CLASS 10 SCIENCE · CHAPTER 9 · STUDY NOTES
Light – Reflection and Refraction
A comprehensive study of light's behavior when it interacts with mirrors and lenses, covering the laws of reflection, refraction, and the mathematical formulas used to predict image formation.
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Light reaches your eye
Trace a light path and measure reflection angles from the normal.LIGHT YOU CAN TRACE
Measure both angles from the normal.
The angles of incidence and reflection are measured from the normal. They are equal; the incident ray, reflected ray and normal lie in one plane.
- Normal
- A line perpendicular to the surface at the point where a ray meets it.
A luminous source emits light. We see a non-luminous object when light reflected from it reaches our eyes. In a uniform transparent medium, the chapter models light as straight rays.
At a reflecting surface, the angle of incidence equals the angle of reflection. Both angles are measured from the normal; the two rays and the normal lie in one plane.
Go deeper: What a plane mirror does
A plane-mirror image is erect, virtual and the same size as the object. Its apparent distance behind the mirror equals the object’s distance in front; the image is laterally inverted. Backward extensions locate it, but light does not travel behind the mirror.
A ray is a model of direction, not a narrow material thread. It works well for the mirror and lens constructions here. Diffraction at very small openings requires a wave description beyond this chapter’s ray treatment.
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Question
A ray makes 25° with a flat mirror surface. A learner labels its incidence angle 25°. Correct the label and predict the reflection angle.
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Find P, F and C on the mirror
Distinguish a curved reflecting surface from the imaginary sphere it belongs to.LIGHT YOU CAN TRACE
The mirror is part of an imaginary sphere.
C is the centre of the imaginary sphere. Parallel paraxial rays meet at F, approximately halfway from P to C.
- Principal focus
- For a concave mirror, the point where paraxial rays parallel to the principal axis converge; for a convex mirror, the point their backward extensions meet.
| Locate | Meaning | Where it lies |
|---|---|---|
| P | Pole | On the reflecting surface |
| C | Centre of the parent sphere | In front of a concave mirror; behind a convex mirror |
| F | Paraxial principal focus | Approximately midway between P and C |
A concave mirror faces inward towards its sphere’s centre; a convex mirror bulges outward. P is the pole on the reflecting surface. C is the centre of the imaginary sphere, not a point on the mirror itself.
The principal axis runs through P and C. For a small-aperture spherical mirror, F is approximately halfway between them: R = 2f. Parallel rays really meet at a concave focus but only appear to come from a convex focus.
Go deeper: Aperture matters
Aperture is the diameter of the reflecting opening. The simple focus relation applies to the small-aperture, paraxial approximation, where rays remain close to the principal axis. A large spherical aperture does not bring every parallel ray to exactly one point.
A source at the concave focus can produce an approximately parallel beam, as in a torch. Concentrating sunlight can produce intense heating; never look at the Sun directly, through a lens or in a reflecting mirror.
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Question
A spherical mirror’s C is 28 cm from P. Where is F in the small-aperture model, and is C part of the reflecting surface?
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Move the object through the focus
Explain changes in image orientation and size by tracing reflected rays.LIGHT YOU CAN TRACE
Move the object. Follow the reflected rays.
Object distance: 200 model units · |f| = 100
A ray parallel to the axis reflects through F. The second ray reflects at the pole, making equal angles with the axis. Where reflected rays meet, a real image forms.
Cartesian: u = −200 · f = -100 · v = -200.0 · |v| = 200.0 · |m| = 1.00
- Real image
- An image located where the outgoing light rays actually converge; it can be received on a screen.
| Object position | Image location | Orientation and relative size |
|---|---|---|
| Beyond C | Between C and F | Real, inverted, smaller |
| At C | At C | Real, inverted, equal-sized |
| Between C and F | Beyond C | Real, inverted, larger |
| At F | No finite image position | Reflected rays parallel |
| Inside F | Behind the mirror | Virtual, erect, larger |
For a concave mirror, an object beyond C forms a smaller inverted real image between C and F. At C the image is at C and equal in size. Moving the object between C and F gives a larger inverted real image beyond C.
At F, reflected rays are parallel: there is no finite image position. Inside F, their backward extensions meet behind the mirror, giving an enlarged, erect virtual image.
Go deeper: Choose rays for a reason
A parallel incident ray reflects through F; a ray through F reflects parallel to the axis. A ray directed through C retraces its path because it strikes normally. At P, incidence and reflection make equal angles with the principal axis. Two rays from the same object point locate its image point.
A very distant object produces an image near F. Magnifying uses, such as viewing teeth or a face, place the object inside F. Only the real-image cases can give a sharp image on a screen; dashed construction lines do not carry light.
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Question
An object is moved from C to a position halfway between C and F. Predict how a screen must move and whether the image will get larger.
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A wider view with a smaller image
Connect diverging reflected rays to an erect virtual image.LIGHT YOU CAN TRACE
A smaller, erect image behind the mirror.
Object distance: 200 model units · |f| = 100
A convex mirror spreads reflected rays. Backward extensions meet behind it, giving an upright, diminished virtual image.
Cartesian: u = −200 · f = 100 · v = 66.7 · |v| = 66.7 · |m| = 0.33
- Virtual image
- An image located where backward extensions meet rather than where outgoing light rays really converge.
For a real object in front of a convex mirror, the image is virtual, erect and diminished. It lies behind the mirror between P and F. A very distant object gives an image close to F.
A convex mirror offers a wider field of view than a plane mirror of comparable size. This makes it useful for vehicle side mirrors, although the image is smaller.
Go deeper: Behind the mirror is an apparent location
The reflected rays spread apart in front of the mirror. Extend them backwards with dashed lines to find the image; those extensions describe where light appears to originate. A screen behind the mirror cannot collect the reflected rays there.
As the object moves farther away, the image becomes smaller and approaches F. A convex mirror does not switch to an enlarged real image merely because the object crosses a distance equal to its focal-length magnitude.
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Question
A convex-mirror drawing puts a tall inverted image in front of the mirror. Which three properties should be corrected for a real object?
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Let signs describe the image
Use a signed coordinate system before substituting into the mirror formula.LIGHT YOU CAN TRACE
Location has a sign. Height has another.
Object distance: 200 model units · |f| = 100
Choose an object position, then read image location and orientation separately. Distances are signed from P; height is signed from the axis. The same calculated image follows the ray model.
- Magnification
- The signed ratio of image height to object height: m = h′/h.
Take P as the origin and incident light travelling from left to right. Distances left of P are negative and right of P positive; heights above the axis are positive and below negative. Thus a real object on the left has u < 0.
For spherical mirrors, 1/v + 1/u = 1/f and m = −v/u. A concave mirror has f < 0; a convex mirror has f > 0. Keep all distances in the same unit and interpret the signs after calculating.
Go deeper: Calculate, then check the picture
An original example uses a concave mirror with u = −24 cm and f = −8 cm: 1/v = −1/8 + 1/24 = −1/12, so v = −12 cm. Then m = −(−12)/(−24) = −0.5. A 6 cm upright object gives h′ = −3 cm: a real, inverted, smaller image.
For the ordinary real-object cases here, negative magnification means inverted and positive means erect. A magnitude above 1 means enlargement; below 1 means reduction. The sign says orientation, not whether an object is “negative sized”.
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Question
A convex mirror has u = −18 cm and f = +9 cm. Find v and m, then describe the image.
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