#apparent depth

Refraction at Spherical Surfaces and Lens Maker's Formula

An object is at a depth of \( 19.95 \, \text{cm} \) in a medium with refractive index \( 1.5 \). What is the apparent de

**Refraction at spherical surface** formula n₁/u + n₂/v = (n₂-n₁)/R governs single surface, extension to two surfaces yields lens maker. Double convex with equal |R| has f = R/[2(n-1)], for R=12 cm, n=1.5, f=12 cm, illustrating dependence on curvature and index. Apparent depth = (real depth/n) . Real depth = 19.95 cm , n = 1.5 . Apparent depth = (19.95/1.5) = 13.3 cm . Substituting values gives 13.3 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.

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Refraction at Spherical Surfaces and Lens Maker's Formula

Why does the apparent depth of an object in a denser medium appear less than its real depth when viewed from air?

**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. When light travels from a denser medium (e.g., water) to a rarer medium (air), it bends away from the normal. This refraction makes the rays appear to diverge from a point closer to the surface than the actual object, reducing the apparent depth compared to the real depth. Substituting values gives Due to refraction bending rays away from the normal, which ma

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Total Internal Reflection and Critical Angle

An object is at a depth of \( 26.6 \, \text{cm} \) in a medium with refractive index \( 1.33 \). What is the apparent de

**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. Apparent depth = (real depth/n) . Real depth = 26.6 cm , n = 1.33 . Apparent depth = (26.6/1.33) = 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 = 1/v + 1/u (mirror), confirming sign conventions and refraction principles.

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

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Total Internal Reflection and Critical Angle

An object is at a depth of \( 13.3 \, \text{cm} \) in a medium with refractive index \( 1.33 \). What is the apparent de

**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. Apparent depth = (real depth/n) . Real depth = 13.3 cm , n = 1.33 . Apparent depth = (13.3/1.33) = 10 cm . 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 (mirror), confirming sign conventions and refraction principles.

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

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Refraction Through Prism and Minimum Deviation

An object at a depth of \( 19.95 \, \text{cm} \) in water (\( n = 1.33 \)) is viewed normally. What is the apparent dept

**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. Apparent depth = (real depth/n) . Real depth = 19.95 cm , n = 1.33 . Apparent depth = (19.95/1.33) ≈ 15 cm . Substituting values gives 15 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.

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Refraction at Plane Surfaces and Snell's Law

An object is at a depth of \( 16 \, \text{cm} \) in water (\( n = 1.33 \)). What is the apparent depth?

**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. Apparent depth = (real depth/n) . Real depth = 16 cm , n = 1.33 . Apparent depth = (16/1.33) ≈ 12.03 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 = 1/v + 1/u (mirror), confirming sign conventions and refraction principles.

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 fish is at a depth of \( 40 \, \text{cm} \) in water (\( n = 1.33 \)). What is its apparent depth when viewed from abo

**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. Apparent depth = (real depth/n) . Real depth = 40 cm , n = 1.33 . Apparent depth = (40/1.33) ≈ 30.08 cm . Substituting values gives 30 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.

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Refraction at Plane Surfaces and Snell's Law

An object is at a depth of \( 24 \, \text{cm} \) in a medium with refractive index \( 1.6 \). What is the apparent depth

**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. Apparent depth = (real depth/n) . Real depth = 24 cm , n = 1.6 . Apparent depth = (24/1.6) = 15 cm . Substituting values gives 15 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 at Plane Surfaces and Snell's Law

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Refraction at Plane Surfaces and Snell's Law

A coin lies at the bottom of a water tank (\( n = 1.33 \)) with a depth of \( 13.3 \, \text{cm} \). What is its apparent

**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. Apparent depth = (real depth/n) . Real depth = 13.3 cm , n = 1.33 . Apparent depth = (13.3/1.33) = 10 cm . 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 (mirror), confirming sign conventions and refraction principles.

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

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Refraction Through Prism and Minimum Deviation

An object is at a depth of \( 15 \, \text{cm} \) in water (\( n = 1.5 \)). What is the apparent depth?

**Prism formula** for small A, δ_m≈(n-1)A, for 30°, n=1.6, δ_m≈0.6×30°=18°, approximate, exact using sin formula. For 45°, n=1.6, sin[(45+δ_m)/2]=1.6×sin22.5°=1.6×0.3827=0.6123, (45+δ_m)/2=37.8°, δ_m=30.6°, showing deviation increases with A and n. Apparent depth = (real depth/n) . Real depth = 15 cm , n = 1.5 . Apparent depth = (15/1.5) = 10 cm . 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 (mirror), confirming sign conventions and refraction principles.

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

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PHYSICS

An object is at a depth of 19.95 cm in a medium with refractive index 1.5 . What is the apparent depth?

Given: An object is at a depth of 19.95 cm in a medium with refractive index 1.5 . What is the apparent depth? Formula: Apparent depth = fracreal depthn. Substitution & Calculation: Real depth = 19.95 cm, n = 1.5 . Apparent depth = 19.95/1.5 = 13.3 cm . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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