#optics question

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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