Refraction Through Prism and Minimum Deviation

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

When light travels from a denser medium to a rarer medium and the angle of incidence exceeds a certain value, what pheno

**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. When light travels from a denser to a rarer medium (e.g., glass to air) and the angle of incidence exceeds the critical angle, total internal reflection occurs. This is because the refracted ray would otherwise require a sine value greater than 1, which is physically impossible. Substituting values gives Total internal reflection, 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 Through Prism and Minimum Deviation

A convex mirror of focal length \( 16 \, \text{cm} \) forms an image \( 8 \, \text{cm} \) behind the mirror. What is 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. Focal length: f = 16 cm (convex mirror). Image distance: v = 8 cm (virtual image). Mirror equation: (1/v) + (1/u) = (1/f) . (1/8) + (1/u) = (1/16) ⇒ (1/u) = (1/16) - (1/8) = (1 - 2/16) = (-1/16) . u = -16 cm . Substituting values gives 12 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 Through Prism and Minimum Deviation

A glass slab (\( n = 1.6 \)) of thickness \( 8 \, \text{cm} \) is placed over a point. What is the apparent shift?

**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. Shift = t ( 1 - (1/n) ) . t = 8 cm , n = 1.6 . Shift = 8 ( 1 - (1/1.6) ) = 8 ( 1 - 0.625 ) = 8 × 0.375 = 3 cm . Substituting values gives 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.

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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 Through Prism and Minimum Deviation

An object is placed \( 10 \, \text{cm} \) from a convex mirror of radius of curvature \( 30 \, \text{cm} \). What is the

**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. f = (R/2) = (30/2) = 15 cm . u = -10 cm . (1/v) + (1/-10) = (1/15) ⇒ (1/v) = (1/15) + (1/10) = (2 + 3/30) = (5/30) = (1/6) . v = 6 cm (virtual image). Substituting values gives 6 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 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 =

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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.

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Refraction Through Prism and Minimum Deviation

A convex lens of focal length \( 18 \, \text{cm} \) has an object placed \( 36 \, \text{cm} \) from it. What is the imag

**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. Focal length: f = 18 cm . Object distance: u = -36 cm . Lens formula: (1/v) - (1/u) = (1/f) . (1/v) - (1/-36) = (1/18) ⇒ (1/v) + (1/36) = (1/18) ⇒ (1/v) = (1/18) - (1/36) = (2 - 1/36) = (1/36) . v = 36 cm (real image). Substituting values gives 36 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 Through Prism and Minimum Deviation

A ray of light passes from glass (\( n = 1.5 \)) to water (\( n = 1.33 \)) at an angle of incidence of \( 30^\circ \). W

**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 . Glass ( n₁ = 1.5 ), water ( n₂ = 1.33 ), i = 30° . 1.5 × sin 30° = 1.33 × sin r . sin 30° = 0.5 ⇒ 1.5 × 0.5 = 1.33 sin r ⇒ 0.75 = 1.33 sin r . sin r = (0.75/1.33) ≈ 0.564 ⇒ r = sin⁻¹(0.564) ≈ 34.3° . Substituting values gives 34°, which matches expected image position

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Refraction Through Prism and Minimum Deviation

In a convex lens, what happens to the image if the object is placed between the focal point and twice the focal length?

**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 convex lens, when the object is between the focal point (F) and twice the focal length (2F), the image is real, inverted, and magnified. It forms beyond 2F on the opposite side, as the rays converge after refraction. Substituting values gives Real, inverted, and magnified, 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),

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Refraction Through Prism and Minimum Deviation

A ray of light passes from air (\( n = 1 \)) to glass (\( n = 1.62 \)) at an angle of incidence of \( 30^\circ \). 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 ), glass ( n₂ = 1.62 ), i = 30° . 1 × sin 30° = 1.62 × sin r . sin 30° = 0.5 ⇒ 0.5 = 1.62 sin r ⇒ sin r = (0.5/1.62) ≈ 0.309 . r = sin⁻¹(0.309) ≈ 18° . Substituting values gives 18°, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v

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