#minimum deviation

Telescope, Human Eye and Defects of Vision

A prism of refracting angle \( 50^\circ \) has a minimum deviation of \( 25^\circ \). What is the refractive index?

**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. Refractive index: n = (sin ( (A + D_m/2) )/sin ( (A/2) )) . A = 50° , D_m = 25° . n = (sin ( (50 + 25/2) )/sin ( (50/2) )) = (sin 37.5°/sin 25°) . sin 37.5° ≈ 0.609 , sin 25° ≈ 0.423 . n = (0.609/0.423) ≈ 1.44 . Substituting values gives 1.44, which matches expected image position and magnification from mirror/lens

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 prism of angle \( 45^\circ \) and refractive index \( 1.6 \) produces what minimum deviation?

**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. For a thin prism: D_m = (n - 1) A . n = 1.6 , A = 45° . D_m = (1.6 - 1) × 45 = 0.6 × 45 = 27° . Substituting values gives 27°, 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 > Telescope, Human Eye and Defects of Vision

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Combination of Lenses and Lens Systems

A prism of angle \( 40^\circ \) and refractive index \( 1.5 \) produces what minimum deviation?

**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. For a thin prism: D_m = (n - 1) A . n = 1.5 , A = 40° . D_m = (1.5 - 1) × 40 = 0.5 × 40 = 20° . Substituting values gives 20°, 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 > Combination of Lenses and Lens Systems

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Combination of Lenses and Lens Systems

A prism of angle \( 30^\circ \) and refractive index \( 1.6 \) produces what minimum deviation?

**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 thin prism: D_m = (n - 1) A . n = 1.6 , A = 30° . D_m = (1.6 - 1) × 30 = 0.6 × 30 = 18° . Substituting values gives 18°, 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 > Combination of Lenses and Lens Systems

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Combination of Lenses and Lens Systems

A prism with refracting angle \( 60^\circ \) has a minimum deviation of \( 40^\circ \). What is the refractive index of

**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. Refractive index: n = (sin ( (A + D_m/2) )/sin ( (A/2) )) . A = 60° , D_m = 40° . n = (sin ( (60 + 40/2) )/sin ( (60/2) )) = (sin 50°/sin 30°) . sin 50° ≈ 0.766 , sin 30° = 0.5 . n = (0.766/0.5) ≈ 1.53 . Substituting

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

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Combination of Lenses and Lens Systems

A prism of angle \( 60^\circ \) has a minimum deviation of \( 36^\circ \). What is the refractive index?

**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. Refractive index: n = (sin ( (A + D_m/2) )/sin ( (A/2) )) . A = 60° , D_m = 36° . n = (sin ( (60 + 36/2) )/sin ( (60/2) )) = (sin 48°/sin 30°) . sin 48° ≈ 0.743 , sin 30° = 0.5 . n = (0.743/0.5) ≈ 1.486 . Substituting values gives 1.49, 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 > Combination of Lenses and Lens Systems

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Refraction Through Prism and Minimum Deviation

A prism of angle \( 50^\circ \) has a minimum deviation of \( 30^\circ \). What is the refractive index?

**Prism formula** for small A, δ_m≈(n-1)A, for 30°, n=1.6, δ_m≈0.6×30°=18°, approximate, exact using sin formula. For 45°, n=1.6, sin[(45+δ_m)/2]=1.6×sin22.5°=1.6×0.3827=0.6123, (45+δ_m)/2=37.8°, δ_m=30.6°, showing deviation increases with A and n. Refractive index: n = (sin ( (A + D_m/2) )/sin ( (A/2) )) . A = 50° , D_m = 30° . n = (sin ( (50 + 30/2) )/sin ( (50/2) )) = (sin 40°/sin 25°) . sin 40° ≈ 0.643 , sin 25° ≈ 0.423 . n = (0.643/0.423) ≈ 1.52 . Substituting values gives 1.52, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f