Refraction and total internal reflection

A straw placed in a glass of water appears bent right at the surface of the liquid.

1. a) Which optical phenomenon produces this effect?
2. b) Explain its physical cause: what happens to light when it passes from water to air?

Light travels through a vacuum at $c = 3\times10^{8}\ \text{m/s}$ and the refractive index of water is $n = 1.33$.

1. a) Write down the relationship between the refractive index $n$, the speed of light in a vacuum $c$ and the speed of light in the medium $v$.
2. b) Calculate the speed of light in water.

In a laboratory practical, a pulse of light is measured to travel through a block of glass at $2\times10^{8}\ \text{m/s}$. Given: $c = 3\times10^{8}\ \text{m/s}$.

1. a) Calculate the refractive index of that glass.
2. b) Explain whether a refractive index can be less than $1$.

A laser beam is incident from air ($n \approx 1$) onto the surface of water in an aquarium ($n = 1.33$) at an angle of incidence of $30^\circ$ to the normal. Given: $\sin 30^\circ = 0.5$. Observe the diagram:

1. a) Write down Snell's law of refraction, $n_1 \times \sin i = n_2 \times \sin r$, identifying each quantity.
2. b) Calculate the angle of refraction inside the water.
3. c) State whether the ray bends towards or away from the normal and why.

Light travels at $2.26\times10^{8}\ \text{m/s}$ in water and at $1.94\times10^{8}\ \text{m/s}$ in quartz. Given: $c = 3\times10^{8}\ \text{m/s}$.

1. a) Calculate the refractive index of each material.
2. b) If the same ray enters both materials from air with the same angle of incidence, in which material is it deviated more? Give reasons for your answer.

In the optics laboratory, a laser beam is incident from air ($n \approx 1$) onto a semicircular glass block of index $n = 1.52$ at an angle of incidence of $60^\circ$. Given: $\sin 60^\circ \approx 0.866$. Observe the diagram:

1. a) Calculate the angle of refraction in the glass.
2. b) Calculate the angle of deviation of the ray, that is, the difference between the angle of incidence and the angle of refraction.

A ray of light travels through the water in a fish tank ($n_1 = 1.33$) and reaches the glass wall ($n_2 = 1.5$) with an angle of incidence of $35^\circ$. Given: $\sen 35^\circ \approx 0.574$.

1. a) Calculate the angle of refraction in the glass by applying Snell's law between the two media.
2. b) State whether the ray bends towards or away from the normal when passing into the glass and justify your answer by comparing the indices.

A gemmologist wants to identify a transparent mineral. They direct a ray from air ($n \approx 1$) at $45^\circ$ and measure an angle of refraction of $28^\circ$. Given: $\sen 45^\circ \approx 0.707$ and $\sen 28^\circ \approx 0.469$.

1. a) Calculate the refractive index of the mineral.
2. b) Identify it from these candidates: water ($n = 1.33$), glass ($n = 1.5$) and diamond ($n = 2.42$). Justify your choice.

A light source submerged in the water of a swimming pool ($n = 1.33$) emits a ray that reaches the surface at an angle of $20^\circ$ to the normal and emerges into the air ($n \approx 1$).

1. a) Without performing calculations, state whether the refracted ray bends towards or away from the normal.
2. b) Justify your answer using Snell's law, explaining what happens when the second medium has a lower refractive index.

A swimmer submerged in a swimming pool ($n_{\text{agua}} = 1.33$) looks up towards the surface. Air has $n \approx 1$. Observe the diagram:

1. a) Calculate the critical angle for the passage of light from water into air, that is, the angle of incidence for which the angle of refraction is $90^\circ$.
2. b) Explain what the swimmer observes when looking at points on the surface that form an angle with the vertical greater than that critical angle.

A block of glass of refractive index $n = 1.5$ is surrounded by air ($n \approx 1$).

1. a) Calculate the critical angle for the glass-air boundary.
2. b) A ray travelling through the glass reaches the surface at an angle of incidence of $30^\circ$: state, giving reasons, whether it emerges into the air or undergoes total internal reflection.
3. c) Answer the same question if the angle of incidence is $50^\circ$.

Diamond has a refractive index of $n = 2.42$. Given: $c = 3\times10^{8}\ \text{m/s}$ and $n_{\text{aire}} \approx 1$.

1. a) Calculate the speed of light inside the diamond.
2. b) Calculate the critical angle for a ray reaching the diamond-air surface from the inside.

A diver points a torch towards the surface of the sea ($n_{\text{agua}} = 1.33$, $n_{\text{aire}} \approx 1$). The ray is incident at $35^\circ$ to the normal. Given: $\sen 35^\circ \approx 0.574$. Observe the diagram:

1. a) Calculate the angle at which the ray emerges into the air.
2. b) Calculate the deviation suffered by the ray when crossing the surface.
3. c) Explain what can happen if the diver gradually increases the angle of incidence.

A jeweller compares the brilliance of diamond ($n = 2.42$) with that of glass ($n = 1.5$). Given: $c = 3\times10^{8}\ \text{m/s}$ and $n_{\text{aire}} \approx 1$.

1. a) Calculate the speed of light in each material.
2. b) Calculate the critical angle of each relative to air.
3. c) Using the previous results, justify why a well-cut diamond reflects back almost all the light that enters it and "sparkles" much more than a piece of glass of the same shape.

