#optics question

Wavefront and Huygens Principle

Why is the backwave not observed in the reflection of light according to the wave model?

**Wavefront** is locus of points in same phase, spherical from point source, plane at large distance because radius large, Huygens principle every point on wavefront acts as secondary source of wavelets, new wavefront envelope of secondary wavelets, allows prediction of new wavefront shape from known wavefront, explains reflection and refraction. The wave model assumes the amplitude of secondary wavelets is zero in the backward direction, though this is an ad-hoc assumption later justified by advanced wave theory. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n,

Ref: NCERT > Physics Book > Wave Optics > Wavefront and Huygens Principle

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Reflection by Spherical Mirrors and Mirror Formula

A glass slab (\( n = 1.5 \)) of thickness \( 7.5 \, \text{cm} \) is placed over a point. What is the apparent shift?

**Spherical mirror reflection** follows law θ_i = θ_r, focal length f = R/2, concave R negative, convex positive. Object beyond center C (u > 2f) forms image between F and C diminished, at C same size, between C and F magnified, beyond F forms at infinity, illustrating mirror equation. Shift = t ( 1 - (1/n) ) . t = 7.5 cm , n = 1.5 . Shift = 7.5 ( 1 - (1/1.5) ) = 7.5 ( 1 - (2/3) ) = 7.5 × (1/3) = 2.5 cm . Substituting values gives 2.5 cm, which matches expected image position and magnification from mirror/lens formula

Ref: NCERT > Physics Book > Ray Optics > Reflection by Spherical Mirrors and Mirror Formula

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Reflection by Spherical Mirrors and Mirror Formula

An object is placed \( 10 \, \text{cm} \) from a convex mirror of focal length \( 15 \, \text{cm} \). What is the image

**Spherical mirror reflection** follows law θ_i = θ_r, focal length f = R/2, concave R negative, convex positive. Object beyond center C (u > 2f) forms image between F and C diminished, at C same size, between C and F magnified, beyond F forms at infinity, illustrating mirror equation. Focal length: f = 15 cm , u = -10 cm . Mirror equation: (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

Ref: NCERT > Physics Book > Ray Optics > Reflection by Spherical Mirrors and Mirror Formula

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Refraction at Plane Surfaces and Snell's Law

A ray of light passes from air (\( n = 1 \)) to water (\( n = 1.33 \)) at an angle of incidence of \( 35^\circ \). What

**Refraction at plane surface** changes direction due to speed change v = c/n, n = c/v. For water n=1.33 to air, n₁>n₂, ray bends away, if θ₁ > C, sinθ₂>1 impossible, TIR occurs. Snell's law quantitative prediction of θ₂ from θ₁ and indices. Snell’s law: n₁ sin i = n₂ sin r . Air ( n₁ = 1 ), water ( n₂ = 1.33 ), i = 35° . 1 × sin 35° = 1.33 × sin r . sin 35° ≈ 0.574 ⇒ 0.574 = 1.33 sin r ⇒ sin r = (0.574/1.33) ≈ 0.432 . r = sin⁻¹(0.432) ≈ 25.6° . Substituting values

Ref: NCERT > Physics Book > Ray Optics > Refraction at Plane Surfaces and Snell's Law

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Refraction Through Prism and Minimum Deviation

A prism of refracting angle \( 30^\circ \) and refractive index \( 1.6 \) 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 = 1.6 , 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 > Refraction Through Prism and Minimum Deviation

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