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Thin Lenses - Lens Formula, Magnification and Power

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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

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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

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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

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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

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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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