#concave lens

Telescope, Human Eye and Defects of Vision

What is the primary reason a concave lens cannot form a real image?

**Astronomical telescope** in normal adjustment M = -f_o/f_e, f_o objective focal length (m), f_e eyepiece focal length, length L = f_o+f_e, final image at infinity, angular magnification ratio of angles subtended by image and object. For f_o=100 cm, f_e=5 cm, M=-20, inverted, used for celestial observation. A concave lens diverges light rays, preventing them from converging to a point on the opposite side. The rays appear to diverge from a virtual focal point on the same side as the object, resulting in a virtual image that cannot be projected, regardless of object position. Substituting values gives It diverges light rays, which matches expected image position

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Telescope, Human Eye and Defects of Vision

A convex lens (\( f = 20 \, \text{cm} \)) and a concave lens (\( f = 40 \, \text{cm} \)) are in contact. What is the eff

**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. f₁ = 20 cm , f₂ = -40 cm . (1/f) = (1/f₁) + (1/f₂) = (1/20) + (1/-40) = (2 - 1/40) = (1/40) . f = 40 cm (converging system). 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.

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Combination of Lenses and Lens Systems

In a concave lens, what is the effect on the image if the object is moved closer to the lens from a distant position?

**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. For a concave lens, the image is always virtual, erect, and diminished. As the object moves closer, the image size increases (though still smaller than the object), and the image moves closer to the lens, but remains on the same side as the object. Substituting values gives Image size increases but remains diminished, which matches expected image position and magnification from mirror/lens formula 1/f

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Combination of Lenses and Lens Systems

A concave lens of focal length \( 25 \, \text{cm} \) forms an image \( 10 \, \text{cm} \) from the lens. What is the obj

**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. Focal length: f = -25 cm (concave lens). Image distance: v = -10 cm (virtual image). Lens formula: (1/v) - (1/u) = (1/f) . (1/-10) - (1/u) = (1/-25) ⇒ (1/u) = (1/-10) - (1/-25) = (-5 + 2/50) = (-3/50) . u = -(50/3) ≈ -16.67 cm . Substituting values gives 16.7 cm, which

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Combination of Lenses and Lens Systems

A convex lens (\( f = 50 \, \text{cm} \)) and a concave lens (\( f = 25 \, \text{cm} \)) are in contact. What is the eff

**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. f₁ = 50 cm , f₂ = -25 cm . (1/f) = (1/f₁) + (1/f₂) = (1/50) + (1/-25) = (1 - 2/50) = (-1/50) . f = -50 cm (diverging system). 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

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Combination of Lenses and Lens Systems

A concave lens of focal length \( 15 \, \text{cm} \) has an object placed \( 30 \, \text{cm} \) from it. What is the ima

**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 = -15 cm (concave lens). Object distance: u = -30 cm . Lens formula: (1/v) - (1/u) = (1/f) . (1/v) - (1/-30) = (1/-15) ⇒ (1/v) + (1/30) = (1/-15) ⇒ (1/v) = (1/-15) - (1/30) = (-2 - 1/30) = (-3/30) = (-1/10) . v = -10 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)

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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 =

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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 refraction principles.

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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 at Spherical Surfaces and Lens Maker's Formula

A concave lens of focal length \( 10 \, \text{cm} \) forms an image \( 5 \, \text{cm} \) from the lens. What is the obje

**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. Focal length: f = -10 cm (concave lens). Image distance: v = -5 cm (virtual image). Lens formula: (1/v) - (1/u) = (1/f) . (1/-5) - (1/u) = (1/-10) ⇒ (1/u) = (1/-5) - (1/-10) = (-2 + 1/10) = (-1/10) . u = -10 cm . Substituting values gives 10 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v -

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Refraction at Spherical Surfaces and Lens Maker's Formula

In a concave lens, why is the image always formed on the same side as the object?

**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. A concave lens diverges light rays, making them appear to originate from a point on the same side as the object when traced backward. This results in a virtual image that cannot be projected on a screen, always forming on the object’s side regardless of its position. Substituting values gives Because rays diverge and appear to come from the same side, which matches expected image position and magnification from mirror/lens

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