A 3 cm tall object is placed 9 cm in front of a concave lens with focal length 6 cm. What is the approximate height of the image formed?
1.2 cm and upright
Using the sign convention, u = −9 cm, f = −6 cm (concave lens) and h = +3 cm.
Lens formula: 1/v − 1/u = 1/f, so 1/v = 1/f + 1/u = −1/6 − 1/9 = (−3 − 2)/18 = −5/18.
Therefore v = −18/5 = −3.6 cm (virtual image on the same side as the object).
Magnification m = v/u = (−3.6)/(−9) = 0.4.
Image height h′ = m × h = 0.4 × 3 = 1.2 cm; since h′ is positive, the image is upright (erect).
Hence, the answer is 1.2 cm and upright.
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