#optics calculation

Optical Instruments - Simple and Compound Microscope

An object of height \( 4 \, \text{cm} \) is placed \( 16 \, \text{cm} \) from a concave mirror of focal length \( 8 \, \

**Microscope principle** objective of short focal length forms real image between F_e and 2F_e of eyepiece, eyepiece magnifies to virtual image at D, total magnification product, for f_o=2 cm, f_e=5 cm, L=20 cm, m_o≈10, M_e≈6, M≈60, illustrating high magnification from two stages. Focal length: f = -8 cm , u = -16 cm . Mirror equation: (1/v) + (1/-16) = (1/-8) ⇒ (1/v) = (1/-8) + (1/16) = (-2 + 1/16) = (-1/16) . v = -16 cm . Magnification: m = -(v/u) = -(-16/-16) = -1 . Image height: h' = m × h = -1 × 4 = -4 cm (inverted). Magnitude =

Ref: NCERT > Physics Book > Ray Optics > Optical Instruments - Simple and Compound Microscope

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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.

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

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Thin Lenses - Lens Formula, Magnification and Power

A lens has a power of \( -3 \, \text{D} \). What is its focal length?

**Convex lens image formation**: object beyond 2F (u>2f) real inverted diminished between F and 2F, at 2F same size at 2F, between F and 2F magnified beyond 2F, at F image at infinity, within F virtual erect magnified same side. For f=15 cm, u=30 cm=2f, m = v/u =30/30=1? Actually v=30 cm, m=-1, same size inverted. Power: P = (1/f) (in meters). P = -3 D ⇒ -3 = (1/f) ⇒ f = -(1/3) ≈ -0.333 m ≈ -33.3 cm . Substituting values gives -33 cm, 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 > Thin Lenses - Lens Formula, Magnification and Power

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Thin Lenses - Lens Formula, Magnification and Power

A convex lens (\( f = 50 \, \text{cm} \)) and a concave lens (\( f = 25 \, \text{cm} \)) are in contact. What is the eff

**Lens formula** 1/f = 1/v - 1/u, f focal length (m), u object distance, v image distance, sign: u negative if object left of lens in Cartesian convention, magnification m = v/u, real image inverted m negative, virtual erect m positive. Power P = 1/f (diopters D), f in meters, +5 D means f=0.2 m=20 cm converging. f₁ = 50 cm , f₂ = -25 cm . (1/f) = (1/f₁) + (1/f₂) = (1/50) + (1/-25) = (1/50) - (2/50) = (1 - 2/50) = (-1/50) . f = -50 cm (diverging system). Substituting values gives -50 cm, which matches expected image position and magnification

Ref: NCERT > Physics Book > Ray Optics > Thin Lenses - Lens Formula, Magnification and Power

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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 \( 50^\circ \). 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 = 50° . 1 × sin 50° = 1.33 × sin r . sin 50° ≈ 0.766 ⇒ 0.766 = 1.33 sin r ⇒ sin r = (0.766/1.33) ≈ 0.576 . r = sin⁻¹(0.576) ≈ 35.1° . Substituting values

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

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Thin Lenses - Lens Formula, Magnification and Power

A lens has a power of \( -5 \, \text{D} \). What is its focal length?

**Convex lens image formation**: object beyond 2F (u>2f) real inverted diminished between F and 2F, at 2F same size at 2F, between F and 2F magnified beyond 2F, at F image at infinity, within F virtual erect magnified same side. For f=15 cm, u=30 cm=2f, m = v/u =30/30=1? Actually v=30 cm, m=-1, same size inverted. Power: P = (1/f) (in meters). P = -5 D ⇒ -5 = (1/f) ⇒ f = -(1/5) = -0.2 m = -20 cm . Substituting values gives -20 cm, 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 > Thin Lenses - Lens Formula, Magnification and Power

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Reflection by Spherical Mirrors and Mirror Formula

An object is placed \( 25 \, \text{cm} \) from a concave mirror of radius of curvature \( 40 \, \text{cm} \). What is th

**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 = (R/2) = (-40/2) = -20 cm . Object distance: u = -25 cm . Mirror equation: (1/v) + (1/-25) = (1/-20) ⇒ (1/v) = (1/-20) + (1/25) = (-5 + 4/100) = (-1/100) . v = -100 cm . Substituting values gives 100 cm, which matches expected image position and magnification from

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

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