Refraction and Total Internal Reflection — Diagnostic Tests

Unit Tests

UT-1: Critical Angle and Dispersion in a Prism

Question:

An equilateral glass prism ($n = 1.52$ for red light, $n = 1.55$ for violet light) is placed in air ($n = 1.00$).

(a) Calculate the critical angle for the glass-air interface for red light.

(b) Light enters one face of the prism at an angle of incidence of $40^\circ$. Calculate the angle of emergence for red light and state whether total internal reflection occurs at the second face.

(c) Calculate the angular dispersion (the angle between the emergent red and violet rays) for the same angle of incidence.

Solution:

(a) Critical angle: $\sin\theta_c = 1/n = 1/1.52 = 0.6579$

$\theta_c = 41.1^\circ$

(b) At the first face (angle of incidence $i_1 = 40^\circ$):

By Snell’s law: $\sin i_1 = n \sin r_1$

$\sin 40^\circ = 1.52 \sin r_1 \Rightarrow \sin r_1 = 0.6428/1.52 = 0.4229$ $r_1 = 25.03^\circ$

Angle of incidence at the second face: $i_2 = 60^\circ - r_1 = 60 - 25.03 = 34.97^\circ$

Since $34.97^\circ \lt 41.1^\circ = \theta_c$Total internal reflection does not occur.

Angle of emergence: $n\sin i_2 = \sin e$

$\sin e = 1.52 \times \sin 34.97^\circ = 1.52 \times 0.5730 = 0.8710$ $e = 60.6^\circ$

(c) For violet light ($n = 1.55$):

$\sin r_1' = \sin 40^\circ/1.55 = 0.6428/1.55 = 0.4147$

$r_1' = 24.51^\circ$

$i_2' = 60 - 24.51 = 35.49^\circ$

$\sin e' = 1.55 \times \sin 35.49^\circ = 1.55 \times 0.5808 = 0.9002$

$e' = 64.2^\circ$

Angular dispersion $= e' - e = 64.2^\circ - 60.6^\circ = 3.6^\circ$

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UT-2: Optical Fibre — Step Index

Question:

A step-index optical fibre has a core of refractive index $n_1 = 1.50$ and cladding of refractive index $n_2 = 1.46$. The core diameter is $50\,\mu\text{m}$.

(a) Calculate the critical angle at the core-cladding boundary.

(b) Calculate the maximum acceptance angle (numerical aperture) for light entering the fibre.

(c) Calculate the maximum number of reflections per metre for a ray entering at the maximum acceptance angle.

Solution:

(a) $\sin\theta_c = n_2/n_1 = 1.46/1.50 = 0.9733$

$\theta_c = 76.7^\circ$

This is the angle measured from the normal to the core-cladding boundary. The angle with the fibre axis is $90^\circ - 76.7^\circ = 13.3^\circ$.

(b) For light entering the fibre end at angle $\theta_a$ from the axis:

Using Snell’s law at the entry face and the critical angle condition:

$n_0 \sin\theta_a = \sqrt{n_1^2 - n_2^2} = \sqrt{2.25 - 2.1316} = \sqrt{0.1184} = 0.3441$

For air ($n_0 = 1$): $\theta_a = \sin^{-1}(0.3441) = 20.1^\circ$

Numerical aperture: $\text{NA} = \sqrt{n_1^2 - n_2^2} = 0.344$

(c) A ray entering at the maximum acceptance angle travels at the steepest angle to the axis inside the fibre.

Angle with the axis inside the fibre: $\alpha = 90^\circ - \theta_c = 13.3^\circ$

The ray travels along the fibre axis with horizontal speed $v\cos\alpha$ and bounces between the walls. The horizontal distance between bounces:

$d_{\text{bounce}} = \frac◆LB◆2r◆RB◆◆LB◆\tan\alpha◆RB◆ = \frac◆LB◆50 \times 10^{-6}◆RB◆◆LB◆\tan 13.3^\circ◆RB◆ = \frac◆LB◆50 \times 10^{-6}◆RB◆◆LB◆0.2363◆RB◆ = 2.12 \times 10^{-4}\,\text{m}$

Number of reflections per metre: $N = 1/(2.12 \times 10^{-4}) = 4720\,\text{reflections}\,\text{m}^{-1}$

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UT-3: Refractive Index from Real and Apparent Depth

Question:

A student measures the apparent depth of a coin at the bottom of a pool of liquid. The real depth is $(40.0 \pm 0.5)\,\text{cm}$ and the apparent depth is $(28.5 \pm 0.5)\,\text{cm}$.

