One Formula Handles Every Mirror If the Signs Are Right
Apply the New Cartesian sign convention to every mirror quantity, solve mirror and magnification numericals, follow a ray through a glass slab to see lateral displacement, and calculate refractive index from the speed of light.
Why does one formula work for concave and convex mirrors alike?
Part 1 gave six separate cases for a concave mirror and one for a convex mirror — seven results to remember. There is a single equation that produces all seven, and it does so because the signs carry the information that the seven cases carried in words.
That is the whole of it. A concave mirror is described by a negative focal length and a convex mirror by a positive one; a real image comes out with a negative image distance and a virtual one with a positive distance. You do not have to decide in advance what kind of image to expect — the arithmetic tells you.
But that convenience depends entirely on using the signs consistently, and a single sign error turns a real image into a virtual one. So the sign convention is not bookkeeping; it is the physics.
The same chapter then turns to what happens when light passes into a new medium instead of bouncing off it. Light bends at the boundary because it travels at a different speed in the new material, and how much it bends is measured by one number for each substance — the refractive index.
This page covers the second part of the CBSE Class 10 Science chapter on light: the New Cartesian sign convention, numericals with the mirror formula and magnification, refraction and the glass slab, and refractive index from the speed of light.
That is the whole of it. A concave mirror is described by a negative focal length and a convex mirror by a positive one; a real image comes out with a negative image distance and a virtual one with a positive distance. You do not have to decide in advance what kind of image to expect — the arithmetic tells you.
But that convenience depends entirely on using the signs consistently, and a single sign error turns a real image into a virtual one. So the sign convention is not bookkeeping; it is the physics.
The same chapter then turns to what happens when light passes into a new medium instead of bouncing off it. Light bends at the boundary because it travels at a different speed in the new material, and how much it bends is measured by one number for each substance — the refractive index.
This page covers the second part of the CBSE Class 10 Science chapter on light: the New Cartesian sign convention, numericals with the mirror formula and magnification, refraction and the glass slab, and refractive index from the speed of light.
What are the sign rules for u, v, f and R?
Everything is measured from the pole, with the object always on the left, and distances measured in the direction the incident light travels are positive.
The four rules of the New Cartesian sign convention.
- The object is always placed to the left of the mirror, so the light travels from left to right
- All distances are measured from the pole as origin
- Distances measured in the same direction as the incident light — to the right — are positive; those measured against it — to the left — are negative
- Heights measured upward from the principal axis are positive; downward, negative
What those rules produce, and this is the list to memorise.
- **The object distance is always negative, because the object is always to the left
- A concave mirror has negative and negative , since its focus and centre of curvature lie to the left, in front of it
- A convex mirror has positive and positive , since its focus and centre lie to the right, behind it
- A real image has negative — it forms in front of the mirror, on the left
- A virtual image has positive — it appears behind the mirror, on the right
- An erect image has positive height; an inverted image has negative height
Worked example — assigning signs before calculating.** A concave mirror of radius of curvature cm has an object placed cm in front of it. Write down , and with their signs.
All three negative, because all three are measured to the left of the pole.
And for a convex mirror of the same radius with the object in the same place:
**Only stays negative, because the object is still on the left even though the mirror's focus is now on the right.
Why is negative even for a convex mirror. The object is a real object, and real objects sit where the light comes from — the left. Nothing about the mirror changes where the object is**, which is why carries the same sign in every problem in this chapter. **If you have written a positive , you have made a mistake before starting.
The habit that prevents almost every error. Write the three values with their signs on a separate line before touching the formula, and draw a quick sketch with P at the origin. The sketch decides the signs and the formula does the arithmetic**, and keeping those two jobs apart is what makes these numericals reliable.
The four rules of the New Cartesian sign convention.
- The object is always placed to the left of the mirror, so the light travels from left to right
- All distances are measured from the pole as origin
- Distances measured in the same direction as the incident light — to the right — are positive; those measured against it — to the left — are negative
- Heights measured upward from the principal axis are positive; downward, negative
What those rules produce, and this is the list to memorise.
