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#radius of curvature

12 public questions tagged with this topic.

Telescope, Human Eye and Defects of Vision

An object is placed \( 8 \, \text{cm} \) from a convex mirror of radius of curvature \( 24 \, \text{cm} \). What is the

**Human eye** least distance D=25 cm, near point, far point infinity for normal eye, accommodation by ciliary muscles changing lens curvature, power ≈60 D total. Defects: myopia far point 25 cm corrected by converging lens, astigmatism cylindrical lens. Focal length: f = (R/2) = (24/2) = 12 cm . Object distance: u = -8 cm . Mirror equation: (1/v) + (1/-8) = (1/12) ⇒ (1/v) = (1/12) + (1/8) = (2 + 3/24) = (5/24) . v = (24/5) = 4.8 cm (virtual image). Substituting values gives 4.8 cm, which matches expected image position and magnification

Ref: NCERT > Physics Book > Ray Optics > Telescope, Human Eye and Defects of Vision

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Combination of Lenses and Lens Systems

A concave mirror of radius of curvature \( 30 \, \text{cm} \) has an object placed \( 45 \, \text{cm} \) from it. What i

**Combination of lenses in contact** effective power P = P₁+P₂, effective focal length 1/F = 1/f₁ + 1/f₂, for thin lenses in contact. If separated by d, 1/F =1/f₁+1/f₂ - d/(f₁ f₂). Power adds algebraically, converging + with diverging - can cancel, used to correct aberrations and design achromatic doublets. Focal length: f = (R/2) = (-30/2) = -15 cm (concave mirror). Object distance: u = -45 cm . Mirror equation: (1/v) + (1/u) = (1/f) . (1/v) + (1/-45) = (1/-15) ⇒ (1/v) = (1/-15) + (1/45) = (-3 + 1/45) = (-2/45) . v = -(45/2) = -22.5 cm (real image). Substituting values

Ref: NCERT > Physics Book > Ray Optics > Combination of Lenses and Lens Systems

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Combination of Lenses and Lens Systems

A convex mirror has a radius of curvature of \( 50 \, \text{cm} \). An object is placed \( 25 \, \text{cm} \) from it. W

**Lens combination** effective focal length calculation uses reciprocal sum, for f₁=10 cm, f₂=20 cm in contact, 1/F=1/10+1/20=3/20, F=6.67 cm, P=15 D, stronger converging than either alone, illustrating power addition. Focal length: f = (R/2) = (50/2) = 25 cm (positive for convex). Object distance: u = -25 cm . Mirror equation: (1/v) + (1/u) = (1/f) . (1/v) + (1/-25) = (1/25) ⇒ (1/v) = (1/25) + (1/25) = (2/25) . v = (25/2) = 12.5 cm (virtual image). Substituting values gives 12.5 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 > Combination of Lenses and Lens Systems

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Refraction at Spherical Surfaces and Lens Maker's Formula

A double convex lens of refractive index \( 1.55 \) has both radii of curvature equal to \( 25 \, \text{cm} \). What is

**Spherical refracting surface** power P = (n₂-n₁)/R, lens power sum of two surfaces. Lens maker derivation combines two refractions, sign of R₂ negative for second surface convex opposite direction, yielding 1/f positive for convex lens. Lens maker’s formula: (1/f) = (n - 1) ( (1/R₁) - (1/R₂) ) . n = 1.55 , R₁ = 25 cm , R₂ = -25 cm . (1/f) = (1.55 - 1) ( (1/25) - (1/-25) ) = 0.55 ( (1/25) + (1/25) ) = 0.55 × (2/25) = (1.1/25) . f = (25/1.1) ≈ 22.73 cm . Substituting values gives 22.7 cm, which matches expected image position and

Ref: NCERT > Physics Book > Ray Optics > Refraction at Spherical Surfaces and Lens Maker's Formula

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Refraction Through Prism and Minimum Deviation

An object is placed \( 10 \, \text{cm} \) from a convex mirror of radius of curvature \( 30 \, \text{cm} \). What is the

**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. f = (R/2) = (30/2) = 15 cm . u = -10 cm . (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 = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror), confirming sign

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

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Total Internal Reflection and Critical Angle

A concave mirror of radius of curvature \( 20 \, \text{cm} \) forms an image \( 20 \, \text{cm} \) from the mirror. What

