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#optics problem

20 public questions tagged with this topic.

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.

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Combination of Lenses and Lens Systems

An object is placed \( 15 \, \text{cm} \) from a convex mirror of focal length \( 30 \, \text{cm} \). What is the magnif

**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 = 30 cm , u = -15 cm . Mirror equation: (1/v) + (1/-15) = (1/30) ⇒ (1/v) = (1/30) + (1/15) = (1 + 2/30) = (3/30) = (1/10) . v = 10 cm . Magnification: m = -(v/u) = -(10/-15) = 0.67 . Substituting values gives 0.67, which matches expected

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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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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 Through Prism and Minimum Deviation

A ray of light passes from water (\( n = 1.33 \)) to glass (\( n = 1.62 \)) at an angle of incidence of \( 45^\circ \).

**Prism minimum deviation** δ_m satisfies n = sin[(A+δ_m)/2]/sin(A/2), A prism angle (degrees), n refractive index, δ_m minimum deviation. For A=60°, n=1.45, sin[(60+δ_m)/2]=1.45×sin30°=0.725, (60+δ_m)/2=46.5°, 60+δ_m=93°, δ_m=33°, illustrating n increase raises δ_m. Snell’s law: n₁ sin i = n₂ sin r . Water ( n₁ = 1.33 ), glass ( n₂ = 1.62 ), i = 45° . 1.33 × sin 45° = 1.62 × sin r . sin 45° = 0.707 ⇒ 1.33 × 0.707 = 1.62 sin r ⇒ 0.941 = 1.62 sin r . sin r = (0.941/1.62) ≈ 0.581 ⇒ r = sin⁻¹(0.581) ≈ 35.5° . Substituting values gives 36°, which matches expected

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Refraction Through Prism and Minimum Deviation

A ray of light passes from air (\( n = 1 \)) to water (\( n = 1.33 \)) at an angle of incidence of \( 60^\circ \). What

**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. Snell’s law: n₁ sin i = n₂ sin r . Air ( n₁ = 1 ), water ( n₂ = 1.33 ), i = 60° . 1 × sin 60° = 1.33 × sin r . sin 60° = 0.866 ⇒ 0.866 = 1.33 sin r ⇒ sin r = (0.866/1.33) ≈ 0.651 . r = sin⁻¹(0.651) ≈ 40.6° . Substituting values gives 41°, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v

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Refraction at Plane Surfaces and Snell's Law

A fish is at a depth of \( 40 \, \text{cm} \) in water (\( n = 1.33 \)). What is its apparent depth when viewed from abo

**Refractive index** n = c/v, water 1.33 means light 1.33 times slower than vacuum. Passing from water to air at 49°, n₁ sinθ₁ =1.33×sin49°≈1.33×0.755=1.004>1, so sinθ₂>1 impossible, total internal reflection occurs, no refraction. Apparent depth = (real depth/n) . Real depth = 40 cm , n = 1.33 . Apparent depth = (40/1.33) ≈ 30.08 cm . Substituting values gives 30 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 Plane Surfaces and Snell's Law

A glass slab (\( n = 1.52 \)) of thickness \( 7.6 \, \text{cm} \) is placed over a dot. What is the apparent shift?

**Snell's law** n₁ sinθ₁ = n₂ sinθ₂ describes refraction at plane interface, n refractive index, θ angle with normal. When light goes from denser n=1.52 glass to rarer air n=1, sinθ₂ = (n₁/n₂) sinθ₁ > sinθ₁, bending away from normal, enabling total internal reflection beyond critical angle. Shift = t ( 1 - (1/n) ) . t = 7.6 cm , n = 1.52 . Shift = 7.6 ( 1 - (1/1.52) ) = 7.6 ( 1 - 0.658 ) = 7.6 × 0.342 ≈ 2.6 cm . Substituting values gives 2.6 cm, which matches expected image position and magnification from mirror/lens formula 1/f =

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Reflection by Spherical Mirrors and Mirror Formula

A concave mirror of focal length \( 10 \, \text{cm} \) forms an image \( 20 \, \text{cm} \) from the mirror. What is the

**Mirror formula** 1/f = 1/v + 1/u governs spherical mirrors, f = R/2, R radius of curvature (m), u object distance (m), v image distance (m), sign convention: distances in front of mirror negative for real is convention but magnitude used, magnification m = -v/u, concave forms real inverted when object beyond F, virtual erect within F. Focal length: f = -10 cm (concave mirror). Image distance: v = -20 cm (real image, same side as object). Mirror equation: (1/v) + (1/u) = (1/f) . (1/-20) + (1/u) = (1/-10) ⇒ (1/u) = (1/-10) + (1/20) = (-2 + 1/20) = (-1/20) . u =

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Refraction at Plane Surfaces and Snell's Law

What is the critical angle for a diamond-air interface if the refractive index of diamond is \( 2.42 \)?

**Refractive index** n = c/v, water 1.33 means light 1.33 times slower than vacuum. Passing from water to air at 49°, n₁ sinθ₁ =1.33×sin49°≈1.33×0.755=1.004>1, so sinθ₂>1 impossible, total internal reflection occurs, no refraction. Critical angle: sin i_c = (n₂/n₁) . Diamond ( n₁ = 2.42 ), air ( n₂ = 1 ). sin i_c = (1/2.42) ≈ 0.413 . i_c = sin⁻¹(0.413) ≈ 24.4° . Substituting values gives 24°, 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 Plane Surfaces and Snell's Law

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