Past One Particular Angle, Glass Stops Letting Light Out Altogether
Define the critical angle and the two conditions for total internal reflection, calculate it from the refractive index, trace rays through equilateral and right-angled prisms, compare a total-reflecting prism with a plane mirror, and explain a diamond's sparkle and a mirage.
Why does an empty test tube dipped in water look like polished silver?
Hold an empty test tube and lower it obliquely into a beaker of water. Its lower part suddenly looks silvery, as though the glass has been coated with metal. Take it out and it is plain glass again.
Nothing has been coated. Light travelling through the water has struck the glass-and-air boundary at a steep angle and been sent entirely back, so the surface behaves like a perfect mirror.
That is total internal reflection, and it happens only under one very specific circumstance. Send light from glass into air at a small angle and most of it passes out, bending away from the normal. Increase the angle and the refracted ray bends further away, until at one particular angle it grazes along the surface — refracted at exactly . Increase the angle even slightly beyond that and no light gets out at all. Every bit of it is reflected back into the glass.
That threshold angle is the critical angle, and it is the whole subject of this part of the chapter.
- **For glass it is about
- For water about
- For diamond only about , which is why a diamond sparkles so much more than a piece of glass cut to the same shape
The effect is not a curiosity. It is a better mirror than a mirror: an ordinary silvered mirror reflects only part of the light falling on it and produces several faint extra images, while total internal reflection sends back essentially all of the light and produces only one image.
So this part builds three things on the critical angle.
- The relation between the critical angle and the refractive index, which makes it calculable
- Ray paths through prisms**, where a triangle of glass can turn a beam through or without any silvering at all
- Applications, from optical fibres carrying telephone calls to the mirage on a hot road
This page covers the second part of the ICSE Class 10 Physics chapter on light: the critical angle, the conditions for total internal reflection, total-reflecting prisms, and the applications of the effect.
Nothing has been coated. Light travelling through the water has struck the glass-and-air boundary at a steep angle and been sent entirely back, so the surface behaves like a perfect mirror.
That is total internal reflection, and it happens only under one very specific circumstance. Send light from glass into air at a small angle and most of it passes out, bending away from the normal. Increase the angle and the refracted ray bends further away, until at one particular angle it grazes along the surface — refracted at exactly . Increase the angle even slightly beyond that and no light gets out at all. Every bit of it is reflected back into the glass.
That threshold angle is the critical angle, and it is the whole subject of this part of the chapter.
- **For glass it is about
- For water about
- For diamond only about , which is why a diamond sparkles so much more than a piece of glass cut to the same shape
The effect is not a curiosity. It is a better mirror than a mirror: an ordinary silvered mirror reflects only part of the light falling on it and produces several faint extra images, while total internal reflection sends back essentially all of the light and produces only one image.
So this part builds three things on the critical angle.
- The relation between the critical angle and the refractive index, which makes it calculable
- Ray paths through prisms**, where a triangle of glass can turn a beam through or without any silvering at all
- Applications, from optical fibres carrying telephone calls to the mirage on a hot road
This page covers the second part of the ICSE Class 10 Physics chapter on light: the critical angle, the conditions for total internal reflection, total-reflecting prisms, and the applications of the effect.
Formula
What is the critical angle and how is it related to the refractive index?
**The critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is exactly .**
Where the relation comes from. Apply Snell's law to light going from the denser medium to air, with the angle of incidence equal to and the angle of refraction equal to . Taking the refractive index of the denser medium with respect to air as and using the reversibility of light,
The two conditions for total internal reflection, and both must hold:
- The light must travel from an optically denser medium into an optically rarer medium
- The angle of incidence in the denser medium must be greater than the critical angle for that pair of media
Neither condition can be dropped. Light going from air into glass can never be totally internally reflected, however large the angle of incidence, because it is entering the denser medium and always bends toward the normal. **And light in glass striking the surface at less than passes out, however clean the glass.
Notice the inverse relation between and .** A larger refractive index gives a smaller and therefore a smaller critical angle — so a denser medium traps light more easily.
Worked example 1 — critical angles of the three standard media.
Water, :
Glass, :
Diamond, :
The ordering matters. Diamond has the largest refractive index and the smallest critical angle, so a ray inside a diamond needs only to exceed to be trapped. Far more of the light entering a diamond bounces around inside before finding a face it can leave through, which is the physical reason for its brilliance.
Worked example 2 — the refractive index from the critical angle. The critical angle for a certain medium is . Find its refractive index.
Worked example 3 — the speed of light from the critical angle. The critical angle for a glass-air boundary is . Find the refractive index of the glass and the speed of light in it.
Two numbers from one angle, and this chain — critical angle to refractive index to speed — is a standard three-mark question.
Worked example 4 — deciding whether a ray escapes. A ray inside glass of refractive index strikes the glass-air surface at . Does it emerge? What if the angle were ?
