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#combined lenses

3 public questions tagged with this topic.

Telescope, Human Eye and Defects of Vision

A convex lens (\( f = 20 \, \text{cm} \)) and a concave lens (\( f = 40 \, \text{cm} \)) are in contact. What is the eff

**Telescope resolving power** depends on objective aperture, magnification limited by objective diffraction, terrestrial telescope adds erecting lens, Galilean uses diverging eyepiece for erect image, reflecting telescope uses concave mirror objective avoiding chromatic aberration. f₁ = 20 cm , f₂ = -40 cm . (1/f) = (1/f₁) + (1/f₂) = (1/20) + (1/-40) = (2 - 1/40) = (1/40) . f = 40 cm (converging system). Substituting values gives 40 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 convention

Ref: NCERT > Physics Book > Ray Optics > Telescope, Human Eye and Defects of Vision

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Combination of Lenses and Lens Systems

A convex lens (\( f = 40 \, \text{cm} \)) and a concave lens (\( f = 20 \, \text{cm} \)) 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 = 40 cm , f₂ = -20 cm . (1/f) = (1/f₁) + (1/f₂) = (1/40) + (1/-20) = (1 - 2/40) = (-1/40) . f = -40 cm (diverging system). Substituting values gives -40 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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Thin Lenses - Lens Formula, Magnification and Power

A convex lens (\( f = 50 \, \text{cm} \)) and a concave lens (\( f = 25 \, \text{cm} \)) are in contact. What is the eff

**Lens formula** 1/f = 1/v - 1/u, f focal length (m), u object distance, v image distance, sign: u negative if object left of lens in Cartesian convention, magnification m = v/u, real image inverted m negative, virtual erect m positive. Power P = 1/f (diopters D), f in meters, +5 D means f=0.2 m=20 cm converging. f₁ = 50 cm , f₂ = -25 cm . (1/f) = (1/f₁) + (1/f₂) = (1/50) + (1/-25) = (1/50) - (2/50) = (1 - 2/50) = (-1/50) . f = -50 cm (diverging system). Substituting values gives -50 cm, which matches expected image position and magnification

Ref: NCERT > Physics Book > Ray Optics > Thin Lenses - Lens Formula, Magnification and Power

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