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21 public questions tagged with this topic.

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 internal reflection, 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 \( 50^\circ \) 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

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, which matches expected image position and magnification from mirror/lens formula 1/f =

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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.

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Refraction Through Prism and Minimum Deviation

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

**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. 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.

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Refraction Through Prism and Minimum Deviation

A prism of angle \( 60^\circ \) has a refractive index of \( 1.4 \). What is the angle of 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 = 60° . D_m = (1.4 - 1) × 60 = 0.4 × 60 = 24° . 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.

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Refraction Through Prism and Minimum Deviation

A prism of angle \( 60^\circ \) and refractive index \( 1.5 \) produces a minimum deviation of \( 30^\circ \). What is t

**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. At minimum deviation: i = (A + D_m/2) . A = 60° , D_m = 30° . i = (60 + 30/2) = (90/2) = 45° . Substituting values gives 45°, 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

In a prism, what condition results in the minimum deviation of light?

**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. In a prism, minimum deviation occurs when the angle of incidence equals the angle of emergence. This symmetry ensures the refracted ray inside the prism is parallel to the base, minimizing the deviation angle. Substituting values gives Angle of incidence equals angle of emergence, 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

Why does a prism disperse white light into a spectrum of colors?

**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. A prism disperses white light because different wavelengths (colors) of light have different refractive indices in the prism material. Shorter wavelengths (e.g., violet) refract more than longer wavelengths (e.g., red), causing the light to split into a spectrum as it exits the prism. Substituting values gives Due to different refractive indices for different wavelengths, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens)

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Total Internal Reflection and Critical Angle

What optical property of a prism allows it to be used in binoculars to invert images?

**Critical angle** C satisfies sinC = n₂/n₁, n₁>n₂, n₂=1 for air, n₁=1.52 for glass gives sinC=1/1.52=0.6579, C≈41.1°, for water n=1.33 C=48.75°. Beyond C, total internal reflection occurs, all light reflected, no refracted ray, used in optical fibers and prisms. Prisms in binoculars use total internal reflection to invert and revert images. By reflecting light multiple times within the prism (e.g., in a Porro prism), the image orientation is corrected from the inverted form produced by the objective lens, maintaining the same size. 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

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Refraction Through Prism and Minimum Deviation

A prism of angle \( 30^\circ \) has a minimum deviation of \( 20^\circ \). What is the refractive index?

**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. Refractive index: n = (sin ( (A + D_m/2) )/sin ( (A/2) )) . A = 30° , D_m = 20° . n = (sin ( (30 + 20/2) )/sin ( (30/2) )) = (sin 25°/sin 15°) . sin 25° ≈ 0.423 , sin 15° ≈ 0.259 . n = (0.423/0.259) ≈ 1.63 . Substituting values gives 1.63, which matches expected image position and magnification from mirror/lens formula 1/f =

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