Quality binoculars use glass Porro prisms ($n = 1.5$) instead of metalled mirrors. At each reflection, the ray travelling through the glass strikes the glass-air face with an angle of incidence of $45^\circ$. Observe the diagram:

1. a) Calculate the glass-air critical angle and verify that total internal reflection occurs at $45^\circ$.
2. b) Explain what advantages this system has over a metallic mirror (think about light loss and degradation).
3. c) Predict what would happen if the prism were submerged in water ($n = 1.33$): calculate the new critical angle and reason whether it would still work.

To measure the refractive index of an unknown liquid, it is placed on a glass block of $n = 1.5$. By directing light from within the glass, it is observed that total reflection begins when the angle of incidence at the glass-liquid interface is $60^\circ$. Given: $\sen 60^\circ \approx 0.866$ and $c = 3\times10^{8}\ \text{m/s}$.

1. a) Calculate the refractive index of the liquid.
2. b) Calculate the speed of light in the liquid.
3. c) Reason whether it would be possible to observe total internal reflection for a ray travelling from the liquid into the glass.

An optical fibre with a core index of $n_1 = 1.50$ and a cladding index of $n_2 = 1.47$ is in air ($n \approx 1$). It can be shown that the maximum angle of entry $\theta_{\text{max}}$ for a ray to be guided by the fibre satisfies $\sen\theta_{\text{max}} = \sqrt{n_1^2 - n_2^2}$. Observe the diagram:

1. a) Calculate the critical angle at the core-cladding interface.
2. b) Calculate the maximum acceptance angle $\theta_{\text{max}}$ of the fibre.
3. c) Interpret the result: explain what the acceptance cone is and why it matters when launching light into the fibre, for example in a medical endoscope.

A light pulse travels along an optical fibre of length $1\ \text{km}$ with a core index of $n_1 = 1.48$ and a cladding index of $n_2 = 1.46$. The fastest ray is the one travelling in a straight line along the axis, and the slowest is the one zigzagging, striking the walls at exactly the critical angle; for the latter, the path length is $L \times n_1/n_2$. Given: $c = 3\times10^{8}\ \text{m/s}$. Observe the diagram:

1. a) Calculate the time taken by the axial ray.
2. b) Calculate the time taken by the critical ray.
3. c) Calculate the temporal pulse broadening at the exit, explain the problem it poses for high-speed data transmission, and how it is resolved using single-mode fibre.

In an optical fibre, light travels through the core without escaping to the outside even if the fibre bends. Observe the diagram:

1. a) Which optical phenomenon keeps the light inside the core?
2. b) What condition must the angle of incidence of the ray on the core wall satisfy for this to occur?
3. c) Must the refractive index of the core be greater or smaller than that of the surrounding cladding? Give reasons for your answer.

An optical fibre has a core of index $n_1 = 1.48$ and a cladding of index $n_2 = 1.46$.

1. a) Calculate the critical angle at the core-cladding boundary.
2. b) A ray in the core reaches that boundary with an angle of incidence of $85^\circ$ with respect to the normal: state, with justification, whether it is transmitted into the cladding or continues to be guided along the core.

A ray of light enters a glass plate of refractive index $n = 1.5$ from air ($n \approx 1$) at an angle of incidence of $45^\circ$. Given: $\sen 45^\circ \approx 0.71$.

1. a) Calculate the angle of refraction by applying Snell's law.
2. b) Draw in your notebook the boundary surface, the normal, the incident ray and the refracted ray, labelling both angles.

You can do it on paper and send a photo with your answer.

A ray of light is incident at $45^\circ$ on a glass sheet with flat, parallel faces, with refractive index $n = 1.5$ and thickness $e = 2\ \text{cm}$, surrounded by air ($n \approx 1$). Data: $\sen 45^\circ \approx 0.707$; once you have calculated the angle of refraction $r$, you can use $\cos r \approx 0.88$ and $\sen(45^\circ - r) \approx 0.29$. Observe the diagram:

1. a) Calculate the angle of refraction $r$ at the first face.
2. b) Applying Snell's law at the second face, show that the emergent ray leaves at $45^\circ$, that is, parallel to the incident ray.
3. c) The lateral displacement of the ray is $d = e \times \sen(i - r)/\cos r$. Calculate its value and explain its meaning with a drawing in your notebook.

A coin rests at the bottom of a swimming pool with a depth of $2\ \text{m}$ ($n_{\text{agua}} = 1.33$, $n_{\text{aire}} \approx 1$). For an observer looking almost vertically downwards, the angles are small and the approximation $\sen\theta \approx \tan\theta$ can be used. Observe the diagram:

1. a) Using Snell's law and this approximation, deduce that the apparent depth is $h' = h/n$.
2. b) Calculate the apparent depth of the coin.
3. c) Explain why the pool appears shallower than it actually is and what practical risk this perception poses.

A ray enters from air ($n_1 \approx 1$) into the water of an aquarium ($n_2 = 1.33$) at an angle of incidence of $50^\circ$ and, without changing direction within the water, then passes through the glass wall ($n_3 = 1.5$), whose faces are parallel to the water surface. Given: $\sen 50^\circ \approx 0.766$. Observe the diagram:

1. a) Calculate the angle of refraction in the water.
2. b) Calculate the angle of refraction in the glass.
3. c) Verify that the result of part b) is the same as if the ray had passed directly from air into glass, and justify it by chaining the two applications of Snell's law.