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Question

A concave lens of focal length \( 25 \, \text{cm} \) forms an image \( 10 \, \text{cm} \) from the
lens. What is the object distance?

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Explanation

**Combination of lenses in contact** effective power P = P₁+P₂, effective focal length 1/F = 1/f₁ + 1/f₂, for thin lenses in contact. If separated by d, 1/F =1/f₁+1/f₂ - d/(f₁ f₂). Power adds algebraically, converging + with diverging - can cancel, used to correct aberrations and design achromatic doublets. Focal length: f = -25 cm (concave lens). Image distance: v = -10 cm (virtual image). Lens formula: (1/v) - (1/u) = (1/f) . (1/-10) - (1/u) = (1/-25) ⇒ (1/u) = (1/-10) - (1/-25) = (-5 + 2/50) = (-3/50) . u = -(50/3) ≈ -16.67 cm . Substituting values gives 16.7 cm, which

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