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#concave mirror

28 public questions tagged with this topic.

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

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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

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Combination of Lenses and Lens Systems

Why does a concave mirror form a real image when the object is placed beyond the focal point?

**Two lenses in contact** behave as single lens with power sum, magnification product m = m₁×m₂. For f₁=20 cm (P₁=5 D), f₂=-20 cm (P₂=-5 D) in contact, P=0, F infinite, afocal system, beam emerges parallel, principle of corrective lenses for myopia/hypermetropia. When the object is beyond the focal point of a concave mirror, the reflected rays converge to a point on the same side as the object. This convergence of actual rays results in a real image that can be projected onto a screen, typically inverted relative to the object. Substituting values gives Due to rays converging to a point after reflection, which matches expected

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Optical Instruments - Simple and Compound Microscope

An object of height \( 5 \, \text{cm} \) is placed \( 30 \, \text{cm} \) from a concave mirror of focal length \( 15 \,

**Compound microscope** M = m_o × M_e, objective magnification m_o = v_o/u_o ≈ L/f_o, L tube length, eyepiece M_e = 1+D/f_e (image at D) or D/f_e (infinity). Objective forms real inverted magnified image at focal plane of eyepiece, eyepiece acts as simple microscope magnifying it. Focal length: f = -15 cm , u = -30 cm . Mirror equation: (1/v) + (1/-30) = (1/-15) ⇒ (1/v) = (1/-15) + (1/30) = (-2 + 1/30) = (-1/30) . v = -30 cm . Magnification: m = -(v/u) = -(-30/-30) = -1 . Image height: h' = m × h = -1 × 5 = -5 cm

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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 =

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Total Internal Reflection and Critical Angle

In a concave mirror, when the object is placed at the center of curvature, where is the image formed?

**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. For a concave mirror, when the object is at the center of curvature (C), the reflected rays converge back to the same point after reflection. This results in a real, inverted image formed at the center of curvature, with the same size as the object. Substituting values gives At the center of curvature, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v

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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

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Optical Instruments - Simple and Compound Microscope

A concave mirror has a radius of curvature of \( 40 \, \text{cm} \). An object of height \( 2 \, \text{cm} \) is placed

**Compound microscope** M = m_o × M_e, objective magnification m_o = v_o/u_o ≈ L/f_o, L tube length, eyepiece M_e = 1+D/f_e (image at D) or D/f_e (infinity). Objective forms real inverted magnified image at focal plane of eyepiece, eyepiece acts as simple microscope magnifying it. Focal length: f = (R/2) = (-40/2) = -20 cm (negative for concave mirror). Object distance: u = -30 cm . Using mirror equation: (1/v) + (1/u) = (1/f) . (1/v) + (1/-30) = (1/-20) ⇒ (1/v) = (1/-20) + (1/30) = (-3 + 2/60) = (-1/60) . v = -60 cm (real image). Magnification: m = -(v/u) = -(-60/-30)

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Optical Instruments - Simple and Compound Microscope

An object of height \( 3 \, \text{cm} \) is placed \( 20 \, \text{cm} \) from a concave mirror of focal length \( 15 \,

**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. f = -15 cm , u = -20 cm . (1/v) + (1/-20) = (1/-15) ⇒ (1/v) = (1/-15) + (1/20) = (-4 + 3/60) = (-1/60) . v = -60 cm . Magnification: m = -(v/u) = -(-60/-20) = -3 . Image height: h' = m × h = -3 × 3 = -9 cm (inverted). Magnitude = 9 cm . Substituting

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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/lens formula

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Telescope, Human Eye and Defects of Vision

Why does a concave mirror used in a reflecting telescope require precise curvature?

**Astronomical telescope** in normal adjustment M = -f_o/f_e, f_o objective focal length (m), f_e eyepiece focal length, length L = f_o+f_e, final image at infinity, angular magnification ratio of angles subtended by image and object. For f_o=100 cm, f_e=5 cm, M=-20, inverted, used for celestial observation. The precise curvature of a concave mirror ensures that all parallel rays from a distant object converge accurately to a single focal point. Any deviation in curvature causes spherical aberration, blurring the image and reducing the telescope’s resolving power. Substituting values gives To ensure rays converge to a single point, which matches expected image position and magnification from mirror/lens

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

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Total Internal Reflection and Critical Angle

In a concave mirror, under what condition is the image formed virtual and magnified?

**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. For a concave mirror, a virtual and magnified image is formed when the object is placed between the focal point (F) and the pole (P). Here, the reflected rays diverge, and their backward extensions converge behind the mirror, producing a virtual, erect, and magnified image. Substituting values gives Object between focal point and pole, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v -

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