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#prism

28 public questions tagged with this topic.

Wave Properties, Frequency and Energy Conservation

What is the shape of the wavefront after a plane wave passes through a thin prism?

**Intensity not depend on speed** when enters denser medium because intensity I ∝ n E₀²? Actually Poynting vector S = E×H, energy density u =½ ε E², for same amplitude E₀ intensity proportional to n, but amplitude changes at interface due to reflection, total energy conserved incident = reflected + transmitted, interference does not destroy energy, it redistributes. A plane wave passing through a prism gets tilted due to the varying thickness, resulting in a tilted plane wavefront. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' =

Ref: NCERT > Physics Book > Wave Optics > Wave Properties, Frequency and Energy Conservation

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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 at Spherical Surfaces and Lens Maker's Formula

What optical principle allows a prism to be used as a reflector in optical devices?

**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. Prisms reflect light via total internal reflection when the angle of incidence exceeds the critical angle at the prism’s internal surfaces. This property, dependent on the prism’s refractive index and angle, enables efficient reflection without loss, as seen in devices like binoculars. Substituting values gives Total i

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

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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

Ref: NCERT > Physics Book > Ray Optics > Refraction Through Prism and Minimum Deviation

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Refraction Through Prism and Minimum Deviation

A prism of angle \( 50^\circ \) and refractive index \( 1.5 \) produces what minimum deviation?

**Refraction through prism** deviation δ = i+e-A, minimum when i=e, symmetrical path, r₁=r₂=A/2, n = sin[(A+δ_m)/2]/sin(A/2). In water n_rel = n_prism/n_water =1.5/1.33=1.128, so δ_m reduces because relative index lower, new δ_m from formula with n_rel. For a thin prism: D_m = (n - 1) A . n = 1.5 , A = 50° . D_m = (1.5 - 1) × 50 = 0.5 × 50 = 25° . Substituting values gives 25°, 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 Through Prism and Minimum Deviation

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Refraction Through Prism and Minimum Deviation

In a prism, when light is incident at a very small angle, what is the approximate relationship between deviation and the

**Refraction through prism** deviation δ = i+e-A, minimum when i=e, symmetrical path, r₁=r₂=A/2, n = sin[(A+δ_m)/2]/sin(A/2). In water n_rel = n_prism/n_water =1.5/1.33=1.128, so δ_m reduces because relative index lower, new δ_m from formula with n_rel. For a thin prism with a small angle of incidence, the deviation is approximately equal to the product of the prism’s refractive index minus one and the prism angle (D ≈ (n - 1)A). This simplification holds because the angles of refraction are small, minimizing higher-order effects. Substituting values gives Deviation ≈ (n - 1) × prism angle, wh

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Refraction Through Prism and Minimum Deviation

A prism of angle \( 50^\circ \) and refractive index \( 1.4 \) produces what minimum deviation?

**Refraction through prism** deviation δ = i+e-A, minimum when i=e, symmetrical path, r₁=r₂=A/2, n = sin[(A+δ_m)/2]/sin(A/2). In water n_rel = n_prism/n_water =1.5/1.33=1.128, so δ_m reduces because relative index lower, new δ_m from formula with n_rel. For a thin prism: D_m = (n - 1) A . n = 1.4 , A = 50° . D_m = (1.4 - 1) × 50 = 0.4 × 50 = 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 > Refraction Through Prism and Minimum Deviation

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