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#mirror optics

9 public questions tagged with this topic.

Combination of Lenses and Lens Systems

A convex mirror of focal length \( 20 \, \text{cm} \) has an object placed \( 40 \, \text{cm} \) from it. What is the im

**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 = 20 cm (convex mirror). Object distance: u = -40 cm . Mirror equation: (1/v) + (1/u) = (1/f) . (1/v) + (1/-40) = (1/20) ⇒ (1/v) = (1/20) + (1/40) = (2 + 1/40) = (3/40) . v = (40/3) ≈ 13.33 cm (virtual image). Substituting values gives 10 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

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

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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

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

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Optical Instruments - Simple and Compound Microscope

An object is placed \( 12 \, \text{cm} \) from a convex mirror of focal length \( 20 \, \text{cm} \). What is the magnif

**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. f = 20 cm , u = -12 cm . (1/v) + (1/-12) = (1/20) ⇒ (1/v) = (1/20) + (1/12) = (3 + 5/60) = (8/60) = (2/15) . v = 7.5 cm . Magnification: m = -(v/u) = -(7.5/-12) = 0.625 . Substituting values gives 0.625, which matches expected image position and magnification from mirror/lens formula 1/f =

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

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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 pos

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

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Refraction at Plane Surfaces and Snell's Law

In a convex mirror, what is the significance of the focal point being behind the mirror?

**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. The focal point of a convex mirror is virtual and behind the mirror because reflected rays diverge and appear to originate from this point when traced backward. This indicates the mirror’s diverging nature, ensuring all images are virtual, erect, and diminished. Substituting values gives It is the point where diverging rays appear to originate, which match

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

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Reflection by Spherical Mirrors and Mirror Formula

An object is placed \( 15 \, \text{cm} \) in front of a concave mirror of focal length \( 10 \, \text{cm} \). Where is t

**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. f = -10 cm , u = -15 cm . (1/v) + (1/-15) = (1/-10) ⇒ (1/v) = (1/-10) + (1/15) = (-3 + 2/30) = (-1/30) . v = -30 cm (real image). Substituting values gives 30 cm, which matches expected image position and magnification from mirror/lens

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 \( 12 \, \text{cm} \) from a concave mirror of focal length \( 6 \, \text{cm} \). 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 = -6 cm , u = -12 cm . Mirror equation: (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 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

In a concave mirror, when an object is placed beyond the center of curvature, what is the nature of the image formed?

**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. For a concave mirror, if the object is placed beyond the center of curvature (C), the image is formed between the focal point (F) and C. The rays converge after reflection, producing a real image that is inverted and diminished in size compared to the object. S

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

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