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#focal length

100 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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Telescope, Human Eye and Defects of Vision

A converging beam meets a convex lens (\( f = 15 \, \text{cm} \)) \( 10 \, \text{cm} \) before the convergence point. Wh

**Human eye** least distance D=25 cm, near point, far point infinity for normal eye, accommodation by ciliary muscles changing lens curvature, power ≈60 D total. Defects: myopia far point 25 cm corrected by converging lens, astigmatism cylindrical lens. Object distance: u = -10 cm (virtual object), f = 15 cm . Lens formula: (1/v) - (1/-10) = (1/15) ⇒ (1/v) + (1/10) = (1/15) . (1/v) = (1/15) - (1/10) = (2 - 3/30) = (-1/30) . v = -30 cm (30 cm to the left). Substituting values gives 30 cm, which matches expected image position and

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

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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 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

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

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Combination of Lenses and Lens Systems

A simple microscope with a lens of focal length \( 10 \, \text{cm} \) forms an image at the least distance of distinct v

**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. Magnification: m = 1 + (D/f) . D = 25 cm , f = 10 cm . m = 1 + (25/10) = 1 + 2.5 = 3.5 . Substituting values gives 3.5, 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 > Combination of Lenses and Lens Systems

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Combination of Lenses and Lens Systems

A convex mirror of focal length \( 20 \, \text{cm} \) has an object placed \( 40 \, \text{cm} \) from it. What is the im

**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 (convex mirror). Object distance: u = -40 cm . Mirror equation: (1/v) + (1/u) = (1/f) . (1/v) + (1/-40) = (1/20) ⇒ (1/v) = (1/20) + (1/40) = (2 + 1/40) = (3/40) . v = (40/3) ≈ 13.33 cm (virtual image). Substituting values gives 10 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

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

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