- **The object distance is always negative, because the object is always to the left
- A concave mirror has negative and negative , since its focus and centre of curvature lie to the left, in front of it
- A convex mirror has positive and positive , since its focus and centre lie to the right, behind it
- A real image has negative — it forms in front of the mirror, on the left
- A virtual image has positive — it appears behind the mirror, on the right
- An erect image has positive height; an inverted image has negative height
Worked example — assigning signs before calculating.** A concave mirror of radius of curvature cm has an object placed cm in front of it. Write down , and with their signs.
All three negative, because all three are measured to the left of the pole.
And for a convex mirror of the same radius with the object in the same place:
**Only stays negative, because the object is still on the left even though the mirror's focus is now on the right.
Why is negative even for a convex mirror. The object is a real object, and real objects sit where the light comes from — the left. Nothing about the mirror changes where the object is**, which is why carries the same sign in every problem in this chapter. **If you have written a positive , you have made a mistake before starting.
The habit that prevents almost every error. Write the three values with their signs on a separate line before touching the formula, and draw a quick sketch with P at the origin. The sketch decides the signs and the formula does the arithmetic**, and keeping those two jobs apart is what makes these numericals reliable.
Formula
How do you use the mirror formula and find the magnification?
One equation for the distances, one for the size.
Here is the object distance, the image distance, the focal length, the object height, the image height and the magnification.
Worked example 1 — a real, magnified image. An object is placed cm in front of a concave mirror of focal length cm. Find the image position, the magnification, and the image of a cm tall object.
**The negative means the image is real and in front of the mirror**, cm from it. Now the magnification:
**A negative means inverted**, and the size is three times the object, so the image height is
— a cm tall inverted image. Check against Part 1: the object at cm lies between C (at cm) and F (at cm), and the table says real, inverted and enlarged. The formula and the table agree.
Worked example 2 — a virtual image from a concave mirror. The same mirror, with the object now cm away.
**Positive means virtual and behind the mirror; positive means erect.** So the image is virtual, erect and three times as large — exactly the shaving-mirror case, since cm is inside the focal length of cm.
Worked example 3 — a convex mirror. An object is cm from a convex mirror of focal length cm.
Virtual, erect and diminished, and the image lies between the pole and the focus — the single convex-mirror case from Part 1, now with numbers.
Notice what the three answers have in common. Nothing was decided in advance: the same two equations produced a real inverted image, a virtual enlarged one and a virtual diminished one, purely from the signs put in. That is the whole advantage of the convention.
The three checks to run on every answer.
- **Negative means real, positive means virtual** — and a convex mirror can never give a negative
- **Negative means inverted, positive means erect — and a real image from a mirror is always inverted
- means enlarged, means diminished
If an answer says a convex mirror has produced a real image, the arithmetic is wrong** — and that check alone catches most sign errors.
Here is the object distance, the image distance, the focal length, the object height, the image height and the magnification.
Worked example 1 — a real, magnified image. An object is placed cm in front of a concave mirror of focal length cm. Find the image position, the magnification, and the image of a cm tall object.
**The negative means the image is real and in front of the mirror**, cm from it. Now the magnification:
**A negative means inverted**, and the size is three times the object, so the image height is
— a cm tall inverted image. Check against Part 1: the object at cm lies between C (at cm) and F (at cm), and the table says real, inverted and enlarged. The formula and the table agree.
Worked example 2 — a virtual image from a concave mirror. The same mirror, with the object now cm away.
**Positive means virtual and behind the mirror; positive means erect.** So the image is virtual, erect and three times as large — exactly the shaving-mirror case, since cm is inside the focal length of cm.
Worked example 3 — a convex mirror. An object is cm from a convex mirror of focal length cm.
Virtual, erect and diminished, and the image lies between the pole and the focus — the single convex-mirror case from Part 1, now with numbers.
Notice what the three answers have in common. Nothing was decided in advance: the same two equations produced a real inverted image, a virtual enlarged one and a virtual diminished one, purely from the signs put in. That is the whole advantage of the convention.
The three checks to run on every answer.
- **Negative means real, positive means virtual** — and a convex mirror can never give a negative
- **Negative means inverted, positive means erect — and a real image from a mirror is always inverted
- means enlarged, means diminished
If an answer says a convex mirror has produced a real image, the arithmetic is wrong** — and that check alone catches most sign errors.
What happens to a ray passing through a rectangular glass slab?
