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#geometric optics

2 public questions tagged with this topic.

Combination of Lenses and Lens Systems

A convex lens (\( f = 30 \, \text{cm} \)) and a concave lens (\( f = 15 \, \text{cm} \)) are in contact. What is the eff

**Two lenses in contact** behave as single lens with power sum, magnification product m = m₁×m₂. For f₁=20 cm (P₁=5 D), f₂=-20 cm (P₂=-5 D) in contact, P=0, F infinite, afocal system, beam emerges parallel, principle of corrective lenses for myopia/hypermetropia. f₁ = 30 cm , f₂ = -15 cm . (1/f) = (1/f₁) + (1/f₂) = (1/30) + (1/-15) = (1 - 2/30) = (-1/30) . f = -30 cm (diverging system). Substituting values gives -30 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror), confirming sign conventions and refraction

Ref: NCERT > Physics Book > Ray Optics > Combination of Lenses and Lens Systems

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Refraction at Plane Surfaces and Snell's Law

A concave lens of focal length \( 25 \, \text{cm} \) has an object placed \( 50 \, \text{cm} \) from it. What is the ima

**Refraction at plane surface** changes direction due to speed change v = c/n, n = c/v. For water n=1.33 to air, n₁>n₂, ray bends away, if θ₁ > C, sinθ₂>1 impossible, TIR occurs. Snell's law quantitative prediction of θ₂ from θ₁ and indices. Focal length: f = -25 cm (concave lens). Object distance: u = -50 cm . Lens formula: (1/v) - (1/u) = (1/f) . (1/v) - (1/-50) = (1/-25) ⇒ (1/v) + (1/50) = (1/-25) ⇒ (1/v) = (1/-25) - (1/50) = (-2 - 1/50) = (-3/50) . v = -(50/3) ≈ -16.67 cm (virtual image). Substituting values gives 16.7 cm, which matches

Ref: NCERT > Physics Book > Ray Optics > Refraction at Plane Surfaces and Snell's Law

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