Total Internal Reflection and Critical Angle

Total Internal Reflection and Critical Angle

In a concave lens, what property of the lens determines the position of the virtual focal point?

**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. The virtual focal point of a concave lens is where diverging rays appear to originate when traced backward. This position is determined by the lens’s focal length, which depends on its curvature and refractive index, defining the extent of divergence. Substituting values gives Focal length, 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),

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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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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),

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Total Internal Reflection and Critical Angle

A converging beam meets a convex lens (\( f = 10 \, \text{cm} \)) \( 5 \, \text{cm} \) before the convergence point. Wha

**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. Object distance: u = -5 cm (virtual object), f = 10 cm . Lens formula: (1/v) - (1/-5) = (1/10) ⇒ (1/v) + (1/5) = (1/10) . (1/v) = (1/10) - (1/5) = (1 - 2/10) = (-1/10) . v = -10 cm (10 cm to the left). Substituting values gives 10 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v

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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.

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Total Internal Reflection and Critical Angle

A convex lens of focal length \( 10 \, \text{cm} \) forms an image at \( 20 \, \text{cm} \) from the lens. What is the o

**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 = 10 cm . Image distance: v = 20 cm (real image). Lens formula: (1/v) - (1/u) = (1/f) . (1/20) - (1/u) = (1/10) ⇒ (1/u) = (1/20) - (1/10) = (1 - 2/20) = (-1/20) . u = -20 cm . Substituting values gives 20 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u

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Total Internal Reflection and Critical Angle

A concave lens of focal length \( 20 \, \text{cm} \) forms an image \( 8 \, \text{cm} \) from the lens. What is the obje

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

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Total Internal Reflection and Critical Angle

What is the critical angle for a dense flint glass (\( n = 1.62 \)) to air interface?

**Total internal reflection** occurs when light travels from denser to rarer medium and incidence > C, condition sinC = 1/n for air interface. For glass n=1.52 C≈41°, water n=1.33 C≈48.8°, so at 49° water-air TIR occurs, explaining why ray does not emerge. Critical angle: sin i_c = (n₂/n₁) . Glass ( n₁ = 1.62 ), air ( n₂ = 1 ). sin i_c = (1/1.62) ≈ 0.617 . i_c = sin⁻¹(0.617) ≈ 38.1° . Substituting values gives 38°, 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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Total Internal Reflection and Critical Angle

A light ray passes from water (\( n = 1.33 \)) to glass (\( n = 1.5 \)) at an angle of incidence of \( 30^\circ \). 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. Snell’s law: n₁ sin i = n₂ sin r . 1.33 sin 30° = 1.5 sin r . sin 30° = 0.5 ⇒ 1.33 × 0.5 = 1.5 sin r ⇒ 0.665 = 1.5 sin r . sin r = (0.665/1.5) ≈ 0.443 ⇒ r = sin⁻¹(0.443) ≈ 26.3° . Substituting values gives 26°, which matches expected image position and magnification from mirror/lens formula 1/f =

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Total Internal Reflection and Critical Angle

An object is placed \( 12 \, \text{cm} \) from a convex mirror of focal length \( 18 \, \text{cm} \). What is the image

**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 = 18 cm , u = -12 cm . Mirror equation: (1/v) + (1/-12) = (1/18) ⇒ (1/v) = (1/18) + (1/12) = (2 + 3/36) = (5/36) . v = (36/5) = 7.2 cm (virtual image). Substituting values gives 7.2 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v +

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Total Internal Reflection and Critical Angle

A ray of light passes from water (\( n = 1.33 \)) to air at an angle of incidence of \( 50^\circ \). What happens?

**Total internal reflection** occurs when light travels from denser to rarer medium and incidence > C, condition sinC = 1/n for air interface. For glass n=1.52 C≈41°, water n=1.33 C≈48.8°, so at 49° water-air TIR occurs, explaining why ray does not emerge. Critical angle: sin i_c = (n₂/n₁) = (1/1.33) ≈ 0.752 ⇒ i_c ≈ 48.75° . Since i = 50° > i_c , total internal reflection occurs. Substituting values gives Total internal reflection, 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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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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