Spherical mirror radius of curvature

[roam]

Homework Statement

A dentist uses a spherical mirror to examine a tooth. The tooth is 1.13 cm in front of the mirror, and the image is formed 10.8 cm behind the mirror. Determine the mirror's radius of curvature.

Homework Equations

1/p+1/q=1/f

f=R/2

The Attempt at a Solution

Since the object is in front of the mirror, p > 0. With the image behind the mirror q < 0. So the radius of curvature is

[tex]\frac{2}{R}=\frac{1}{p} + \frac{1}{q} = \frac{1}{1.13} - \frac{1}{10.8} = \frac{10.8-1.13}{10.8}[/tex]

So [tex]R = 2 \frac{10.8}{9.67} = 2.233[/tex]

Why is my answer is wrong? I think I used the correct equations. I tried the same question with different numbers and the computer still marks my answer wrong. Any explanation would be appreciated.

 

[kuruman]

[itex]\frac{1}{1.13}-\frac{1}{10.8}\neq\frac{10.8-1.13}{10.8}[/itex]

 

[PeterO]

Homework Statement

A dentist uses a spherical mirror to examine a tooth. The tooth is 1.13 cm in front of the mirror, and the image is formed 10.8 cm behind the mirror. Determine the mirror's radius of curvature.

Homework Equations

1/p+1/q=1/f

f=R/2

The Attempt at a Solution

Since the object is in front of the mirror, p > 0. With the image behind the mirror q < 0. So the radius of curvature is

[tex]\frac{2}{R}=\frac{1}{p} + \frac{1}{q} = \frac{1}{1.13} - \frac{1}{10.8} = \frac{10.8-1.13}{10.8}[/tex]

So [tex]R = 2 \frac{10.8}{9.67} = 2.233[/tex]

Why is my answer is wrong? I think I used the correct equations. I tried the same question with different numbers and the computer still marks my answer wrong. Any explanation would be appreciated.

You mucked up the subtraction of two fractions - in particular, the denominator.