#apex angle

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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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, which matches expected image position and magnification from mirror/lens formula 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 \( 1.4 \) 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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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.

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

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