**The critical angle is .
- At , the angle is less than the critical angle, so the ray is refracted out, bending away from the normal. A small part is also reflected back
- At , the angle is greater than the critical angle, so the ray is totally internally reflected and none of it leaves the glass
One difference between total internal reflection and ordinary reflection worth stating. At a silvered mirror, part of the light is always absorbed. At total internal reflection, practically the whole of the light is returned**, which is why the effect is called total — and why it makes a better reflector than any mirror.
A boundary case. Exactly at the critical angle the refracted ray grazes along the surface at , so it is neither properly out nor properly reflected. Total internal reflection begins only beyond that angle, which is why the condition is written with a strict "greater than".
Where the relation comes from. Apply Snell's law to light going from the denser medium to air, with the angle of incidence equal to and the angle of refraction equal to . Taking the refractive index of the denser medium with respect to air as and using the reversibility of light,
The two conditions for total internal reflection, and both must hold:
- The light must travel from an optically denser medium into an optically rarer medium
- The angle of incidence in the denser medium must be greater than the critical angle for that pair of media
Neither condition can be dropped. Light going from air into glass can never be totally internally reflected, however large the angle of incidence, because it is entering the denser medium and always bends toward the normal. **And light in glass striking the surface at less than passes out, however clean the glass.
Notice the inverse relation between and .** A larger refractive index gives a smaller and therefore a smaller critical angle — so a denser medium traps light more easily.
Worked example 1 — critical angles of the three standard media.
Water, :
Glass, :
Diamond, :
The ordering matters. Diamond has the largest refractive index and the smallest critical angle, so a ray inside a diamond needs only to exceed to be trapped. Far more of the light entering a diamond bounces around inside before finding a face it can leave through, which is the physical reason for its brilliance.
Worked example 2 — the refractive index from the critical angle. The critical angle for a certain medium is . Find its refractive index.
Worked example 3 — the speed of light from the critical angle. The critical angle for a glass-air boundary is . Find the refractive index of the glass and the speed of light in it.
Two numbers from one angle, and this chain — critical angle to refractive index to speed — is a standard three-mark question.
Worked example 4 — deciding whether a ray escapes. A ray inside glass of refractive index strikes the glass-air surface at . Does it emerge? What if the angle were ?
**The critical angle is .
- At , the angle is less than the critical angle, so the ray is refracted out, bending away from the normal. A small part is also reflected back
- At , the angle is greater than the critical angle, so the ray is totally internally reflected and none of it leaves the glass
One difference between total internal reflection and ordinary reflection worth stating. At a silvered mirror, part of the light is always absorbed. At total internal reflection, practically the whole of the light is returned**, which is why the effect is called total — and why it makes a better reflector than any mirror.
A boundary case. Exactly at the critical angle the refracted ray grazes along the surface at , so it is neither properly out nor properly reflected. Total internal reflection begins only beyond that angle, which is why the condition is written with a strict "greater than".
What happens to a ray inside an equilateral or a right-angled prism?
Send a ray into a prism perpendicular to one face and it travels straight to the next face, where the geometry of the prism decides the angle of incidence — and therefore whether it escapes or is totally reflected.
The general principle. A ray entering a face at right angles is not bent at all, so it continues along its original direction inside the glass. The angle at which it meets the second face is then fixed entirely by the angles of the prism. Compare that angle with the critical angle of for glass and you know at once what happens.
And when a ray is reflected, its deviation is
where is the angle of incidence at the reflecting face.
**Case 1 — the equilateral prism, angles , , .**
A ray entering one face normally travels to a second face. Because each angle of the prism is , **the ray meets that second face at an angle of incidence of .
- , so the ray is totally internally reflected
- Its deviation is
- It then meets the third face normally and emerges undeviated**, having been turned through in all
**Case 2 — the right-angled isosceles prism, angles , , . This is the most useful prism of all, and it works in two quite different ways.
Turning a beam through . Let the ray enter normally through one of the two perpendicular faces.** It travels straight to the hypotenuse, which it meets at an angle of incidence of .
- **, so the ray is totally internally reflected
- Its deviation is
- It leaves normally through the other perpendicular face
Turning a beam through . Now let the ray enter normally through the hypotenuse.** It travels to one of the perpendicular faces, meeting it at .
- It is totally internally reflected and turned through
- It travels to the second perpendicular face, meeting that one at as well, and is totally internally reflected again, turned through another
- It emerges back through the hypotenuse, travelling parallel to the incident ray but in the opposite direction — **a total deviation of
Two reflections, each of , giving in all. That is the arrangement used to send a beam back on itself.
Case 3 — the right-angled prism with angles , , , which behaves differently depending on which face the ray enters.
Entering normally through the shorter perpendicular face** — the one opposite the angle — the ray meets the hypotenuse at an angle of incidence of .
