1. 2nd period: TX 146 3rd period: VA 144
2nd period: TX 146
3rd period: VA 144

2. Inverted
Reduced in size
Smaller
Real image

3. At center of curvature
Inverted
Same size
Real image

4. -Located beyond center -inverted -larger -real image

5. NO IMAGE

6. ON OPPOSITE SIDE OF MIRROR
UPRIGHT
Bigger
VIRTUAL

7. Mirror Equation – Concave Mirrors
Also called the magnification equation
Relationship between the object distance (do) and the focal length (f).
The magnification equation relates to the ratio of the image distance and object distance to the ratio of image height and object height.

8. The magnification equation relates to the ratio of the image distance and object distance to the ratio of image height(hi) and object height (ho). .

9. A 4 cm tall light bulb is placed a distance of 45
A 4 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.
What do we know and don’t know.
ho = 4.0 cm
do = 45.7 cm
f = 15.2 cm
di = ???
hi = ???

10. 1/f = 1/do + 1/di .07 = .02 + 1/di .05 = 1/di Di = 20
Want to find di first in order to find the next portion of the equation
1/f = 1/do + 1/di
.07 = /di
.05 = 1/di
Di = 20

11. Now that we know the value is 22.8 cm
Hi/ho = -di/do
Hi/4 = - 20/45.7
Hi = x 4
Hi= cm
Negative value means that the image is inverted.

12. A 4. 0-cm tall light bulb is placed a distance of 8
A 4.0-cm tall light bulb is placed a distance of 8.3 cm from a concave mirror having a focal length of 15.2 cm. (NOTE: this is the same object and the same mirror, only this time the object is placed closer to the mirror.) Determine the image distance and the image size.
Di=-18.3 cm (virtual image, behind mirror) hi=8.8cm