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#refractive index

61 public questions tagged with this topic.

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 formula 1/f = 1/v -

Ref: NCERT > Physics Book > Ray Optics > 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

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/lens

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

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

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

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

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

An object is at a depth of \( 19.95 \, \text{cm} \) in a medium with refractive index \( 1.5 \). What is the apparent de

**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. Apparent depth = (real depth/n) . Real depth = 19.95 cm , n = 1.5 . Apparent depth = (19.95/1.5) = 13.3 cm . Substituting values gives 13.3 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.

Ref: NCERT > Physics Book > Ray Optics > Refraction at Spherical Surfaces and Lens Maker's Formula

What is the critical angle for a diamond (\( n = 2.42 \)) to water (\( n = 1.33 \)) interface?

**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. Critical angle: sin i_c = (n₂/n₁) . Diamond ( n₁ = 2.42 ), water ( n₂ = 1.33 ). sin i_c = (1.33/2.42) ≈ 0.55 . i_c = sin⁻¹(0.55) ≈ 33.4° . Substituting values gives 33°, 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 > Refraction at Spherical Surfaces and Lens Maker's Formula

A double convex lens has radii of curvature \( 30 \, \text{cm} \) and \( -30 \, \text{cm} \) with refractive index \( 1.

**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. Lens maker’s formula: (1/f) = (n - 1) ( (1/R₁) - (1/R₂) ) . n = 1.6 , R₁ = 30 cm , R₂ = -30 cm . (1/f) = (1.6 - 1) ( (1/30) - (1/-30) ) = 0.6 ( (1/30) + (1/30) ) = 0.6 × (2/30) = (1.2/30) = (1/25) . f = 25 cm . Substituting values gives 25 cm, which matches expected image position and

Ref: NCERT > Physics Book > Ray Optics > Refraction at Spherical Surfaces and Lens Maker's Formula

What is the critical angle for a crown glass (\( n = 1.52 \)) to air interface?

**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. Critical angle: sin i_c = (n₂/n₁) . Crown glass ( n₁ = 1.52 ), air ( n₂ = 1 ). sin i_c = (1/1.52) ≈ 0.658 . i_c = sin⁻¹(0.658) ≈ 41.1° . Substituting values gives 41°, 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 > Refraction at Spherical Surfaces and Lens Maker's Formula

A double convex lens of refractive index \( 1.5 \) has radii of curvature \( 18 \, \text{cm} \) and \( -18 \, \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 = 1.5 , R₁ = 18 cm , R₂ = -18 cm . (1/f) = (1.5 - 1) ( (1/18) - (1/-18) ) = 0.5 ( (1/18) + (1/18) ) = 0.5 × (2/18) = (1/18) . f = 18 cm . Substituting values gives 18 cm, which matches expected image

Ref: NCERT > Physics Book > Ray Optics > Refraction at Spherical Surfaces and Lens Maker's Formula