1. MIRROR EQUATIONS

2. Ray diagrams provide useful
information about the
characteristics of an image.
They do not provide any
quantitative information. eg: the
distance and image size.
We need to use the Mirror equation
and the Magnification equation.

3. Mirror Equations

5. Sample problem #1
10.0 cm tall candle is placed a distance of cm from a concave mirror having a focal length of 7.0cm
Determine the image distance and the image size.

6. 1/f = 1/do + 1/di 1/(7.0cm) = 1/(14.0 cm) + 1/di • 14 = di 1 - 1 = 1
= 1 di
• •
= 1 di
1 = 1 di

7. hi/ho = - di/do hi/10.0cm = - 14.0cm/14.0cm hi = - 1 x 10.0cm
 the image height is cm.
The negative value for image height indicate that the image is an inverted.

8. Sample Problem #2 Given: ho = 4.0 cm do = 45.7 cm f = 15.2 cm
A 4.0-cm tall light bulb is placed a distance of 45.7 cm from a concave mirror having a focal length of 15.2 cm. Determine the image distance and the image size.
Given:
ho = 4.0 cm do = 45.7 cm f = 15.2 cm
Unknowns:
di = ? hi = ?

9. di = 22.8 cm 1/f = 1/do + 1/di 1/(15.2 cm) = 1/(45.7 cm) + 1/di
 the image distance is 22.8 cm.

10. To determine the image height, the magnification equation is needed
hi/ho = - di/do
hi /(4.0 cm) = - (22.8 cm)/(45.7 cm)
hi = - (4.0 cm) * (22.8 cm)/(45.7 cm)
hi = cm
 the image height is cm.

11. Sample Problem #3 Given: ho = 4.0 cm do = 8.3 cm f = 15.2 cm Unknowns:
The same light bulb is placed a distance of 8.3 cm from a concave mirror having a focal length of 15.2 cm. Determine the image distance and the image size.
Given:
ho = 4.0 cm do = 8.3 cm f = 15.2 cm
Unknowns:
di = ? hi = ?

12. di = -18.3 cm 1/f = 1/do + 1/di 1/(15.2 cm) = 1/(8.3 cm) + 1/di
 the image distance is -18.3cm.
The negative distance indicates that the image is virtual and located behind the mirrror.

13. To determine the image height, the magnification equation is needed
hi/ho = - di/do
hi /(4.0 cm) = - (-18.2 cm)/(8.3 cm)
hi = - (4.0 cm) * (-18.2 cm)/(8.3 cm)
hi = 8.8 cm
 the image height is 8.8 cm.
The positive value for image height indicate that the image is an upright or erect..

14. The sign conventions for the given quantities in the mirror and magnification equations
f is +ve if the mirror is a concave mirror
f is -ve if the mirror is a convex mirror
di is + ve if the image is a real image and located on the object's side of the mirror.
di is - ve if the image is a virtual image and located behind the mirror.
hi is + ve if the image is an upright image
hi is - ve if the image an inverted image