It bends towards the normal on entering, bends away from the normal on leaving, and emerges parallel to its original direction but shifted sideways.
The two laws of refraction.
- The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane
- The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media:
That constant is the refractive index of medium 2 with respect to medium 1, and the relation is called Snell's law.
Which way the ray bends.
- Going from a rarer to a denser medium — air into glass — the ray bends towards the normal, so
- Going from a denser to a rarer medium — glass into air — the ray bends away from the normal, so
- A ray along the normal () does not bend at all
Worked example. A ray strikes a glass surface from air at an angle of incidence of . Taking the refractive index of glass as , find the angle of refraction.
The ray has bent towards the normal, from to about , as entering a denser medium requires.
Now follow it through the whole slab. At the second surface the ray passes from glass back into air, and the geometry of a rectangular slab makes the angle of incidence inside the glass equal to the previous angle of refraction. So the second bending is exactly the reverse of the first:
- At the first surface the ray bends towards the normal
- At the second surface it bends away from the normal by the same amount
- The emergent ray is therefore parallel to the incident ray
The two bendings cancel in direction but not in position. The ray comes out travelling the same way it went in, but along a line displaced sideways from the original path. That sideways shift is called lateral displacement.
What the displacement depends on. A thicker slab gives a larger shift, a larger angle of incidence gives a larger shift, and a more refractive glass gives a larger shift. A ray entering along the normal is not displaced at all, because it never bent.
Why a coin in a bowl of water appears raised. The light leaving the coin bends away from the normal on entering the air, and your eye traces it back along a straight line — which meets at a point higher than the coin actually is. The same effect makes a pencil in a glass of water look bent, and a swimming pool look shallower than it is. All three are one refraction seen from three angles.
The two laws of refraction.
- The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane
- The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media:
That constant is the refractive index of medium 2 with respect to medium 1, and the relation is called Snell's law.
Which way the ray bends.
- Going from a rarer to a denser medium — air into glass — the ray bends towards the normal, so
- Going from a denser to a rarer medium — glass into air — the ray bends away from the normal, so
- A ray along the normal () does not bend at all
Worked example. A ray strikes a glass surface from air at an angle of incidence of . Taking the refractive index of glass as , find the angle of refraction.
The ray has bent towards the normal, from to about , as entering a denser medium requires.
Now follow it through the whole slab. At the second surface the ray passes from glass back into air, and the geometry of a rectangular slab makes the angle of incidence inside the glass equal to the previous angle of refraction. So the second bending is exactly the reverse of the first:
- At the first surface the ray bends towards the normal
- At the second surface it bends away from the normal by the same amount
- The emergent ray is therefore parallel to the incident ray
The two bendings cancel in direction but not in position. The ray comes out travelling the same way it went in, but along a line displaced sideways from the original path. That sideways shift is called lateral displacement.
What the displacement depends on. A thicker slab gives a larger shift, a larger angle of incidence gives a larger shift, and a more refractive glass gives a larger shift. A ray entering along the normal is not displaced at all, because it never bent.
Why a coin in a bowl of water appears raised. The light leaving the coin bends away from the normal on entering the air, and your eye traces it back along a straight line — which meets at a point higher than the coin actually is. The same effect makes a pencil in a glass of water look bent, and a swimming pool look shallower than it is. All three are one refraction seen from three angles.
How do you calculate refractive index from the speed of light?
Divide the speed of light in vacuum by its speed in the material.
with m/s. Refractive index has no unit, because it is a ratio of two speeds.
Worked example 1. Light travels through glass at m/s. Find the refractive index of glass.
Worked example 2. The refractive index of water is . Find the speed of light in water.
Worked example 3. Diamond has a refractive index of . How fast does light travel in it?
— less than half its speed in vacuum, which is why diamond bends light so strongly.
Comparing two media directly. The refractive index of medium 2 with respect to medium 1 is
Worked example 4. Find the refractive index of glass with respect to water, given and .
So glass bends light only slightly when it enters from water, and much more when it enters from air — which is why a glass rod becomes almost invisible in a liquid of nearly the same refractive index.
Optically denser and optically rarer. Of two media, the one with the higher refractive index is optically denser, and light travels more slowly in it.