- **, so the ray is totally internally reflected**, with a deviation of
Entering normally through the longer perpendicular face — the one opposite the angle — the ray meets the hypotenuse at an angle of incidence of only .
- **, so the ray is not totally internally reflected. It refracts out of the hypotenuse, bending away from the normal, with only a faint partial reflection back inside
That contrast is the most instructive thing in this section. The same prism, the same glass, the same normal incidence — and the outcome is completely different because the geometry delivers a different angle to the reflecting face. Total internal reflection is not a property of the prism but of the angle at which the ray arrives.
Worked example — checking a prism for total internal reflection.** Would a , , prism made of a material of refractive index still turn a beam through ?
**The ray arrives at the hypotenuse at , which is now less than the critical angle of . So it would refract out instead, and the prism would fail. A total-reflecting prism only works if its material has a critical angle below **, which means a refractive index above about — and ordinary glass at comfortably satisfies it.
The general principle. A ray entering a face at right angles is not bent at all, so it continues along its original direction inside the glass. The angle at which it meets the second face is then fixed entirely by the angles of the prism. Compare that angle with the critical angle of for glass and you know at once what happens.
And when a ray is reflected, its deviation is
where is the angle of incidence at the reflecting face.
**Case 1 — the equilateral prism, angles , , .**
A ray entering one face normally travels to a second face. Because each angle of the prism is , **the ray meets that second face at an angle of incidence of .
- , so the ray is totally internally reflected
- Its deviation is
- It then meets the third face normally and emerges undeviated**, having been turned through in all
**Case 2 — the right-angled isosceles prism, angles , , . This is the most useful prism of all, and it works in two quite different ways.
Turning a beam through . Let the ray enter normally through one of the two perpendicular faces.** It travels straight to the hypotenuse, which it meets at an angle of incidence of .
- **, so the ray is totally internally reflected
- Its deviation is
- It leaves normally through the other perpendicular face
Turning a beam through . Now let the ray enter normally through the hypotenuse.** It travels to one of the perpendicular faces, meeting it at .
- It is totally internally reflected and turned through
- It travels to the second perpendicular face, meeting that one at as well, and is totally internally reflected again, turned through another
- It emerges back through the hypotenuse, travelling parallel to the incident ray but in the opposite direction — **a total deviation of
Two reflections, each of , giving in all. That is the arrangement used to send a beam back on itself.
Case 3 — the right-angled prism with angles , , , which behaves differently depending on which face the ray enters.
Entering normally through the shorter perpendicular face** — the one opposite the angle — the ray meets the hypotenuse at an angle of incidence of .
- **, so the ray is totally internally reflected**, with a deviation of
Entering normally through the longer perpendicular face — the one opposite the angle — the ray meets the hypotenuse at an angle of incidence of only .
- **, so the ray is not totally internally reflected. It refracts out of the hypotenuse, bending away from the normal, with only a faint partial reflection back inside
That contrast is the most instructive thing in this section. The same prism, the same glass, the same normal incidence — and the outcome is completely different because the geometry delivers a different angle to the reflecting face. Total internal reflection is not a property of the prism but of the angle at which the ray arrives.
Worked example — checking a prism for total internal reflection.** Would a , , prism made of a material of refractive index still turn a beam through ?
**The ray arrives at the hypotenuse at , which is now less than the critical angle of . So it would refract out instead, and the prism would fail. A total-reflecting prism only works if its material has a critical angle below **, which means a refractive index above about — and ordinary glass at comfortably satisfies it.
Why is a total-reflecting prism better than a plane mirror?
Because it returns practically all the light, produces only one image, and has no silvering that can tarnish or peel.
How a right-angled prism is used as a reflector. Placed so that the light enters normally through a perpendicular face, it deviates the beam through ; placed so that the light enters normally through the hypotenuse, it deviates the beam through . Both without any reflecting coating at all — the glass-air boundary does the work.
The comparison, point by point.
- Brightness. A silvered mirror absorbs some light in the silvering and reflects the rest, so the reflected beam is noticeably weaker. A prism reflects essentially the whole of the incident light, because the reflection is total
- Number of images. A back-silvered mirror gives a faint image from partial reflection at the front glass surface, a bright one from the silvering, and further faint ones from repeated internal reflections. A prism gives a single image, since the light enters and leaves at normal incidence and only the one total reflection occurs
- Durability. The silvering of a mirror can tarnish, scratch or peel away with time. A prism has no coating to lose, so its reflecting property is permanent
- Sharpness. Because there is only one reflection and no multiple images, the image from a prism is sharper and free of the ghosting a thick mirror produces
Which is why prisms rather than mirrors are used inside optical instruments where brightness and sharpness matter and where the instrument must survive years of handling.
