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: Object size = 4.5 cm, Object distance u = -12 cm, Focal length f = +15 cm (convex mirror)
Step 1: Finding image location using mirror formula Using 1/v + 1/u = 1/f 1/v + 1/(-12) = 1/15 1/v = 1/15 + 1/12 1/v = (4+5)/60 = 9/60 v = +6.67 cm The image is formed 6.67 cm behind the mirror.
Step 2: Finding magnification m = -v/u = -(6.67)/(-12) = +0.56 The magnification is +0.56 (positive indicates virtual and erect image, less than 1 indicates diminished).
Step 3: Effect of moving needle farther As the needle is moved farther from the mirror, the image moves closer to the focus and becomes smaller. The image remains virtual, erect and diminished, located between the pole and the focus.
Explanation
This question tests application of the mirror formula for convex mirrors. The convex mirror has a positive focal length (+15 cm). Using the mirror equation, we find the image distance v = +6.67 cm, indicating a virtual image formed behind the mirror. The magnification m = +0.56 shows the image is erect and diminished. As per Exercise 9.15 in the textbook, a convex mirror always produces a virtual image that is diminished and located between the focus and the pole, regardless of object position.
Solution Steps
Step 1: Apply mirror formula 1/v + 1/u = 1/f with u = -12 cm, f = +15 cm
Step 2: Calculate v = +6.67 cm (image location)
Step 3: Calculate magnification m = -v/u = +0.56
Step 4: Describe behavior as object moves farther - image approaches focus, remains virtual and diminished