And here is the point that catches everybody. Optical density is not the same as mass density. Kerosene has a refractive index of about and water about , so kerosene is optically denser than water — yet kerosene floats on water, so it is less dense in mass.
Two different properties, two different orderings, and a question asking which of two liquids is optically denser is asking about refractive index, not about which one floats. Never infer one from the other.
One more consequence worth noticing. A higher refractive index means a slower speed, which means a larger bend. So the substances that bend light most — diamond among them — are the ones in which light is slowed most. The bending is not a separate fact from the slowing; it is caused by it.
with m/s. Refractive index has no unit, because it is a ratio of two speeds.
Worked example 1. Light travels through glass at m/s. Find the refractive index of glass.
Worked example 2. The refractive index of water is . Find the speed of light in water.
Worked example 3. Diamond has a refractive index of . How fast does light travel in it?
— less than half its speed in vacuum, which is why diamond bends light so strongly.
Comparing two media directly. The refractive index of medium 2 with respect to medium 1 is
Worked example 4. Find the refractive index of glass with respect to water, given and .
So glass bends light only slightly when it enters from water, and much more when it enters from air — which is why a glass rod becomes almost invisible in a liquid of nearly the same refractive index.
Optically denser and optically rarer. Of two media, the one with the higher refractive index is optically denser, and light travels more slowly in it.
And here is the point that catches everybody. Optical density is not the same as mass density. Kerosene has a refractive index of about and water about , so kerosene is optically denser than water — yet kerosene floats on water, so it is less dense in mass.
Two different properties, two different orderings, and a question asking which of two liquids is optically denser is asking about refractive index, not about which one floats. Never infer one from the other.
One more consequence worth noticing. A higher refractive index means a slower speed, which means a larger bend. So the substances that bend light most — diamond among them — are the ones in which light is slowed most. The bending is not a separate fact from the slowing; it is caused by it.
Exam tip
What layout keeps a mirror or refraction numerical safe?
Write the given quantities with their signs, then the formula, then the substitution, then the answer with its unit and its interpretation. Four lines, and the signs are the part that earns or loses the marks.
- **Write , and with signs on their own line** before using any formula. is always negative
- **Concave mirror: negative. Convex mirror: positive. Get this wrong and every later number is wrong
- Use rather than rearranging mid-calculation, and take the reciprocal only at the very end
- Interpret the answer in words**: *negative , so the image is real and in front of the mirror*. The interpretation is a separate mark
- **Use for mirrors and state whether the image is erect or inverted from its sign
- Check against the Part 1 table: a convex mirror can never give a real image, and a real mirror image is always inverted
- For refraction, measure angles from the normal, and say which way the ray bends and why
- Give refractive index without a unit**, and keep m/s to one significant figure unless told otherwise
The misconception to name. Optical density is not mass density. Kerosene is optically denser than water yet floats on it. A question asking which medium is optically denser wants the higher refractive index, and the answer has nothing to do with which liquid is heavier.
- **Write , and with signs on their own line** before using any formula. is always negative
- **Concave mirror: negative. Convex mirror: positive. Get this wrong and every later number is wrong
- Use rather than rearranging mid-calculation, and take the reciprocal only at the very end
- Interpret the answer in words**: *negative , so the image is real and in front of the mirror*. The interpretation is a separate mark
- **Use for mirrors and state whether the image is erect or inverted from its sign
- Check against the Part 1 table: a convex mirror can never give a real image, and a real mirror image is always inverted
- For refraction, measure angles from the normal, and say which way the ray bends and why
- Give refractive index without a unit**, and keep m/s to one significant figure unless told otherwise
The misconception to name. Optical density is not mass density. Kerosene is optically denser than water yet floats on it. A question asking which medium is optically denser wants the higher refractive index, and the answer has nothing to do with which liquid is heavier.
Did you know
Why does a glass rod vanish in the right liquid?
Stand a glass stirring rod in a beaker of water and you can see it clearly. Stand the same rod in a liquid whose refractive index happens to match the glass, and the part below the surface disappears — you can see the top of the rod above the liquid and nothing at all below it.
The reason is that you only see a transparent object because light bends as it enters and leaves. Bending is what makes edges visible, and bending happens only when the two media have different refractive indices. Match them and
The ray passes straight through with no deviation at all, exactly as if the boundary were not there — so there is nothing for your eye to detect.