- A periscope uses two right-angled prisms, one at each end, each turning the light through
- Binoculars and prism telescopes use prisms to fold a long light path into a short body and to correct the inversion of the image
- A camera viewfinder uses a prism to present an upright image to the eye
Worked example — a periscope's two prisms. A periscope has a right-angled isosceles prism at the top and another at the bottom. Trace the beam.
- Light from the scene enters the upper prism normally through a vertical perpendicular face and travels horizontally to the hypotenuse
- **It meets the hypotenuse at , exceeding the critical angle of , and is totally internally reflected through , now travelling vertically down the tube
- It enters the lower prism normally**, meets its hypotenuse at , and is totally internally reflected through another
- It emerges horizontally into the observer's eye, having been displaced vertically by the length of the tube
**Two total reflections, each of , and the net effect is to see over an obstacle without any loss of brightness.
One honest qualification about "no loss". Light entering and leaving the prism faces does suffer a small partial reflection at those faces, since any glass-air boundary reflects a little. So a prism is not perfectly lossless either — but the loss happens at the entry and exit faces, not at the reflecting face, and it is far smaller than the loss at a silvered surface. The reflection itself is total; the transmission through the faces is not.**
How a right-angled prism is used as a reflector. Placed so that the light enters normally through a perpendicular face, it deviates the beam through ; placed so that the light enters normally through the hypotenuse, it deviates the beam through . Both without any reflecting coating at all — the glass-air boundary does the work.
The comparison, point by point.
- Brightness. A silvered mirror absorbs some light in the silvering and reflects the rest, so the reflected beam is noticeably weaker. A prism reflects essentially the whole of the incident light, because the reflection is total
- Number of images. A back-silvered mirror gives a faint image from partial reflection at the front glass surface, a bright one from the silvering, and further faint ones from repeated internal reflections. A prism gives a single image, since the light enters and leaves at normal incidence and only the one total reflection occurs
- Durability. The silvering of a mirror can tarnish, scratch or peel away with time. A prism has no coating to lose, so its reflecting property is permanent
- Sharpness. Because there is only one reflection and no multiple images, the image from a prism is sharper and free of the ghosting a thick mirror produces
Which is why prisms rather than mirrors are used inside optical instruments where brightness and sharpness matter and where the instrument must survive years of handling.
- A periscope uses two right-angled prisms, one at each end, each turning the light through
- Binoculars and prism telescopes use prisms to fold a long light path into a short body and to correct the inversion of the image
- A camera viewfinder uses a prism to present an upright image to the eye
Worked example — a periscope's two prisms. A periscope has a right-angled isosceles prism at the top and another at the bottom. Trace the beam.
- Light from the scene enters the upper prism normally through a vertical perpendicular face and travels horizontally to the hypotenuse
- **It meets the hypotenuse at , exceeding the critical angle of , and is totally internally reflected through , now travelling vertically down the tube
- It enters the lower prism normally**, meets its hypotenuse at , and is totally internally reflected through another
- It emerges horizontally into the observer's eye, having been displaced vertically by the length of the tube
**Two total reflections, each of , and the net effect is to see over an obstacle without any loss of brightness.
One honest qualification about "no loss". Light entering and leaving the prism faces does suffer a small partial reflection at those faces, since any glass-air boundary reflects a little. So a prism is not perfectly lossless either — but the loss happens at the entry and exit faces, not at the reflecting face, and it is far smaller than the loss at a silvered surface. The reflection itself is total; the transmission through the faces is not.**
Where is total internal reflection used in practice?
Wherever light has to be turned, guided or trapped without losing brightness — and in several natural effects as well.
Optical fibres. A fibre is a very thin thread of glass or plastic, with a core of higher refractive index surrounded by a cladding of lower refractive index.
- Light entering one end at a suitable angle strikes the core-cladding boundary at an angle greater than the critical angle and is totally internally reflected
- It is reflected again and again along the length of the fibre, following the fibre even round bends
- Because each reflection is total, almost no light is lost at the reflections themselves
Two uses, both worth naming. In communication, signals are sent as pulses of light along fibres, carrying telephone, television and internet traffic. In medicine, an endoscope made of a bundle of fibres is passed into the body, one set of fibres carrying light in to illuminate an organ and another carrying the image back out.
Total-reflecting prisms, as described in the previous section — in periscopes, binoculars, prism telescopes and camera viewfinders.
The brilliance of a diamond. A diamond's refractive index of gives it a critical angle of only about .
- **Almost any ray that enters the stone strikes an internal face beyond and is totally internally reflected
- It bounces from face to face many times before finally finding a face it can leave through
- A cut diamond's faces are angled deliberately so that most of the trapped light emerges through the top, which is why it appears to blaze with light
A piece of glass cut to exactly the same shape looks dull by comparison**, because its critical angle of lets far more of the light escape at the first internal face it meets.
A mirage on a hot road. On a hot day the air just above the road surface is much hotter, and therefore rarer, than the air above it.