(a) Calculate the refractive index of the liquid and its uncertainty.

(b) The student then observes a mark on the bottom of a glass block of refractive index $1.52$ through the liquid. The glass block is $3.0\,\text{cm}$ thick. Calculate the apparent thickness of the glass block as viewed through the liquid.

(c) Explain why the apparent depth formula only applies for near-normal viewing angles.

Solution:

(a) $n = \text{real depth}/\text{apparent depth} = 40.0/28.5 = 1.404$

Uncertainty: $\frac◆LB◆\Delta n◆RB◆◆LB◆n◆RB◆ = \frac◆LB◆\Delta d_{\text{real}}◆RB◆◆LB◆d_{\text{real}}◆RB◆ + \frac◆LB◆\Delta d_{\text{app}}◆RB◆◆LB◆d_{\text{app}}◆RB◆ = \frac{0.5}{40.0} + \frac{0.5}{28.5} = 0.0125 + 0.01754 = 0.0300$

$\Delta n = 0.0300 \times 1.404 = 0.042$

$n = 1.40 \pm 0.04$

(b) For near-normal incidence, the apparent depth of an object in a medium of refractive index $n_1$ viewed from a medium of refractive index $n_2$ is $d_{\text{apparent}} = d \times n_2/n_1$.

The light path is: glass ($n_g = 1.52$Thickness $d_g = 3.0\,\text{cm}$) $\to$ liquid ($n_l = 1.40$) $\to$ air ($n_a = 1.00$).

For the glass block viewed from air through the intervening liquid, the apparent thickness is:

$d_{\text{app}} = \frac{d_g}{n_g} = \frac{3.0}{1.52} = 1.97\,\text{cm}$

The apparent depth formula uses the refractive index of the object’s medium relative to the observer’s medium. The liquid layer does not change the apparent thickness of the glass block itself; it only affects the apparent depth of objects within or below the liquid. The total apparent depth of the bottom of the glass from the liquid surface would be $d_l/n_l + d_g/n_g$But the apparent thickness of the glass alone is $d_g/n_g = 1.97\,\text{cm}$.

(c) The apparent depth formula $d_{\text{app}} = d/n$ is derived using the small-angle (paraxial) approximation. At larger angles, Snell’s law gives a curved relationship between real and apparent position, and the simple ratio no longer holds. The apparent depth depends on the viewing angle, and the image position varies with angle — a phenomenon known as aberration. Additionally, at large angles, the formula breaks down because it assumes all rays from a point converge to a single image point, which is only true for paraxial rays.

Integration Tests

IT-1: Refraction at a Curved Interface (with Wave Properties)

Question:

A glass sphere of radius $R = 10\,\text{cm}$ and refractive index $n = 1.50$ is in air. A narrow beam of light enters the sphere parallel to a diameter, at a distance $h = 6.0\,\text{cm}$ from the diameter.

(a) Calculate the angle of refraction at the first surface.

(b) Calculate the angle of incidence at the second (inner) surface and determine whether the ray undergoes total internal reflection.

(c) If not, calculate the angle of emergence and the total deviation of the ray.

Solution:

(a) The angle of incidence at the first surface:

$\sin i_1 = h/R = 6.0/10 = 0.60 \Rightarrow i_1 = 36.9^\circ$

By Snell’s law: $\sin i_1 = n\sin r_1$

$\sin r_1 = 0.60/1.50 = 0.40 \Rightarrow r_1 = 23.6^\circ$

(b) The ray travels inside the sphere and hits the far surface. The geometry gives:

The angle of incidence at the second surface $i_2 = r_1 = 23.6^\circ$ (by the isosceles triangle formed by the two radii).

Critical angle: $\theta_c = \sin^{-1}(1/1.50) = 41.1^\circ$

Since $i_2 = 23.6^\circ \lt \theta_c$TIR does not occur.

(c) At the second surface: $n\sin i_2 = \sin e_2$

$\sin e_2 = 1.50 \times \sin 23.6^\circ = 1.50 \times 0.400 = 0.600$ $e_2 = 36.9^\circ$

The ray emerges parallel to the original direction (as expected for a sphere). The total deviation is:

$\delta = (i_1 - r_1) + (e_2 - i_2) = (36.9 - 23.6) + (36.9 - 23.6) = 26.6^\circ$

For TIR to occur: $r_1 \ge \theta_c = 41.1^\circ$Which requires $i_1$ such that $\sin i_1 \ge 1.50 \times \sin 41.1^\circ = 1.50 \times 0.657 = 0.986$I.e. $i_1 \ge 80.4^\circ$ or $h \ge 9.86\,\text{cm}$. Only rays very close to the edge undergo TIR.

