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Thin Lenses - Lens Formula, Magnification and Power

This category covers the fundamentals of thin lenses, including how to apply the lens formula, calculate magnification, and determine lens power. It is useful for students studying optics and related physics topics.

30 questions

Thin Lenses - Lens Formula, Magnification and Power

A telescope has an objective of focal length \( 180 \, \text{cm} \) and an eyepiece of focal length \( 6 \, \text{cm} \)

**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. Magnifying power: m = (f_o/f_e) . f_o = 180 cm , f_e = 6 cm . m = (180/6) = 30 . Substituting values gives 30, 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

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

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Thin Lenses - Lens Formula, Magnification and Power

A convex lens of focal length \( 25 \, \text{cm} \) forms an image of an object placed \( 50 \, \text{cm} \) from it. Wh

**Convex lens image formation**: object beyond 2F (u>2f) real inverted diminished between F and 2F, at 2F same size at 2F, between F and 2F magnified beyond 2F, at F image at infinity, within F virtual erect magnified same side. For f=15 cm, u=30 cm=2f, m = v/u =30/30=1? Actually v=30 cm, m=-1, same size inverted. Focal length: f = 25 cm . 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) = (1/50) . v = 50

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

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Thin Lenses - Lens Formula, Magnification and Power

A lens has a power of \( -4 \, \text{D} \). What is its focal length?

**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. Power: P = (1/f) (in meters). P = -4 D ⇒ -4 = (1/f) ⇒ f = -(1/4) = -0.25 m = -25 cm . Substituting values gives -25 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f

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

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Thin Lenses - Lens Formula, Magnification and Power

A simple microscope uses a convex lens of focal length \( 5 \, \text{cm} \). What is the magnification when the image is

**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. Magnification: m = 1 + (D/f) . D = 25 cm , f = 5 cm . m = 1 + (25/5) = 1 + 5 = 6 . Substituting values gives 6, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u

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

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Thin Lenses - Lens Formula, Magnification and Power

A lens has a power of \( -3 \, \text{D} \). What is its focal length?

**Convex lens image formation**: object beyond 2F (u>2f) real inverted diminished between F and 2F, at 2F same size at 2F, between F and 2F magnified beyond 2F, at F image at infinity, within F virtual erect magnified same side. For f=15 cm, u=30 cm=2f, m = v/u =30/30=1? Actually v=30 cm, m=-1, same size inverted. Power: P = (1/f) (in meters). P = -3 D ⇒ -3 = (1/f) ⇒ f = -(1/3) ≈ -0.333 m ≈ -33.3 cm . Substituting values gives -33 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f =

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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

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