle \frac{2-1}{18\,cm} = \frac{1}{d_i} \\ \displaystyle \frac{1}{18\,cm} = \frac{1}{d_i} \\ \displaystyle d_i = \frac{18\,cm}{1} \\ \displaystyle d_i = 18\,cm {/eq}

The image is located 18 cm in front of the mirror, that is at the radius of curvature. It is real.

The magnification is computed by

{eq}\displaystyle M = -\frac{d_i}{p} \\ \displaystyle M = -\frac{18\,cm}{18\,cm} \\ \displaystyle M = -1 {/eq}

The image is same size since its magnification's magnitude is 1 and it is inverted.

In summary, the image is

- located at center of curvature
- real
- same size
- inverted

**(c)**

{eq}\displaystyle \frac{1}{f} = \frac{1}{p} + \frac{1}{d_i} \\ \displaystyle \frac{1}{9\,cm} = \frac{1}{31\,cm} + \frac{1}{d_i} \\ \displaystyle \frac{1}{9\,cm} - \frac{1}{31\,cm} = \frac{1}{d_i} \\ \displaystyle \left (\frac{1}{9\,cm} \cdot \frac{31}{31} \right ) - \left (\frac{1}{31\,cm} \cdot \frac{9}{9} \right ) = \frac{1}{d_i} \\ \displaystyle \left (\frac{31}{279\,cm} \right ) - \left (\frac{9}{279\,cm} \right ) = \frac{1}{d_i} \\ \displaystyle \frac{31-9}{279\,cm} = \frac{1}{d_i} \\ \displaystyle \frac{22}{279\,cm} = \frac{1}{d_i} \\ \displaystyle d_i = \frac{279\,cm}{22} \\ \displaystyle d_i = 12.68\,cm {/eq}

The magnification is

{eq}\displaystyle M = - \frac{d_i}{p} \\ \displaystyle M = - \frac{12.68\,cm}{31.00\,cm} \\ \displaystyle M = - 0.409 {/eq}

The image is

- located 12.68 cm in front of the mirror
- real
- magnified -0.409x
- inverted

#### Learn more about this topic:

from Physics 101: Help and Review

Chapter 14 / Lesson 17