The same reasoning explains why clean window glass is nearly invisible while a cracked or frosted pane is obvious. A crack is a thin layer of air inside the glass, and every air-to-glass boundary bends light. You see the crack and not the glass, because the crack has a mismatch and the glass does not.
And it explains a familiar irritation. A drop of water on a mobile screen is visible not because water is opaque but because its refractive index differs from air and from the screen. Wipe it away and the boundary goes with it.
The reverse case is diamond. Its refractive index of is one of the highest among common transparent materials, so the mismatch with air is enormous and the bending is extreme. Light entering a cut diamond is bent so strongly that much of it bounces around inside before leaving — which is the reason for the sparkle, and the reason a cutter's angles matter so much.
One check you can actually perform. Fill a glass with water and put a spoon in it: the handle appears bent at the surface. Now look at the spoon through the side of the glass rather than from above, and the apparent bend changes. The amount of apparent displacement depends on the angle you look from, which is precisely what a refraction calculation predicts — and the clearest evidence that the bending is real and not an illusion of the eye.
The reason is that you only see a transparent object because light bends as it enters and leaves. Bending is what makes edges visible, and bending happens only when the two media have different refractive indices. Match them and
The ray passes straight through with no deviation at all, exactly as if the boundary were not there — so there is nothing for your eye to detect.
The same reasoning explains why clean window glass is nearly invisible while a cracked or frosted pane is obvious. A crack is a thin layer of air inside the glass, and every air-to-glass boundary bends light. You see the crack and not the glass, because the crack has a mismatch and the glass does not.
And it explains a familiar irritation. A drop of water on a mobile screen is visible not because water is opaque but because its refractive index differs from air and from the screen. Wipe it away and the boundary goes with it.
The reverse case is diamond. Its refractive index of is one of the highest among common transparent materials, so the mismatch with air is enormous and the bending is extreme. Light entering a cut diamond is bent so strongly that much of it bounces around inside before leaving — which is the reason for the sparkle, and the reason a cutter's angles matter so much.
One check you can actually perform. Fill a glass with water and put a spoon in it: the handle appears bent at the surface. Now look at the spoon through the side of the glass rather than from above, and the apparent bend changes. The amount of apparent displacement depends on the angle you look from, which is precisely what a refraction calculation predicts — and the clearest evidence that the bending is real and not an illusion of the eye.
Exam relevance
How are mirror and refraction numericals set in JEE and NEET?
This is foundation work whose formulas are used unchanged in Class 11 and in both competitive papers.
Where the mirror formula leads. Class 11 Ray Optics derives it, extends it to refraction at a spherical surface, and then to lenses and to combinations. The New Cartesian convention is carried over exactly as you learn it here, and JEE Main problems routinely chain two or three elements together — the image from one becoming the object for the next. A sign error at the first step destroys the whole chain, which is why the convention matters more than the arithmetic.
Where Snell's law leads. Class 11 uses it for total internal reflection and the critical angle, for the optical fibre, for the prism and for the mirage. The condition for total internal reflection follows directly from Snell's law when the ray goes from denser to rarer, and JEE Main sets numericals on the critical angle every year in some form.
Where the refractive index leads. It becomes in Class 11 Wave Optics as well, where the slowing of light explains the bending physically, and the relation between wavelength, frequency and speed in a medium is examined. The frequency does not change when light enters a medium, but the wavelength and speed both do — a standard assertion-reason item.
Where the glass slab leads. Lateral displacement is examined directly, and the result that the emergent ray is parallel to the incident ray is used in prism and slab combinations. The reason — equal and opposite bendings at parallel surfaces — is the part that carries forward.
Question types to expect. At this level: mirror-formula numericals, magnification, refractive index from speeds. In competitive papers: multi-element numericals, critical-angle problems, and assertion-reason items on optical against mass density.
The single trap that costs marks. Writing as positive, or giving a concave mirror a positive focal length. **The object is always on the left, so is always negative**, and a concave mirror's focus is on the same side as the object, so its is negative too. In JEE the same convention governs lenses, where the signs differ — which is exactly why it must be applied and not guessed.
