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#lens formula

17 public questions tagged with this topic.

Telescope, Human Eye and Defects of Vision

A double convex lens has radii of curvature \( 20 \, \text{cm} \) each and refractive index \( 1.5 \). What is its focal

**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. Lens maker’s formula: (1/f) = (n - 1) ( (1/R₁) - (1/R₂) ) . n = 1.5 , R₁ = 20 cm , R₂ = -20 cm (sign convention). (1/f) = (1.5 - 1) ( (1/20) - (1/-20) ) = 0.5 ( (1/20) + (1/20) ) = 0.5 × (2/20) = (1/20) . f = 20 cm . Substituting values gives 20 cm, which matches expected image position and magnification from mirror

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 of focal length \( 20 \, \text{cm} \) forms an image at \( 40 \, \text{cm} \) from the lens. What is the o

**Lens combination** effective focal length calculation uses reciprocal sum, for f₁=10 cm, f₂=20 cm in contact, 1/F=1/10+1/20=3/20, F=6.67 cm, P=15 D, stronger converging than either alone, illustrating power addition. Focal length: f = 20 cm . Image distance: v = 40 cm (real image). Lens formula: (1/v) - (1/u) = (1/f) . (1/40) - (1/u) = (1/20) ⇒ (1/u) = (1/40) - (1/20) = (1 - 2/40) = (-1/40) . u = -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 convent

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Combination of Lenses and Lens Systems

A concave lens of focal length \( 30 \, \text{cm} \) produces an image \( 15 \, \text{cm} \) from the lens. What is the

**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. Focal length: f = -30 cm (concave lens). Image distance: v = -15 cm (virtual image). Lens formula: (1/v) - (1/u) = (1/f) . (1/-15) - (1/u) = (1/-30) ⇒ (1/u) = (1/-15) - (1/-30) = (-2 + 1/30) = (-1/30) . u = -30 cm . Substituting values gives 30 cm, which matches expected image position and magnification from mirror/lens formula 1/f =

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

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Combination of Lenses and Lens Systems

In a simple microscope, why is the image formed larger when the object is placed closer to the lens than the focal point

**Lens combination** effective focal length calculation uses reciprocal sum, for f₁=10 cm, f₂=20 cm in contact, 1/F=1/10+1/20=3/20, F=6.67 cm, P=15 D, stronger converging than either alone, illustrating power addition. In a simple microscope, placing the object between the lens and focal point results in a virtual, erect, and magnified image. The closer the object is to the lens (inside F), the greater the divergence of rays, increasing the apparent size of the virtual image seen by the observer. Substituting values gives Due to increased divergence of rays, which matches expected image positi

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Refraction at Spherical Surfaces and Lens Maker's Formula

A converging beam meets a concave lens (\( f = 20 \, \text{cm} \)) \( 5 \, \text{cm} \) before the convergence point. Wh

**Spherical refracting surface** power P = (n₂-n₁)/R, lens power sum of two surfaces. Lens maker derivation combines two refractions, sign of R₂ negative for second surface convex opposite direction, yielding 1/f positive for convex lens. Object distance: u = -5 cm (virtual object), f = -20 cm . Lens formula: (1/v) - (1/-5) = (1/-20) ⇒ (1/v) + (1/5) = (1/-20) . (1/v) = (1/-20) - (1/5) = (-1 - 4/20) = (-5/20) = (-1/4) . v = -4 cm (4 cm to the left). Substituting values gives 4 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u

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Refraction at Spherical Surfaces and Lens Maker's Formula

A converging beam meets a concave lens (\( f = 15 \, \text{cm} \)) \( 6 \, \text{cm} \) before the convergence point. Wh

**Refraction at spherical surface** formula n₁/u + n₂/v = (n₂-n₁)/R governs single surface, extension to two surfaces yields lens maker. Double convex with equal |R| has f = R/[2(n-1)], for R=12 cm, n=1.5, f=12 cm, illustrating dependence on curvature and index. Object distance: u = -6 cm (virtual object), f = -15 cm . Lens formula: (1/v) - (1/-6) = (1/-15) ⇒ (1/v) + (1/6) = (1/-15) . (1/v) = (1/-15) - (1/6) = (-2 - 5/30) = (-7/30) . v = -(30/7) ≈ -4.29 cm (4.29 cm to the left). Substituting values gives 4.3 cm, which matches expected image position and magnification from mirror/lens

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Refraction Through Prism and Minimum Deviation

A convex lens of focal length \( 18 \, \text{cm} \) has an object placed \( 36 \, \text{cm} \) from it. What is the imag

