An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
Given: Object-size, h = +7.0 cm Object-distance, u = -27 cm Focal length, f = -18 cm (concave mirror)
To find: Image-distance (v), Image-size (h'), Nature of image
Step 1: Finding Image-distance (v) Using the mirror formula: 1/v + 1/u = 1/f
1/v = 1/f - 1/u 1/v = 1/(-18) - 1/(-27) 1/v = -1/18 + 1/27 1/v = (-3 + 2)/54 = -1/54
v = -54 cm
The screen should be placed at 54 cm in front of the mirror to obtain a sharp focussed image.
Step 2: Finding Magnification (m) m = -v/u = -(-54)/(-27) = -2
Step 3: Finding Image-size (h') m = h'/h -2 = h'/7.0 h' = -14 cm
Nature of Image: The image is real, inverted and enlarged (magnification = 2 times).
Explanation
This is a standard numerical problem on mirror formula. The object is placed beyond the centre of curvature (27 cm > 2f = 36 cm would be C, but here object is at 27 cm which is between f and 2f). Actually, since f = 18 cm, 2f = 36 cm. The object at 27 cm is between f and 2f, so the image formed will be beyond 2f, which is real, inverted and enlarged. The calculations confirm this - image distance is 54 cm (beyond 2f), magnification is -2 (negative indicates inverted, magnitude 2 indicates enlarged).
Solution Steps
Step 1: Apply mirror formula 1/v + 1/u = 1/f to find image distance v = -54 cm
Step 2: Calculate magnification m = -v/u = -2
Step 3: Find image size using h' = m × h = -14 cm
Step 4: Conclude nature - real (v negative), inverted (h' negative), enlarged (|m| > 1)