Exercises
A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.
Given: h = 4.5 cm, u = -12 cm, f = +15 cm (convex mirror).
Mirror equation: 1/v + 1/u = 1/f
1/v = 1/f - 1/u = 1/15 - 1/(-12) = 1/15 + 1/12 = (4 + 5)/60 = 9/60
v = +6.67 cm (approximately 20/3 cm)
The image is formed 6.67 cm behind the mirror (virtual image).
Magnification: m = -v/u = -(6.67)/(-12) = +0.556
Image height: h' = m × h = 0.556 × 4.5 = +2.5 cm
The image is virtual, erect, and diminished (size = 2.5 cm).
As the needle is moved farther from the mirror (u increases in magnitude), the virtual image moves toward the focus F (which is behind the mirror at 15 cm). The image always remains virtual, erect, and diminished. Even when the object is at infinity, the image is at the focus, 15 cm behind the mirror.
Answer: Image at 6.67 cm behind the mirror (virtual, erect, diminished); magnification = +0.56. As needle moves farther, image approaches but never crosses the focal point.