A second trap. Confusing optical density with mass density. Kerosene is optically denser than water and lighter than it, and that pairing is set as an assertion-reason item precisely because the two words sound the same.
Board versus competitive emphasis. The CBSE paper marks the signed given values, the formula, the substitution and the interpretation; a competitive paper marks the final number. The transferable habit is the signed line at the top — it costs one line and prevents the error that ruins entire questions.
Where the mirror formula leads. Class 11 Ray Optics derives it, extends it to refraction at a spherical surface, and then to lenses and to combinations. The New Cartesian convention is carried over exactly as you learn it here, and JEE Main problems routinely chain two or three elements together — the image from one becoming the object for the next. A sign error at the first step destroys the whole chain, which is why the convention matters more than the arithmetic.
Where Snell's law leads. Class 11 uses it for total internal reflection and the critical angle, for the optical fibre, for the prism and for the mirage. The condition for total internal reflection follows directly from Snell's law when the ray goes from denser to rarer, and JEE Main sets numericals on the critical angle every year in some form.
Where the refractive index leads. It becomes in Class 11 Wave Optics as well, where the slowing of light explains the bending physically, and the relation between wavelength, frequency and speed in a medium is examined. The frequency does not change when light enters a medium, but the wavelength and speed both do — a standard assertion-reason item.
Where the glass slab leads. Lateral displacement is examined directly, and the result that the emergent ray is parallel to the incident ray is used in prism and slab combinations. The reason — equal and opposite bendings at parallel surfaces — is the part that carries forward.
Question types to expect. At this level: mirror-formula numericals, magnification, refractive index from speeds. In competitive papers: multi-element numericals, critical-angle problems, and assertion-reason items on optical against mass density.
The single trap that costs marks. Writing as positive, or giving a concave mirror a positive focal length. **The object is always on the left, so is always negative**, and a concave mirror's focus is on the same side as the object, so its is negative too. In JEE the same convention governs lenses, where the signs differ — which is exactly why it must be applied and not guessed.
A second trap. Confusing optical density with mass density. Kerosene is optically denser than water and lighter than it, and that pairing is set as an assertion-reason item precisely because the two words sound the same.
Board versus competitive emphasis. The CBSE paper marks the signed given values, the formula, the substitution and the interpretation; a competitive paper marks the final number. The transferable habit is the signed line at the top — it costs one line and prevents the error that ruins entire questions.
Key takeaways
What should you know before moving on to lenses?
One convention, two formulas for mirrors, two laws for refraction and one ratio.
- New Cartesian convention: object on the left, distances from the pole, rightward positive, upward positive
- ** is always negative; concave mirror and negative, convex mirror and positive
- Negative means real and in front; positive means virtual and behind
- Mirror formula** , and magnification
- **Negative means inverted, positive means erect**; means enlarged
- A convex mirror can never give a real image, and a real mirror image is always inverted — use both as checks
- Laws of refraction: the three lines lie in one plane, and
- Rarer to denser bends towards the normal; denser to rarer bends away; along the normal there is no bending
- A rectangular slab gives an emergent ray parallel to the incident ray, displaced sideways — lateral displacement
- ****, with no unit, and
- Higher refractive index means optically denser and slower light — and optical density is not mass density
The sharpest self-test is the pair of concave cases. Take cm with the object first at cm and then at cm, work both out, and check that one gives a real inverted image and the other a virtual erect one — without looking up the table.
- New Cartesian convention: object on the left, distances from the pole, rightward positive, upward positive
- ** is always negative; concave mirror and negative, convex mirror and positive
- Negative means real and in front; positive means virtual and behind
- Mirror formula** , and magnification
- **Negative means inverted, positive means erect**; means enlarged
- A convex mirror can never give a real image, and a real mirror image is always inverted — use both as checks
- Laws of refraction: the three lines lie in one plane, and
- Rarer to denser bends towards the normal; denser to rarer bends away; along the normal there is no bending
- A rectangular slab gives an emergent ray parallel to the incident ray, displaced sideways — lateral displacement
- ****, with no unit, and
- Higher refractive index means optically denser and slower light — and optical density is not mass density
The sharpest self-test is the pair of concave cases. Take cm with the object first at cm and then at cm, work both out, and check that one gives a real inverted image and the other a virtual erect one — without looking up the table.