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#mirror equation

13 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 concave mirror of focal length \( 7 \, \text{cm} \) has an object placed \( 14 \, \text{cm} \) from it. What is the im

**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 = -7 cm (concave mirror). Object distance: u = -14 cm . Mirror equation: (1/v) + (1/u) = (1/f) . (1/v) + (1/-14) = (1/-7) ⇒ (1/v) = (1/-7) + (1/14) = (-2 + 1/14) = (-1/14) . v = -14 cm (real image). Substituting values gives 14 cm, which matches expected

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

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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 convex mirror of focal length \( 16 \, \text{cm} \) forms an image \( 8 \, \text{cm} \) behind the mirror. What is 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. Focal length: f = 16 cm (convex mirror). Image distance: v = 8 cm (virtual image). Mirror equation: (1/v) + (1/u) = (1/f) . (1/8) + (1/u) = (1/16) ⇒ (1/u) = (1/16) - (1/8) = (1 - 2/16) = (-1/16) . u = -16 cm . Substituting values gives 12 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens)

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

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Total Internal Reflection and Critical Angle

An object is placed \( 16 \, \text{cm} \) from a convex mirror of focal length \( 24 \, \text{cm} \). What is the image

**TIR application** requires n₁>n₂ and θ₁>C. Critical angle formula derived from Snell's law with θ₂=90°, n₁ sinC = n₂ sin90° = n₂, so sinC = n₂/n₁. For glass-air sinC=1/1.52, C≈41°, for water-air 48.6°, determining cutoff for transmission. Focal length: f = 24 cm , u = -16 cm . Mirror equation: (1/v) + (1/-16) = (1/24) ⇒ (1/v) = (1/24) + (1/16) = (2 + 3/48) = (5/48) . v = (48/5) = 9.6 cm (virtual image). Substituting values gives 9.6 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),

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

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Optical Instruments - Simple and Compound Microscope

An object of height \( 3 \, \text{cm} \) is placed \( 15 \, \text{cm} \) from a concave mirror of focal length \( 10 \,

**Simple microscope** magnification M = 1 + D/f when image at D=25 cm near point, M = D/f when image at infinity (relaxed eye), f focal length (cm), D least distance of distinct vision 25 cm. For f=4 cm, image at 25 cm, M=1+25/4=7.25, angular magnification ratio of angle subtended by image to that by object at D. Focal length: f = -10 cm , u = -15 cm . Mirror equation: (1/v) + (1/-15) = (1/-10) ⇒ (1/v) = (1/-10) + (1/15) = (-3 + 2/30) = (-1/30) . v = -30 cm . Magnification: m = -(v/u) = -(-30/-15) = -2 . Image

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

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Refraction at Plane Surfaces and Snell's Law

A concave mirror of focal length \( 6 \, \text{cm} \) has an object placed \( 12 \, \text{cm} \) from it. What is the im

**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. Focal length: f = -6 cm (concave mirror). Object distance: u = -12 cm . Mirror equation: (1/v) + (1/u) = (1/f) . (1/v) + (1/-12) = (1/-6) ⇒ (1/v) = (1/-6) + (1/12) = (-2 + 1/12) = (-1/12) . v = -12 cm (real image). Substituting values gives 12 cm, which matches expected image position and magnification from mirror/lens formula

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

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Refraction at Plane Surfaces and Snell's Law

A convex mirror of focal length \( 18 \, \text{cm} \) produces an image \( 6 \, \text{cm} \) behind the mirror. What is

**Refractive index** n = c/v, water 1.33 means light 1.33 times slower than vacuum. Passing from water to air at 49°, n₁ sinθ₁ =1.33×sin49°≈1.33×0.755=1.004>1, so sinθ₂>1 impossible, total internal reflection occurs, no refraction. Focal length: f = 18 cm (convex mirror). Image distance: v = 6 cm (virtual image). Mirror equation: (1/v) + (1/u) = (1/f) . (1/6) + (1/u) = (1/18) ⇒ (1/u) = (1/18) - (1/6) = (1 - 3/18) = (-2/18) = (-1/9) . u = -9 cm . Substituting values gives 12 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f

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Reflection by Spherical Mirrors and Mirror Formula

A concave mirror has a focal length of \( 12 \, \text{cm} \). An object is placed \( 18 \, \text{cm} \) from it. What is

**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 = -12 cm (concave mirror). Object distance: u = -18 cm . Mirror equation: (1/v) + (1/u) = (1/f) . (1/v) + (1/-18) = (1/-12) ⇒ (1/v) = (1/-12) + (1/18) = (-3 + 2/36) = (-1/36) . v = -36 cm (real image). Substituting values gives 36 cm, which matches expected image

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 \( 20 \, \text{cm} \) from a concave mirror of focal length \( 8 \, \text{cm} \). What is the image

**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 = -8 cm , u = -20 cm . Mirror equation: (1/v) + (1/-20) = (1/-8) ⇒ (1/v) = (1/-8) + (1/20) = (-5 + 2/40) = (-3/40) . v = -(40/3) ≈ -13.33 cm (real image). Substituting values gives 13.3 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v -

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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