- Light from the sky travelling almost horizontally passes from denser air above into rarer air below, bending progressively away from the normal
- At a shallow enough angle it is totally internally reflected by the air layers and turned upward
- The eye receives it from a low direction and interprets it as coming from the road, so a patch of sky appears as a shimmering pool of water
Notice that both conditions are met: the light is going from denser to rarer, and the angle is large. A mirage is total internal reflection in air rather than in glass, which is a good answer to "can total internal reflection happen without a solid?"
Several everyday sparkles.
- An empty test tube dipped obliquely in water looks silvery, as in the opening section
- A crack inside a glass block shines brightly, because the thin layer of air in the crack is rarer than the glass around it
- An air bubble in water sparkles, for the same reason
- A shallow pool of clear water seen from below the surface shows a mirror-like surface beyond a certain angle
Worked example — will light stay inside a fibre? An optical fibre has a core of refractive index and a cladding of refractive index . Find the critical angle at the core-cladding boundary.
Here neither medium is air, so use the relative refractive index of the core with respect to the cladding:
**So a ray must strike the boundary at more than about to be trapped** — a much larger angle than the for glass and air, because the two media are so close in refractive index. That is why light must be launched into a fibre along a direction close to its axis, and it explains why a fibre cannot be bent too sharply without leaking light.
One caution about what total internal reflection is not. It is not the ordinary reflection you get from any surface. Ordinary reflection happens at every boundary, at every angle, and returns only part of the light. Total internal reflection happens only from denser to rarer and only beyond the critical angle, and returns all of it. Confusing the two makes it impossible to state either condition correctly.
Optical fibres. A fibre is a very thin thread of glass or plastic, with a core of higher refractive index surrounded by a cladding of lower refractive index.
- Light entering one end at a suitable angle strikes the core-cladding boundary at an angle greater than the critical angle and is totally internally reflected
- It is reflected again and again along the length of the fibre, following the fibre even round bends
- Because each reflection is total, almost no light is lost at the reflections themselves
Two uses, both worth naming. In communication, signals are sent as pulses of light along fibres, carrying telephone, television and internet traffic. In medicine, an endoscope made of a bundle of fibres is passed into the body, one set of fibres carrying light in to illuminate an organ and another carrying the image back out.
Total-reflecting prisms, as described in the previous section — in periscopes, binoculars, prism telescopes and camera viewfinders.
The brilliance of a diamond. A diamond's refractive index of gives it a critical angle of only about .
- **Almost any ray that enters the stone strikes an internal face beyond and is totally internally reflected
- It bounces from face to face many times before finally finding a face it can leave through
- A cut diamond's faces are angled deliberately so that most of the trapped light emerges through the top, which is why it appears to blaze with light
A piece of glass cut to exactly the same shape looks dull by comparison**, because its critical angle of lets far more of the light escape at the first internal face it meets.
A mirage on a hot road. On a hot day the air just above the road surface is much hotter, and therefore rarer, than the air above it.
- Light from the sky travelling almost horizontally passes from denser air above into rarer air below, bending progressively away from the normal
- At a shallow enough angle it is totally internally reflected by the air layers and turned upward
- The eye receives it from a low direction and interprets it as coming from the road, so a patch of sky appears as a shimmering pool of water
Notice that both conditions are met: the light is going from denser to rarer, and the angle is large. A mirage is total internal reflection in air rather than in glass, which is a good answer to "can total internal reflection happen without a solid?"
Several everyday sparkles.
- An empty test tube dipped obliquely in water looks silvery, as in the opening section
- A crack inside a glass block shines brightly, because the thin layer of air in the crack is rarer than the glass around it
- An air bubble in water sparkles, for the same reason
- A shallow pool of clear water seen from below the surface shows a mirror-like surface beyond a certain angle
Worked example — will light stay inside a fibre? An optical fibre has a core of refractive index and a cladding of refractive index . Find the critical angle at the core-cladding boundary.
Here neither medium is air, so use the relative refractive index of the core with respect to the cladding:
**So a ray must strike the boundary at more than about to be trapped** — a much larger angle than the for glass and air, because the two media are so close in refractive index. That is why light must be launched into a fibre along a direction close to its axis, and it explains why a fibre cannot be bent too sharply without leaking light.
One caution about what total internal reflection is not. It is not the ordinary reflection you get from any surface. Ordinary reflection happens at every boundary, at every angle, and returns only part of the light. Total internal reflection happens only from denser to rarer and only beyond the critical angle, and returns all of it. Confusing the two makes it impossible to state either condition correctly.
Exam tip
Which steps protect the marks in a total internal reflection question?
State both conditions, compute the critical angle, and then compare it with the actual angle of incidence. Every question in this part is decided by that comparison.
