#depth calculation

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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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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Surface Tension, Capillarity and Excess Pressure

A bubble of radius 3mm is blown at 30cm depth in water (ρ\=1000kg/m3, S\=0.0727N/m). What is the total pressure inside?

Pi = Pa+ρgh+2Sr. Pa = 1.01×105Pa, ρgh = 1000×10×0.3 = 3000Pa. 2Sr = 2×0.07273×10−3 = 48.47Pa. Pi = 1.01×105+3000+48.47 = 1.04048×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.04 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Equation of Continuity and Flow Rate

What is the gauge pressure at 2.5m depth in a tank of mercury (ρ\=13.6×103kg/m3)? (Take g\=10m/s2)

Gauge pressure: Pg = ρgh. ρ = 13.6×103kg/m3, g = 10m/s2, h = 2.5m. Pg = 13.6×103×10×2.5 = 3.4×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.4 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Equation of Continuity and Flow Rate

What is the total pressure at a depth of 1.8m in oxygen liquid (ρ\=1.43kg/m3) with atmospheric pressure 1.01×105Pa? (Tak

Total pressure: P = Pa+ρgh. Pa = 1.01×105Pa, ρ = 1.43kg/m3, g = 9.8m/s2, h = 1.8m. P = 1.01×105+1.43×9.8×1.8 = 1.01×105+25.21 = 1.0102521×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.01 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Bernoulli's Theorem and Applications

What is the total pressure at a depth of 2.2m in hydrogen liquid (ρ\=9.0×10−2kg/m3) with atmospheric pressure 1.01×105Pa

Total pressure: P = Pa+ρgh. Pa = 1.01×105Pa, ρ = 9.0×10−2kg/m3, g = 9.8m/s2, h = 2.2m. P = 1.01×105+9.0×10−2×9.8×2.2 = 1.01×105+1.94 = 1.0100194×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.01 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Equation of Continuity and Flow Rate

What is the gauge pressure at a depth of 0.9m in whole blood (ρ\=1.06×103kg/m3)? (Take g\=10m/s2)

Gauge pressure: Pg = ρgh. ρ = 1.06×103kg/m3, g = 10m/s2, h = 0.9m. Pg = 1.06×103×10×0.9 = 9540Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 9540 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Potential Energy of Mass Systems

At what depth below Earth’s surface is g reduced to 8.33m/s2? (g0\=9.8m/s2,RE\=6.4×106m)

g(d) = g0(1−d/RE). 8.33 = 9.8(1−d/RE). 1−d/RE = 0.85. d/RE = 0.15. d = 0.15×6.4×106 = 9.6×105m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 9.6 × 10⁵ m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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