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#lens power

9 public questions tagged with this topic.

Combination of Lenses and Lens Systems

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

**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. Power: P = (1/f) (in meters). P = +3 D ⇒ 3 = (1/f) ⇒ f = (1/3) = 0.333 m ≈ 33.33 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 = 1/v + 1/u (mirror), confirming sign conventions and refract

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

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Combination of Lenses and Lens Systems

A concave lens has a power of \( -2 \, \text{D} \). What is its focal length?

**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. Power: P = (1/f) (in meters). P = -2 D ⇒ -2 = (1/f) ⇒ f = -(1/2) = -0.5 m = -50 cm . Substituting values gives -50 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 refract

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

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Refraction at Spherical Surfaces and Lens Maker's Formula

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

**Lens maker's formula** 1/f = (n-1)(1/R₁ - 1/R₂), n refractive index, R₁,R₂ radii of curvature (m), sign convention R positive if surface convex towards incident light. For double convex R₁=12 cm, R₂=-12 cm, n=1.5, 1/f=(0.5)(1/12 -1/(-12))=(0.5)(2/12)=1/12, f=12 cm, converging. 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 = 1/v + 1/u (mirror), confirming sign conventions and refraction principles.

Ref: NCERT > Physics Book > Ray Optics > Refraction at Spherical Surfaces and Lens Maker's Formula

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Total Internal Reflection and Critical Angle

A lens has a power of \( +2.5 \, \text{D} \). What is its focal length in centimeters?

**Total internal reflection** occurs when light travels from denser to rarer medium and incidence > C, condition sinC = 1/n for air interface. For glass n=1.52 C≈41°, water n=1.33 C≈48.8°, so at 49° water-air TIR occurs, explaining why ray does not emerge. Power: P = (1/f) (in meters). P = +2.5 D ⇒ 2.5 = (1/f) ⇒ f = (1/2.5) = 0.4 m = 40 cm . 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 principles.

Ref: NCERT > Physics Book > Ray Optics > Total Internal Reflection and Critical Angle

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Refraction at Plane Surfaces and Snell's Law

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

**Snell's law** n₁ sinθ₁ = n₂ sinθ₂ describes refraction at plane interface, n refractive index, θ angle with normal. When light goes from denser n=1.52 glass to rarer air n=1, sinθ₂ = (n₁/n₂) sinθ₁ > sinθ₁, bending away from normal, enabling total internal reflection beyond critical angle. 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 = 1/v + 1/u (mirror), confirming sign conventions and refraction principles.

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

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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 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 =

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 \( -5 \, \text{D} \). 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 = -5 D ⇒ -5 = (1/f) ⇒ f = -(1/5) = -0.2 m = -20 cm . Substituting values gives -20 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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