- Give both conditions together — denser to rarer, and angle of incidence greater than the critical angle
- **Use **, and check that a larger gives a smaller
- Remember the three values: about for water, for glass, for diamond
- For a boundary between two media neither of which is air, use the relative refractive index, as in the optical-fibre example
- In a prism, find the angle of incidence at the reflecting face from the prism's angles, not by guessing
- **Use for each total reflection
- Mark the normal at every face in a ray diagram, and show the angles against it
- Show that explicitly before asserting that total internal reflection occurs
- Give three points of comparison when asked why a prism beats a mirror — brightness, single image, and no silvering to tarnish
- Name applications specifically — optical fibre, endoscope, periscope, binoculars, diamond, mirage
The misconception to name. Total internal reflection cannot occur when light goes from a rarer to a denser medium. Light travelling from air into glass bends toward the normal at every angle of incidence and always gets in, so there is no angle, however large, at which it is turned back. A question asking whether light can be totally internally reflected at an air-to-glass boundary has the answer no, and the reason is the direction of travel rather than the size of the angle.
A second trap.** Using as the critical angle for every boundary. It is the critical angle for glass and air only. For a glass-water boundary or a core-cladding boundary the relative refractive index is much smaller, so the critical angle is much larger — about in the fibre example. Check which two media the boundary separates before quoting a number.
- Give both conditions together — denser to rarer, and angle of incidence greater than the critical angle
- **Use **, and check that a larger gives a smaller
- Remember the three values: about for water, for glass, for diamond
- For a boundary between two media neither of which is air, use the relative refractive index, as in the optical-fibre example
- In a prism, find the angle of incidence at the reflecting face from the prism's angles, not by guessing
- **Use for each total reflection
- Mark the normal at every face in a ray diagram, and show the angles against it
- Show that explicitly before asserting that total internal reflection occurs
- Give three points of comparison when asked why a prism beats a mirror — brightness, single image, and no silvering to tarnish
- Name applications specifically — optical fibre, endoscope, periscope, binoculars, diamond, mirage
The misconception to name. Total internal reflection cannot occur when light goes from a rarer to a denser medium. Light travelling from air into glass bends toward the normal at every angle of incidence and always gets in, so there is no angle, however large, at which it is turned back. A question asking whether light can be totally internally reflected at an air-to-glass boundary has the answer no, and the reason is the direction of travel rather than the size of the angle.
A second trap.** Using as the critical angle for every boundary. It is the critical angle for glass and air only. For a glass-water boundary or a core-cladding boundary the relative refractive index is much smaller, so the critical angle is much larger — about in the fibre example. Check which two media the boundary separates before quoting a number.
Did you know
Why does a fish see the whole sky squeezed into a circle above it?
Sit underwater and look up. The entire world above the surface — the whole sky and everything on the bank — appears inside a bright circular window directly overhead. Outside that circle the surface looks like a mirror, showing you the bottom of the pool reflected.
The window is total internal reflection seen from the inside, and its size is set by the critical angle.
Follow the rays the other way. Light from any point above the water enters the water and bends toward the normal. **A ray arriving almost along the surface — at grazing incidence — refracts into the water at exactly the critical angle of about . Nothing from above the water can come in at a steeper angle than that.
So every ray from the whole of the world above the water arrives inside a cone of half-angle . Looking up, the fish sees all of it compressed into a circle, and that circle is called the fish's window.
And outside the circle?** Those directions correspond to angles beyond , from which no light can arrive from above. What the fish sees there is light from inside the water, totally internally reflected off the underside of the surface — so the rest of the surface acts as a mirror showing the pool bed.
The size of the window depends on the depth in a simple way. Because the cone has a fixed half-angle, its radius at the surface grows in proportion to the depth: the deeper the fish, the wider its window. A fish just below the surface sees a small bright disc; one near the bottom sees a large one.
The same geometry has a practical consequence for anyone standing on a bank. Because all the light from above is squeezed into that cone, a fish can see you long before you are directly overhead — even when you are standing well back from the edge and appear, from your own point of view, to be nowhere near the water.
And the mirror region explains something you can test in a bath. Look along the underside of the water surface at a shallow angle and it looks perfectly reflective, showing the bottom of the bath. Look straight up and it is transparent. The changeover happens at the critical angle, and it is sharp rather than gradual.
**One last thought about why the effect is called total. At any ordinary reflecting surface some light is absorbed or transmitted, so "perfect reflection" is an engineering goal that is never quite reached. Beyond the critical angle it is reached exactly, and by nothing more than a boundary between two transparent media.** That is why an optical fibre can carry a signal for kilometres with almost no loss at the reflections, and why the best reflector in an optical instrument is a piece of plain glass rather than a coated one.
The window is total internal reflection seen from the inside, and its size is set by the critical angle.
Follow the rays the other way. Light from any point above the water enters the water and bends toward the normal. **A ray arriving almost along the surface — at grazing incidence — refracts into the water at exactly the critical angle of about . Nothing from above the water can come in at a steeper angle than that.
