Practice question
Question
A concave lens of focal length \( 15 \, \text{cm} \) has an object placed \( 30 \, \text{cm} \) from
it. What is the image distance?
Explanation
**Lens combination** effective focal length calculation uses reciprocal sum, for f₁=10 cm, f₂=20 cm in contact, 1/F=1/10+1/20=3/20, F=6.67 cm, P=15 D, stronger converging than either alone, illustrating power addition. Focal length: f = -15 cm (concave lens). Object distance: u = -30 cm . Lens formula: (1/v) - (1/u) = (1/f) . (1/v) - (1/-30) = (1/-15) ⇒ (1/v) + (1/30) = (1/-15) ⇒ (1/v) = (1/-15) - (1/30) = (-2 - 1/30) = (-3/30) = (-1/10) . v = -10 cm (virtual image). Substituting values gives 10 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens)
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