**Prism minimum deviation** δ_m satisfies n = sin[(A+δ_m)/2]/sin(A/2), A prism angle (degrees), n refractive index, δ_m minimum deviation. For A=60°, n=1.45, sin[(60+δ_m)/2]=1.45×sin30°=0.725, (60+δ_m)/2=46.5°, 60+δ_m=93°, δ_m=33°, illustrating n increase raises δ_m. Focal length: f = 18 cm . Object distance: u = -36 cm . Lens formula: (1/v) - (1/u) = (1/f) . (1/v) - (1/-36) = (1/18) ⇒ (1/v) + (1/36) = (1/18) ⇒ (1/v) = (1/18) - (1/36) = (2 - 1/36) = (1/36) . v = 36 cm (real image). Substituting values gives 36 cm, which matches expected image position and magnification from mirror/lens formula

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Refraction at Spherical Surfaces and Lens Maker's Formula

A double convex lens of refractive index \( 1.55 \) has radii of curvature \( 22 \, \text{cm} \) and \( -22 \, \text{cm}

**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. Lens maker’s formula: (1/f) = (n - 1) ( (1/R₁) - (1/R₂) ) . n = 1.55 , R₁ = 22 cm , R₂ = -22 cm . (1/f) = (1.55 - 1) ( (1/22) - (1/-22) ) = 0.55 ( (1/22) + (1/22) ) = 0.55 × (2/22) = (1.1/22) = (1/20) . f = 20 cm . Substituting values gives 20 cm, which matches

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Total Internal Reflection and Critical Angle

A converging beam meets a convex lens (\( f = 10 \, \text{cm} \)) \( 5 \, \text{cm} \) before the convergence point. Wha

**Critical angle** C satisfies sinC = n₂/n₁, n₁>n₂, n₂=1 for air, n₁=1.52 for glass gives sinC=1/1.52=0.6579, C≈41.1°, for water n=1.33 C=48.75°. Beyond C, total internal reflection occurs, all light reflected, no refracted ray, used in optical fibers and prisms. Object distance: u = -5 cm (virtual object), f = 10 cm . Lens formula: (1/v) - (1/-5) = (1/10) ⇒ (1/v) + (1/5) = (1/10) . (1/v) = (1/10) - (1/5) = (1 - 2/10) = (-1/10) . v = -10 cm (10 cm to the left). Substituting values gives 10 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v

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Total Internal Reflection and Critical Angle

A concave lens of focal length \( 20 \, \text{cm} \) forms an image \( 8 \, \text{cm} \) from the lens. What is the obje

**TIR application** requires n₁>n₂ and θ₁>C. Critical angle formula derived from Snell's law with θ₂=90°, n₁ sinC = n₂ sin90° = n₂, so sinC = n₂/n₁. For glass-air sinC=1/1.52, C≈41°, for water-air 48.6°, determining cutoff for transmission. Focal length: f = -20 cm (concave lens). Image distance: v = -8 cm (virtual image). Lens formula: (1/v) - (1/u) = (1/f) . (1/-8) - (1/u) = (1/-20) ⇒ (1/u) = (1/-8) - (1/-20) = (-5 + 2/40) = (-3/40) . u = -(40/3) ≈ -13.33 cm . Substituting values gives 13.3 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v -

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Refraction at Plane Surfaces and Snell's Law

A converging beam of light meets a concave lens of focal length \( 25 \, \text{cm} \) at \( 10 \, \text{cm} \) before th

**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. u = -10 cm (virtual object), f = -25 cm . (1/v) - (1/-10) = (1/-25) ⇒ (1/v) + (1/10) = (1/-25) . (1/v) = (1/-25) - (1/10) = (-2 - 5/50) = (-7/50) . v = -(50/7) ≈ -7.14 cm (7.14 cm to the left). Substituting values gives 7.14 cm, which matches expected image position and magnification from mirror/lens formula 1/f

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Refraction at Plane Surfaces and Snell's Law

In a convex lens, why does the image transition from virtual to real as the object moves from inside to outside the foca

**Refractive index** n = c/v, water 1.33 means light 1.33 times slower than vacuum. Passing from water to air at 49°, n₁ sinθ₁ =1.33×sin49°≈1.33×0.755=1.004>1, so sinθ₂>1 impossible, total internal reflection occurs, no refraction. Inside the focal point, a convex lens diverges rays, forming a virtual image on the same side. Beyond the focal point, the lens converges rays to a point on the opposite side, forming a real image. This transition occurs as the object crosses the focal point, changing the ray behavior. Substituting values gives Due to change from divergence to convergence, which mat

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