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Combination of Lenses and Lens Systems

A convex lens of focal length \( 12 \, \text{cm} \) forms an image at \( 24 \, \text{cm} \) from the lens. What is the o

**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 = 12 cm . Image distance: v = 24 cm (real image). Lens formula: (1/v) - (1/u) = (1/f) . (1/24) - (1/u) = (1/12) ⇒ (1/u) = (1/24) - (1/12) = (1 - 2/24) = (-1/24) . u = -24 cm . Substituting values gives 24 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v

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Combination of Lenses and Lens Systems

A convex mirror of focal length \( 24 \, \text{cm} \) forms an image \( 8 \, \text{cm} \) behind the mirror. What is the

**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 = 24 cm (convex mirror). Image distance: v = 8 cm (virtual image). Mirror equation: (1/v) + (1/u) = (1/f) . (1/8) + (1/u) = (1/24) ⇒ (1/u) = (1/24) - (1/8) = (1 - 3/24) = (-2/24) = (-1/12) . u = -12 cm . 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

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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 mirror of focal length \( 7 \, \text{cm} \) has an object placed \( 14 \, \text{cm} \) from it. What is the im

**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 = -7 cm (concave mirror). Object distance: u = -14 cm . Mirror equation: (1/v) + (1/u) = (1/f) . (1/v) + (1/-14) = (1/-7) ⇒ (1/v) = (1/-7) + (1/14) = (-2 + 1/14) = (-1/14) . v = -14 cm (real image). Substituting values gives 14 cm, which matches expected

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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 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 convex mirror has a radius of curvature of \( 50 \, \text{cm} \). An object is placed \( 25 \, \text{cm} \) from it. W

**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 = (R/2) = (50/2) = 25 cm (positive for convex). Object distance: u = -25 cm . Mirror equation: (1/v) + (1/u) = (1/f) . (1/v) + (1/-25) = (1/25) ⇒ (1/v) = (1/25) + (1/25) = (2/25) . v = (25/2) = 12.5 cm (virtual image). Substituting values gives 12.5 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v +

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Refraction at Spherical Surfaces and Lens Maker's Formula

An object is placed \( 18 \, \text{cm} \) from a convex mirror of focal length \( 12 \, \text{cm} \). What is the image

**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. Focal length: f = 12 cm , u = -18 cm . Mirror equation: (1/v) + (1/-18) = (1/12) ⇒ (1/v) = (1/12) + (1/18) = (3 + 2/36) = (5/36) . v = (36/5) = 7.2 cm (virtual image). Substituting values gives 7.2 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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Refraction at Spherical Surfaces and Lens Maker's Formula

A converging beam meets a convex lens (\( f = 20 \, \text{cm} \)) \( 8 \, \text{cm} \) before the convergence point. Wha

**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 = -8 cm (virtual object). Focal length: f = 20 cm . Lens formula: (1/v) - (1/-8) = (1/20) ⇒ (1/v) + (1/8) = (1/20) . (1/v) = (1/20) - (1/8) = (2 - 5/40) = (-3/40) . v = -(40/3) ≈ -13.33 cm (13.33 cm to the left). 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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Optical Instruments - Simple and Compound Microscope

An object of height \( 5 \, \text{cm} \) is placed \( 30 \, \text{cm} \) from a concave mirror of focal length \( 15 \,

**Compound microscope** M = m_o × M_e, objective magnification m_o = v_o/u_o ≈ L/f_o, L tube length, eyepiece M_e = 1+D/f_e (image at D) or D/f_e (infinity). Objective forms real inverted magnified image at focal plane of eyepiece, eyepiece acts as simple microscope magnifying it. Focal length: f = -15 cm , u = -30 cm . Mirror equation: (1/v) + (1/-30) = (1/-15) ⇒ (1/v) = (1/-15) + (1/30) = (-2 + 1/30) = (-1/30) . v = -30 cm . Magnification: m = -(v/u) = -(-30/-30) = -1 . Image height: h' = m × h = -1 × 5 = -5 cm

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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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Optical Instruments - Simple and Compound Microscope

An object is placed \( 12 \, \text{cm} \) from a convex mirror of focal length \( 20 \, \text{cm} \). What is the magnif

**Compound microscope** M = m_o × M_e, objective magnification m_o = v_o/u_o ≈ L/f_o, L tube length, eyepiece M_e = 1+D/f_e (image at D) or D/f_e (infinity). Objective forms real inverted magnified image at focal plane of eyepiece, eyepiece acts as simple microscope magnifying it. f = 20 cm , u = -12 cm . (1/v) + (1/-12) = (1/20) ⇒ (1/v) = (1/20) + (1/12) = (3 + 5/60) = (8/60) = (2/15) . v = 7.5 cm . Magnification: m = -(v/u) = -(7.5/-12) = 0.625 . Substituting values gives 0.625, which matches expected image position and magnification from mirror/lens formula 1/f =

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Refraction at Spherical Surfaces and Lens Maker's Formula

What happens to the image formed by a convex mirror if the object is moved closer to the mirror from a distant position?

**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. A convex mirror always forms a virtual, erect, and diminished image. As the object moves closer, the image size increases slightly but remains diminished (less than the object size), and the image distance increases, approaching the focal length as a limit, though it never exceeds it. Substituting values gives Image size increases but remains diminished, which matches expected image position and magnification from mirror/lens

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