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IT-2: Optical Fibre Signal Attenuation (with DC Circuits)

Question:

An optical fibre of length $5.0\,\text{km}$ has an attenuation of $0.30\,\text{dB}\,\text{km}^{-1}$. A laser source couples $5.0\,\text{mW}$ of optical power into the fibre.

(a) Calculate the power at the output end of the fibre.

(b) The fibre has a numerical aperture of 0.22 and core diameter $62.5\,\mu\text{m}$. Calculate the maximum acceptance angle and the solid angle of acceptance.

(c) If the fibre is bent to a radius of curvature of $5.0\,\text{cm}$Estimate whether significant power loss occurs due to bending.

Solution:

(a) Total attenuation: $A = 0.30 \times 5.0 = 1.5\,\text{dB}$

$A = 10\log_{10}(P_{\text{in}}/P_{\text{out}})$ $1.5 = 10\log_{10}(5.0/P_{\text{out}})$ $\log_{10}(5.0/P_{\text{out}}) = 0.15$ $5.0/P_{\text{out}} = 10^{0.15} = 1.413$ $P_{\text{out}} = 5.0/1.413 = 3.54\,\text{mW}$

(b) Maximum acceptance angle: $\theta_a = \sin^{-1}(\text{NA}) = \sin^{-1}(0.22) = 12.7^\circ$

Solid angle of acceptance (approximate): $\Omega \approx \pi\theta_a^2 = \pi(0.22)^2 = 0.152\,\text{sr}$ (for small angles, where $\theta$ is in radians: $\theta_a = 0.22\,\text{rad}$)

(c) Bending loss becomes significant when the bend radius approaches a critical value. For a step-index fibre:

$R_{\text{critical}} \approx \frac◆LB◆3n_1^2\lambda◆RB◆◆LB◆4\pi(n_1^2 - n_2^2)^{3/2}◆RB◆$

Without the wavelength, we can estimate: for typical telecom fibres, bend losses become significant below $R \approx 10$—$30\,\text{mm}$ for single-mode fibres. For multimode fibres with larger cores, the critical radius is smaller.

At $R = 5.0\,\text{cm} = 50\,\text{mm}$A multimode fibre of $62.5\,\mu\text{m}$ core diameter would experience minimal bending loss. However, tight bends at $5\,\text{mm}$ radius would cause significant loss.

The key point is that bending changes the angle of incidence at the core-cladding boundary. Some rays that previously satisfied the TIR condition no longer do, and they leak into the cladding. This is macrobending loss.

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IT-3: Prism as a Reflecting Element (with Superposition)

Question:

A right-angled isosceles glass prism ($n = 1.55$) is used as a reflector, with light entering one of the short faces and hitting the hypotenuse.

(a) Calculate the critical angle for the glass-air interface.

(b) Show that total internal reflection occurs at the hypotenuse when light enters the short face at normal incidence.

(c) Calculate the maximum angle of incidence on the short face for which TIR still occurs at the hypotenuse.

Solution:

(a) $\sin\theta_c = 1/n = 1/1.55 = 0.6452$

$\theta_c = 40.2^\circ$

(b) At normal incidence on the short face, the light enters undeviated ($r_1 = 0$). It hits the hypotenuse at $45^\circ$ angle of incidence.

Since $45^\circ \gt 40.2^\circ = \theta_c$TIR occurs. The ray is reflected through $90^\circ$.

(c) Let the angle of incidence on the short face be $i$.

By Snell’s law: $\sin i = n\sin r$Where $r$ is the angle of refraction.

The ray hits the hypotenuse at angle $i_2 = 45^\circ - r$.

For TIR: $i_2 \ge \theta_c = 40.2^\circ$

$45^\circ - r \ge 40.2^\circ$ $r \le 4.8^\circ$

$\sin i = n\sin r \le 1.55 \times \sin 4.8^\circ = 1.55 \times 0.0837 = 0.1297$ $i \le 7.5^\circ$

The maximum angle of incidence for TIR at the hypotenuse is $7.5^\circ$. Beyond this, some light is transmitted through the hypotenuse and the prism no longer acts as a perfect reflector. This limits the acceptance angle of prismatic reflectors.