So every ray from the whole of the world above the water arrives inside a cone of half-angle . Looking up, the fish sees all of it compressed into a circle, and that circle is called the fish's window.
And outside the circle?** Those directions correspond to angles beyond , from which no light can arrive from above. What the fish sees there is light from inside the water, totally internally reflected off the underside of the surface — so the rest of the surface acts as a mirror showing the pool bed.
The size of the window depends on the depth in a simple way. Because the cone has a fixed half-angle, its radius at the surface grows in proportion to the depth: the deeper the fish, the wider its window. A fish just below the surface sees a small bright disc; one near the bottom sees a large one.
The same geometry has a practical consequence for anyone standing on a bank. Because all the light from above is squeezed into that cone, a fish can see you long before you are directly overhead — even when you are standing well back from the edge and appear, from your own point of view, to be nowhere near the water.
And the mirror region explains something you can test in a bath. Look along the underside of the water surface at a shallow angle and it looks perfectly reflective, showing the bottom of the bath. Look straight up and it is transparent. The changeover happens at the critical angle, and it is sharp rather than gradual.
**One last thought about why the effect is called total. At any ordinary reflecting surface some light is absorbed or transmitted, so "perfect reflection" is an engineering goal that is never quite reached. Beyond the critical angle it is reached exactly, and by nothing more than a boundary between two transparent media.** That is why an optical fibre can carry a signal for kilometres with almost no loss at the reflections, and why the best reflector in an optical instrument is a piece of plain glass rather than a coated one.
Exam relevance
How is total internal reflection tested in JEE and NEET?
This is foundation work for Class 12 Ray Optics and Optical Instruments, a chapter examined heavily in both JEE Main and NEET Physics.
**Where leads.** Class 12 keeps it unchanged and generalises it to a boundary between any two media as , with medium the denser one. The optical-fibre calculation you do here is exactly that general form, and JEE Main sets numericals on the critical angle for glass-water and core-cladding boundaries where the air formula would give a wrong answer.
Where the prism ray tracing leads. Class 12 adds the general prism formulas — the relation between the angle of the prism, the two refractions and the deviation, and the condition for minimum deviation. It also sets the standard question of whether a ray inside a prism escapes or is totally reflected at the second face, which is decided by the same comparison with the critical angle that you make here. JEE Advanced problems often hinge on a ray being trapped at a face the candidate expected it to leave through.
Where the optical fibre leads. Class 12 defines the acceptance angle and the numerical aperture of a fibre, both derived by combining Snell's law at the entry face with the critical-angle condition at the core-cladding boundary. Your observation that light must be launched close to the axis is that result in words, and NEET asks for the principle of the fibre and the role of the cladding.
Where the prism-versus-mirror comparison leads. It is reused in Class 12 when discussing the construction of optical instruments, and the reason — total reflection returning all the light — is the answer expected.
Where the atmospheric applications lead. The mirage appears in Class 12 alongside atmospheric refraction, and the explanation required is the same two-condition argument: denser to rarer, and an angle beyond the critical value.
Question types to expect. At this level: critical-angle numericals, stating the two conditions, prism ray diagrams for the three standard shapes, and named applications. In competitive papers: critical angle at a general boundary, ray tracing in prisms with a refraction at the first face as well, acceptance angle of a fibre, and assertion-reason items on why total internal reflection cannot occur from rarer to denser.
The single trap that costs marks. Using for every boundary. That value belongs to glass and air only — a glass-water boundary has a much larger critical angle, and the fibre example gives about . At JEE level the same error appears as using the absolute refractive index where the relative one is needed.
A second trap. Asserting total internal reflection for light going from air into glass. It is impossible in that direction, because the ray always bends toward the normal and always enters. Both JEE and NEET set assertion-reason questions on precisely this, and the reason to give is the direction of travel, not the size of the angle.
Board versus competitive emphasis. The ICSE paper marks the labelled prism diagram, the two conditions, the substituted critical angle and a named application; a competitive paper marks an angle or a numerical aperture. The transferable habit is computing the critical angle first and writing it down before looking at the ray at all — because every question in this family is then settled by one comparison.
**Where leads.** Class 12 keeps it unchanged and generalises it to a boundary between any two media as , with medium the denser one. The optical-fibre calculation you do here is exactly that general form, and JEE Main sets numericals on the critical angle for glass-water and core-cladding boundaries where the air formula would give a wrong answer.
Where the prism ray tracing leads. Class 12 adds the general prism formulas — the relation between the angle of the prism, the two refractions and the deviation, and the condition for minimum deviation. It also sets the standard question of whether a ray inside a prism escapes or is totally reflected at the second face, which is decided by the same comparison with the critical angle that you make here. JEE Advanced problems often hinge on a ray being trapped at a face the candidate expected it to leave through.
