# At an intersection of hospital hallways, a convex spherical mirror is mounted high on a wall to...

## Question:

At an intersection of hospital hallways, a convex spherical mirror is mounted high on a wall to help people avoid collisions. The magnitude of the mirror s radius of curvature is 0.550 m.

(a) Locate the image of a patient located 10.0 m from the mirror.

(b) Indicate whether the image is upright or inverted.

(c) Determine the magnification of the image.

## Convex Mirror:

The focal length and radius of curvature for the convex mirror is behind the mirror that's why it is negative. By sign rules, if the center of curvature is on the opposite side of the outgoing light, the radius of curvature is negative. Otherwise, positive.

The focal length is given by the formula:

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

Given:

R = -0.550 cm (negative since it is a convex mirror)

s = 10.0 cm

Solution:

Part A.

Since the focal length is half the radius of curvature, focal length is:

{eq}f = \frac{R}{2} = \frac{-0.550 \ m}{2} = -0.275 \ m {/eq}

The image distance is equal to:

{eq}\frac{1}{f} = \frac{1}{s} + \frac{1}{s'} \\ \frac{1}{s'} = \frac{1}{f} - \frac{1}{s} \\ \frac{1}{s'} = \frac{1}{-0.275 \ m} - \frac{1}{10 \ m} \\ s' = -0.27 \ m {/eq}

Part B.

The image is upright.

Part C.

Magnification is given by the formula:

{eq}m = \frac{-s'}{s} \\ m = \frac{-(-0.27 \ m)}{10 \ m} \\ m = 0.027 {/eq}