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#total internal reflection

22 public questions tagged with this topic.

What is the angle of refraction when light is incident at the critical angle from a denser medium to a rarer medium?

**Convex lens focusing** plane wave into point because lens introduces phase delay proportional to thickness, converting plane wavefront to spherical converging to focal point, property ensures rays parallel to axis meet at focus, spherical aberration minimized for paraxial rays, lensmaker's formula determines focal length. At the critical angle, the angle of refraction is 90° , as the refracted ray travels along the boundary between the two media. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation gives 90°, illustrating i

Ref: NCERT > Physics Book > Wave Optics > Optical Phenomena and Applications

Why does the angle of refraction increase beyond 90° become impossible when light moves from a denser to a rarer medium?

**Critical angle** sinC = n₂/n₁, n₁ denser, n₂ rarer 1 for air, for C=36.9° n₁=1/sin36.9°=1/0.6=1.67, for C=39° n=1/sin39°=1/0.629=1.59, for n=1.85 C=arcsin(1/1.85)=arcsin0.5405=32.7°, total internal reflection occurs only when light travels from denser to rarer and incidence > C, because Snell's law would require sinθ₂>1 impossible. Beyond the critical angle, the sine of the refraction angle exceeds 1, which is mathematically impossible, leading to total internal reflection. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculati

Ref: NCERT > Physics Book > Wave Optics > Refraction, Refractive Index and Critical Angle

What explains the absence of a refracted ray when the angle of incidence exceeds the critical angle?

**Critical angle** sinC = n₂/n₁, n₁ denser, n₂ rarer 1 for air, for C=36.9° n₁=1/sin36.9°=1/0.6=1.67, for C=39° n=1/sin39°=1/0.629=1.59, for n=1.85 C=arcsin(1/1.85)=arcsin0.5405=32.7°, total internal reflection occurs only when light travels from denser to rarer and incidence > C, because Snell's law would require sinθ₂>1 impossible. Beyond the critical angle, the refracted ray would require a sine greater than 1, which is impossible, leading to total internal reflection. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation g

Ref: NCERT > Physics Book > Wave Optics > Refraction, Refractive Index and Critical Angle

What is the critical angle for light passing from a medium with refractive index 1.9 to air (refractive index 1.0)?

**Critical angle** sinC = n₂/n₁, n₁ denser, n₂ rarer 1 for air, for C=36.9° n₁=1/sin36.9°=1/0.6=1.67, for C=39° n=1/sin39°=1/0.629=1.59, for n=1.85 C=arcsin(1/1.85)=arcsin0.5405=32.7°, total internal reflection occurs only when light travels from denser to rarer and incidence > C, because Snell's law would require sinθ₂>1 impossible. sin i_c = (n₂/n₁) , where n₁ = 1.9 , n₂ = 1.0 . sin i_c = (1.0/1.9) ≈ 0.526 , i_c = sin⁻¹(0.526) ≈ 31.8° . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation

Ref: NCERT > Physics Book > Wave Optics > Refraction, Refractive Index and Critical Angle

What is the refractive index of a medium if the critical angle for light passing into air is \( 42^\circ \)?

**Refractive index** n = c/v, c=3×10⁸ m/s vacuum, v speed in medium, for v=2.25×10⁸ m/s n=3/2.25=1.33, for v=1.8×10⁸ n=1.67, for n=1.3 v=3×10⁸/1.3=2.31×10⁸ m/s, for n=1.6 v=1.875×10⁸ m/s, for n=1.2 λ_medium = λ_vacuum/n =600/1.2=500 nm, for n=1.75 λ=700/1.75=400 nm, for n=1.5 λ=750/1.5=500 nm, wavelength in medium λ' = λ/n. sin i_c = (n₂/n₁) , where n₂ = 1.0 (air), i_c = 42° . sin 42° ≈ 0.669 , n₁ = (1.0/0.669) ≈ 1.49 . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a

Ref: NCERT > Physics Book > Wave Optics > Refraction, Refractive Index and Critical Angle

What is the refractive index of a medium if the critical angle for light passing into air is \( 48.6^\circ \)?

**Frequency of light** does not change on refraction, c = f λ, when enters denser medium speed decreases, wavelength decreases λ' = v/f = c/(n f) = λ/n, frequency same, for λ=480 nm f=c/λ=3×10⁸/480×10⁻⁹=6.25×10¹⁴ Hz, for 630 nm f=4.76×10¹⁴ Hz, for 540 nm f=5.56×10¹⁴ Hz, refractive index determines bending. sin i_c = (n₂/n₁) , where n₂ = 1.0 (air), i_c = 48.6° . sin 48.6° ≈ 0.75 , n₁ = (1.0/0.75) ≈ 1.33 . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A =

Ref: NCERT > Physics Book > Wave Optics > Refraction, Refractive Index and Critical Angle

What is the condition for total internal reflection to occur?