Where the optical fibre leads. Class 12 defines the acceptance angle and the numerical aperture of a fibre, both derived by combining Snell's law at the entry face with the critical-angle condition at the core-cladding boundary. Your observation that light must be launched close to the axis is that result in words, and NEET asks for the principle of the fibre and the role of the cladding.
Where the prism-versus-mirror comparison leads. It is reused in Class 12 when discussing the construction of optical instruments, and the reason — total reflection returning all the light — is the answer expected.
Where the atmospheric applications lead. The mirage appears in Class 12 alongside atmospheric refraction, and the explanation required is the same two-condition argument: denser to rarer, and an angle beyond the critical value.
Question types to expect. At this level: critical-angle numericals, stating the two conditions, prism ray diagrams for the three standard shapes, and named applications. In competitive papers: critical angle at a general boundary, ray tracing in prisms with a refraction at the first face as well, acceptance angle of a fibre, and assertion-reason items on why total internal reflection cannot occur from rarer to denser.
The single trap that costs marks. Using for every boundary. That value belongs to glass and air only — a glass-water boundary has a much larger critical angle, and the fibre example gives about . At JEE level the same error appears as using the absolute refractive index where the relative one is needed.
A second trap. Asserting total internal reflection for light going from air into glass. It is impossible in that direction, because the ray always bends toward the normal and always enters. Both JEE and NEET set assertion-reason questions on precisely this, and the reason to give is the direction of travel, not the size of the angle.
Board versus competitive emphasis. The ICSE paper marks the labelled prism diagram, the two conditions, the substituted critical angle and a named application; a competitive paper marks an angle or a numerical aperture. The transferable habit is computing the critical angle first and writing it down before looking at the ray at all — because every question in this family is then settled by one comparison.
Key takeaways
What must you be able to do from this part?
One angle, two conditions and three prisms.
- The critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is
- ****, so — a larger refractive index gives a smaller critical angle
- The two conditions for total internal reflection: the light must go from denser to rarer, and the angle of incidence must be greater than the critical angle
- Critical angles: about for water, for glass, for diamond
- ** means ; means and m s
- A ray in glass at escapes; at it is totally internally reflected
- For a boundary between two media, neither air, use the relative refractive index** — a core of in a cladding of gives
- Deviation on reflection
- Equilateral prism: a ray entering normally meets the next face at , is totally reflected and deviated by
- **, , prism**: entering normally through a perpendicular face gives one reflection at and a ** deviation; entering normally through the hypotenuse gives two reflections and a deviation
- , , prism**: entering through the shorter perpendicular face gives at the hypotenuse and total reflection; entering through the longer one gives only and the ray refracts out
- **A total-reflecting prism needs a critical angle below **, so a material of would fail
- A prism beats a mirror on brightness, on giving a single image instead of several, and on having no silvering to tarnish
- Applications: optical fibres for communication and endoscopy, total-reflecting prisms in periscopes and binoculars, the brilliance of diamond, the mirage, a shining crack in glass and a silvery test tube in water
- A mirage is total internal reflection in air, with the hot rarer layer at the road surface
- Total internal reflection cannot occur from rarer to denser, at any angle
The most satisfying self-test needs a glass of water and your own eyes. Hold the glass above eye level and look up through the side at the underside of the water surface — find the angle at which it stops being transparent and starts acting as a mirror, and then check whether your estimate is anywhere near .
- The critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is
- ****, so — a larger refractive index gives a smaller critical angle
- The two conditions for total internal reflection: the light must go from denser to rarer, and the angle of incidence must be greater than the critical angle
- Critical angles: about for water, for glass, for diamond
- ** means ; means and m s
- A ray in glass at escapes; at it is totally internally reflected
- For a boundary between two media, neither air, use the relative refractive index** — a core of in a cladding of gives
- Deviation on reflection
- Equilateral prism: a ray entering normally meets the next face at , is totally reflected and deviated by
- **, , prism**: entering normally through a perpendicular face gives one reflection at and a ** deviation; entering normally through the hypotenuse gives two reflections and a deviation
- , , prism**: entering through the shorter perpendicular face gives at the hypotenuse and total reflection; entering through the longer one gives only and the ray refracts out
- **A total-reflecting prism needs a critical angle below **, so a material of would fail
- A prism beats a mirror on brightness, on giving a single image instead of several, and on having no silvering to tarnish
- Applications: optical fibres for communication and endoscopy, total-reflecting prisms in periscopes and binoculars, the brilliance of diamond, the mirage, a shining crack in glass and a silvery test tube in water
- A mirage is total internal reflection in air, with the hot rarer layer at the road surface
- Total internal reflection cannot occur from rarer to denser, at any angle
The most satisfying self-test needs a glass of water and your own eyes. Hold the glass above eye level and look up through the side at the underside of the water surface — find the angle at which it stops being transparent and starts acting as a mirror, and then check whether your estimate is anywhere near .