**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. Focal length: f = (R/2) = (-20/2) = -10 cm (concave mirror). Image distance: v = -20 cm (real image). Mirror equation: (1/v) + (1/u) = (1/f) . (1/-20) + (1/u) = (1/-10) ⇒ (1/u) = (1/-10) + (1/20) = (-2 + 1/20) = (-1/20) . u = -20 cm . Substituting values gives 20 cm, which matches expected image position and magnification from mirror/

Ref: NCERT > Physics Book > Ray Optics > Total Internal Reflection and Critical Angle

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Reflection by Spherical Mirrors and Mirror Formula

A convex mirror of radius of curvature \( 36 \, \text{cm} \) has an object placed \( 27 \, \text{cm} \) from it. What is

**Concave mirror image formation** depends on object position: beyond C real inverted diminished between F and C, at C real inverted same size at C, between C and F real inverted magnified beyond C, at F image at infinity, within F virtual erect magnified behind mirror. Focal length: f = (R/2) = (36/2) = 18 cm (convex mirror). Object distance: u = -27 cm . Mirror equation: (1/v) + (1/u) = (1/f) . (1/v) + (1/-27) = (1/18) ⇒ (1/v) = (1/18) + (1/27) = (3 + 2/54) = (5/54) . v = (54/5) = 10.8 cm (virtual image). Substituting values gives 9 cm, which

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 \( 25 \, \text{cm} \) 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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Reflection by Spherical Mirrors and Mirror Formula

A convex mirror of radius of curvature \( 40 \, \text{cm} \) forms an image \( 10 \, \text{cm} \) behind the mirror. Wha

**Mirror formula** 1/f = 1/v + 1/u governs spherical mirrors, f = R/2, R radius of curvature (m), u object distance (m), v image distance (m), sign convention: distances in front of mirror negative for real is convention but magnitude used, magnification m = -v/u, concave forms real inverted when object beyond F, virtual erect within F. Focal length: f = (R/2) = (40/2) = 20 cm (convex mirror). Image distance: v = 10 cm (virtual image). Mirror equation: (1/v) + (1/u) = (1/f) . (1/10) + (1/u) = (1/20) ⇒ (1/u) = (1/20) - (1/10) = (1 - 2/20) = (-1/20) . u =

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

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Torque on Current Loop, Magnetic Moment and Galvanometer

An electron moves at \( 6.5 \times 10^6 \, \text{m/s} \) perpendicular to a field of \( 0.2 \, \text{T} \). What is the

**Torque on current loop** in magnetic field B is τ = N I A × B, magnitude τ = N I A B sinθ, N turns, I current (A), A area (m²) = l×b for rectangular, θ angle between normal to plane and B. Maximum when plane parallel to B (θ=90°), zero when perpendicular (θ=0°), magnetic moment m = N I A direction along normal via right-hand rule. r = (mv/qB) . r = (9.1 × 10⁻³¹ × 6.5 × 10⁶/1.6 × 10⁻¹⁹ × 0.2) = (5.915 × 10⁻²⁴/3.2 × 10⁻²⁰) = 1.848 × 10⁻⁴ m ≈ 0.0185 cm . Using F = q v

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Torque on Current Loop, Magnetic Moment and Galvanometer

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A proton moves at \( 3.5 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.08 \, \text{T} \). What is the ra

**Motion of charged particle perpendicular to uniform B** is circular because force F ⊥ v, no work done, speed constant, radius r = m v/(q B), period T = 2π m/(q B) independent of v. If v has component parallel to B, helical path results, pitch = v_parallel·T. r = (mv/qB) . r = (1.67 × 10⁻²⁷ × 3.5 × 10⁷/1.6 × 10⁻¹⁹ × 0.08) = (5.845 × 10⁻²⁰/1.28 × 10⁻²⁰) = 4.5664 ≈ 4.57 m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N I

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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Magnetic Force on Moving Charge - Lorentz Force and Motion

An electron moves with a speed of \( 3.5 \times 10^6 \, \text{m/s} \) perpendicular to a magnetic field of \( 0.25 \, \t

**Lorentz force** on charge q moving with velocity v in magnetic field B is F = q v × B, magnitude F = q v B sinθ, θ angle between v and B (degrees), unit N. Direction perpendicular to both v and B via right-hand rule. When v ⊥ B, motion circular with radius r = m v/(q B), centripetal force provided by magnetic force. Radius r = (mv/qB) . r = (9.1 × 10⁻³¹ × 3.5 × 10⁶/1.6 × 10⁻¹⁹ × 0.25) = (3.185 × 10⁻²⁴/4 × 10⁻²⁰) = 7.9625 × 10⁻⁵ m = 7.96 × 10⁻³ cm . Using F = q v

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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