**Refractive index** n = c/v, c=3×10⁸ m/s vacuum, v speed in medium, for v=2.25×10⁸ m/s n=3/2.25=1.33, for v=1.8×10⁸ n=1.67, for n=1.3 v=3×10⁸/1.3=2.31×10⁸ m/s, for n=1.6 v=1.875×10⁸ m/s, for n=1.2 λ_medium = λ_vacuum/n =600/1.2=500 nm, for n=1.75 λ=700/1.75=400 nm, for n=1.5 λ=750/1.5=500 nm, wavelength in medium λ' = λ/n. Total internal reflection occurs when the angle of incidence exceeds the critical angle, and light travels from a denser to a rarer medium. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation

Ref: NCERT > Physics Book > Wave Optics > Refraction, Refractive Index and Critical Angle

What happens to the refracted ray when the angle of incidence exceeds the critical angle?

**Frequency of light** does not change on refraction, c = f λ, when enters denser medium speed decreases, wavelength decreases λ' = v/f = c/(n f) = λ/n, frequency same, for λ=480 nm f=c/λ=3×10⁸/480×10⁻⁹=6.25×10¹⁴ Hz, for 630 nm f=4.76×10¹⁴ Hz, for 540 nm f=5.56×10¹⁴ Hz, refractive index determines bending. When the angle of incidence exceeds the critical angle, no refraction occurs, and the ray undergoes total internal reflection. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation gives Undergoes total

Ref: NCERT > Physics Book > Wave Optics > Refraction, Refractive Index and Critical Angle

Why does total internal reflection occur only when light travels from a denser to a rarer medium?

**Refractive index** n = c/v, c=3×10⁸ m/s vacuum, v speed in medium, for v=2.25×10⁸ m/s n=3/2.25=1.33, for v=1.8×10⁸ n=1.67, for n=1.3 v=3×10⁸/1.3=2.31×10⁸ m/s, for n=1.6 v=1.875×10⁸ m/s, for n=1.2 λ_medium = λ_vacuum/n =600/1.2=500 nm, for n=1.75 λ=700/1.75=400 nm, for n=1.5 λ=750/1.5=500 nm, wavelength in medium λ' = λ/n. In a denser medium, the critical angle exists due to a lower speed, beyond which refraction cannot occur, leading to reflection; this doesn’t happen in the reverse direction. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and

Ref: NCERT > Physics Book > Wave Optics > Refraction, Refractive Index and Critical Angle

What optical principle allows a prism to be used as a reflector in optical devices?

**Lens maker's formula** 1/f = (n-1)(1/R₁ - 1/R₂), n refractive index, R₁,R₂ radii of curvature (m), sign convention R positive if surface convex towards incident light. For double convex R₁=12 cm, R₂=-12 cm, n=1.5, 1/f=(0.5)(1/12 -1/(-12))=(0.5)(2/12)=1/12, f=12 cm, converging. Prisms reflect light via total internal reflection when the angle of incidence exceeds the critical angle at the prism’s internal surfaces. This property, dependent on the prism’s refractive index and angle, enables efficient reflection without loss, as seen in devices like binoculars. Substituting values gives Total i

Ref: NCERT > Physics Book > Ray Optics > Refraction at Spherical Surfaces and Lens Maker's Formula

In optical fibers, what ensures that light remains confined within the core during transmission?

**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. In optical fibers, the core has a higher refractive index than the cladding, enabling total internal reflection. When light strikes the core-cladding boundary at an angle greater than the critical angle, it reflects back into the core, ensuring confinement and minimal loss. Substituting values gives Higher refractive index of core than cladding, which matches

Ref: NCERT > Physics Book > Ray Optics > Refraction at Spherical Surfaces and Lens Maker's Formula

What is the critical angle for a crown glass (\( n = 1.52 \)) to air interface?

**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. Critical angle: sin i_c = (n₂/n₁) . Crown glass ( n₁ = 1.52 ), air ( n₂ = 1 ). sin i_c = (1/1.52) ≈ 0.658 . i_c = sin⁻¹(0.658) ≈ 41.1° . Substituting values gives 41°, 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 refr

Ref: NCERT > Physics Book > Ray Optics > Refraction at Spherical Surfaces and Lens Maker's Formula