Worked examples · Sheet 2

Total internal reflection and critical angles

Theory used: Total internal reflection and critical angle Snell's law Direction of the deviation

Exercise 1

A small lamp is at the bottom of a swimming pool, at depth $h = 1.20 m$ below the surface of the water ( $n = 1.33$ ), which is flat and still. Seen from above, the rays of the lamp leave the water only through a bright disk centered on the vertical through the lamp. Compute the radius of the disk and its area.

The rays of the lamp

S

that strike the surface beyond the critical angle

θ_{c}

are totally reflected; only those that fall within the disk of radius

r

leave

sin θ_{c} = \frac{1}{n}, r = h tan θ_{c} = \frac{h}{n ^{2} - 1}, A = π r^{2}

θ_{c} r A = arcsin \frac{1}{1.33} \approx 48. 8^{\circ} = \frac{1.20}{1.3 3 ^{2} - 1} m = \frac{1.20}{0.877} m \approx 1.37 m = π \cdot 1.3 7^{2} m^{2} \approx 5.88 m^{2}

The light passes from water to air, that is, toward a less optically dense medium, so the critical angle $θ_{c}$ exists, with $sin θ_{c} = \frac{1}{n}$ . A ray that leaves the lamp at angle $θ$ from the vertical strikes the surface at the same angle $θ$ , because the normal to the surface is vertical, and reaches it at the distance $h tan θ$ from the vertical through the lamp. The rays with $θ < θ_{c}$ leave, those with $θ > θ_{c}$ are totally reflected; hence the edge of the bright disk is at the distance $r = h tan θ_{c}$ .

For the second form of the formula we use $tan θ_{c} = \frac{sin θ _{c}}{1 - sin ^{2} θ _{c}}$ with $sin θ_{c} = \frac{1}{n}$ , which gives $tan θ_{c} = \frac{1}{n ^{2} - 1}$ . The radius of the disk is proportional to the depth: a deeper lamp illuminates a wider disk, but the light is spread over a larger area.

Exercise 2

A glass prism has a cross section in the shape of an isosceles right triangle. A ray enters perpendicular to one of the legs and strikes the hypotenuse from inside. Compute the minimum index of refraction of the glass for total internal reflection to occur at the hypotenuse with the prism in air. Then determine whether a prism with $n = 1.50$ also produces total internal reflection when it is immersed in water ( $n_{a} = 1.33$ ).

θ_{1} = 4 5^{\circ}, θ_{1} \geq θ_{c} ⟺ sin 4 5^{\circ} \geq \frac{n _{ext}}{n} ⟺ n \geq 2 n_{ext}

in air: n_{m i n} in water: sin θ_{c} = 2 \cdot 1.00 \approx 1.41 = \frac{1.33}{1.50} \approx 0.887 ⟹ θ_{c} \approx 62. 5^{\circ} > 4 5^{\circ}

The ray enters perpendicular to the leg, so by Snell’s law it is not deviated. The leg and the hypotenuse form an angle of $4 5^{\circ}$ , hence the inner ray, perpendicular to the leg, forms $4 5^{\circ}$ with the normal to the hypotenuse: the angle of incidence on the hypotenuse is $4 5^{\circ}$ . There is total internal reflection if this angle is at least equal to the critical angle, that is, if $sin 4 5^{\circ} \geq \frac{n _{ext}}{n}$ , where $n_{ext}$ is the index of the external medium. In air $n \geq 2 \approx 1.41$ is required, and ordinary glass with $n = 1.50$ satisfies the condition.

In water the ratio of the indices is $\frac{1.50}{1.33} \approx 1.13$ , less than $2$ . The glass-water critical angle is about $62. 5^{\circ}$ , greater than $4 5^{\circ}$ , so the ray partly exits through the hypotenuse and the immersed prism does not produce total internal reflection.

Exercise 3

An optical fiber has a core with index $n_{1} = 1.50$ and a cladding with index $n_{2} = 1.40$ . A ray propagating in air strikes the entrance face of the fiber, which is flat and perpendicular to the axis, forming the angle $θ_{a}$ with the axis. Compute the critical angle between core and cladding and the maximum value of $θ_{a}$ for which the ray remains guided in the core.

The ray enters at angle

θ_{a}

, is refracted at angle

θ_{r}

and strikes the cladding with

θ = 9 0^{\circ} - θ_{r}

: it is guided if

θ \geq θ_{c}

sin θ_{c} = \frac{n _{2}}{n _{1}}, sin θ_{a} = n_{1} sin θ_{r}, θ_{r} = 9 0^{\circ} - θ

θ \geq θ_{c} ⟺ sin θ_{a} \leq n_{1} cos θ_{c} = n_{1} 1 - \frac{n _{2}^{2}}{n _{1}^{2}} = n_{1}^{2} - n_{2}^{2}

θ_{c} sin θ_{a, m a x} = arcsin \frac{1.40}{1.50} \approx 69. 0^{\circ} = 1.5 0^{2} - 1.4 0^{2} = 0.29 \approx 0.539 ⟹ θ_{a, m a x} \approx 32. 6^{\circ}

At the entrance the normal to the face is parallel to the axis, so the ray forms with the normal the angle $θ_{a}$ in air and the angle $θ_{r}$ in the core, related by Snell’s law. The refracted ray then strikes the lateral surface, whose normal is perpendicular to the axis, hence the angle of incidence on the cladding is $θ = 9 0^{\circ} - θ_{r}$ .

The ray remains in the core if total internal reflection occurs at this surface, that is, if $θ \geq θ_{c}$ . A larger angle $θ_{a}$ gives a larger $θ_{r}$ and therefore a smaller $θ$ ; the condition is hence equivalent to $θ_{r} \leq 9 0^{\circ} - θ_{c}$ , that is, to $sin θ_{a} \leq n_{1} cos θ_{c}$ . Writing the cosine using $sin θ_{c} = \frac{n _{2}}{n _{1}}$ gives $sin θ_{a} \leq n_{1}^{2} - n_{2}^{2}$ . With the data the fiber accepts the rays within $32. 6^{\circ}$ of the axis; the limit corresponds to $θ_{r} \approx 21. 0^{\circ}$ , the complement of the critical angle. The quantity $n_{1}^{2} - n_{2}^{2}$ is called the numerical aperture of the fiber.