Practice question
Question
An object of height \( 3 \, \text{cm} \) is placed \( 15 \, \text{cm} \) from a concave mirror of focal
length \( 10 \, \text{cm} \). What is the height of the image?
Explanation
**Simple microscope** magnification M = 1 + D/f when image at D=25 cm near point, M = D/f when image at infinity (relaxed eye), f focal length (cm), D least distance of distinct vision 25 cm. For f=4 cm, image at 25 cm, M=1+25/4=7.25, angular magnification ratio of angle subtended by image to that by object at D. Focal length: f = -10 cm , u = -15 cm . Mirror equation: (1/v) + (1/-15) = (1/-10) ⇒ (1/v) = (1/-10) + (1/15) = (-3 + 2/30) = (-1/30) . v = -30 cm . Magnification: m = -(v/u) = -(-30/-15) = -2 . Image
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