Refraction and image formation
A straw in a glass of water seems to bend right at the surface of the liquid; a cut diamond sparkles with intense flashes that a piece of glass never manages; an optical signal travels thousands of kilometres across the ocean floor with hardly any loss of intensity. All three phenomena are the same refraction, taken to its most interesting consequences.
Refractive index and Snell's law
When we point a laser pointer at a tank of water, we see the ray bend towards the vertical as it enters, and when we measure the angles we find that the ratio $i/r$ changes with the angle of incidence, but the ratio of their sines, $\sin i / \sin r$, stays strictly constant: which physical property of the medium determines how much the light is slowed down and deviated?
Light travels through a vacuum at $c = 3\cdot10^{8}\ \text{m/s}$ and the refractive index of water is $n = 1,33$. a) Write down the relationship between the refractive…
Light travels through a vacuum at $c = 3\cdot10^{8}\ \text{m/s}$ and the refractive index of water is $n = 1,33$.
- 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 $v$.
- 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 $2\cdot10^{8}\ \text{m/s}$. Given: $c = 3\cdot10^{8}\ \text{m/s}$. a)…
In a laboratory practical, a pulse of light is measured to travel through a block of glass at $2\cdot10^{8}\ \text{m/s}$. Given: $c = 3\cdot10^{8}\ \text{m/s}$.
- a) Calculate the refractive index of that glass.
- 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:…
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: $\sen 30^\circ = 0,5$. Observe the diagram:
- a) Write down Snell's law of refraction, $n_1 \cdot \sen i = n_2 \cdot \sen r$, identifying each quantity.
- b) Calculate the angle of refraction inside the water.
- c) State whether the ray bends towards or away from the normal and why.
Critical angle and total internal reflection
A straw placed in a glass of water appears bent right at the surface of the liquid. a) Which optical phenomenon produces this effect? b) Explain its physical cause: what…
A straw placed in a glass of water appears bent right at the surface of the liquid.
- a) Which optical phenomenon produces this effect?
- b) Explain its physical cause: what happens to light when it passes from water to air?
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 $60^\circ$.…
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: $\sen 60^\circ \approx 0,866$. Observe the diagram:
- a) Calculate the angle of refraction in the glass.
- b) Calculate the angle of deviation of the ray, that is, the difference between the angle of incidence and the angle of refraction.
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…
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:
- a) Calculate the glass-air critical angle and verify that total internal reflection occurs at $45^\circ$.
- b) Explain what advantages this system has over a metallic mirror (think about light loss and degradation).
- 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.
Optical fibre: total internal reflection in action
In an optical fibre, light travels through the core without escaping to the outside even if the fibre bends. Observe the diagram: a) Which optical phenomenon keeps the…
In an optical fibre, light travels through the core without escaping to the outside even if the fibre bends. Observe the diagram:
- a) Which optical phenomenon keeps the light inside the core?
- b) What condition must the angle of incidence of the ray on the core wall satisfy for this to occur?
- c) Must the refractive index of the core be greater or smaller than that of the surrounding cladding? Give reasons for your answer.
A block of glass of refractive index $n = 1,5$ is surrounded by air ($n \approx 1$). a) Calculate the critical angle for the glass-air boundary. b) A ray travelling…
A block of glass of refractive index $n = 1,5$ is surrounded by air ($n \approx 1$).
- a) Calculate the critical angle for the glass-air boundary.
- 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.
- c) Answer the same question if the angle of incidence is $50^\circ$.
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…
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:
- a) Calculate the critical angle at the core-cladding interface.
- b) Calculate the maximum acceptance angle $\theta_{\text{max}}$ of the fibre.
- 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.
Plane parallel-sided slabs
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: $\sen 45^\circ \approx 0,71$.…
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: $\sen 45^\circ \approx 0,71$.
- a) Calculate the angle of refraction by applying Snell's law.
- 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 swimmer submerged in a swimming pool ($n_{\text{agua}} = 1,33$) looks up towards the surface. Air has $n \approx 1$. Observe the diagram: a) Calculate the critical…
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:
- 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 $90^\circ$.
- 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 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…
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:
- a) Using Snell's law and this approximation, deduce that the apparent depth is $h' = h/n$.
- b) Calculate the apparent depth of the coin.
- c) Explain why the pool appears shallower than it actually is and what practical risk this perception poses.
Back to the beginning: The straw looks broken because the rays coming from the water bend away from the normal as they leave into the air, distorting the apparent position of the submerged object, and the diamond sparkles so strongly because its very high index ($n \approx 2{,}42$) gives a very small critical angle ($L \approx 24{,}4^\circ$), trapping the light in multiple total internal reflections before letting it escape through its upper faces.