1. Introduction to Light
Light is a form of energy that travels in straight lines and enables us to see objects.
We see an object only when light from the object enters our eyes. Light can travel through air, glass and water. Under normal conditions light travels in straight lines – this idea is the basis of ray diagrams and is called rectilinear propagation of light.
Ray and Beam of Light
- A ray of light is the path along which light travels in a straight line. It is shown as a straight line with an arrowhead indicating the direction of travel.
- A beam of light is a bundle of parallel rays travelling together, for example the light from a torch or a projector.
Two Important Phenomena Used in This Chapter
| Phenomenon | What Happens | Everyday Example |
|---|---|---|
| Reflection | Light bounces back from a shiny surface into the same medium. | Seeing your face in a plane mirror. |
| Refraction | Light changes direction when it passes from one transparent medium into another. | A pencil appearing bent in a glass of water. |
Optical Terms (Basic)
Throughout this chapter you will come across a few standard terms: the incident ray (light falling on a surface), the reflected ray (light bouncing back), the refracted ray (light bending into a new medium), and the normal (an imaginary line perpendicular to the surface at the point where light strikes it). Each is explained in detail in the sections below.
2. Reflection of Light
Reflection of light is the bouncing back of light into the same medium when it falls on a shiny (polished) surface.
Terms Used in Reflection
| Term | Meaning |
|---|---|
| Incident ray | The ray of light that falls on the reflecting surface. |
| Point of incidence | The point on the surface where the incident ray strikes. |
| Reflected ray | The ray that bounces back after reflection. |
| Normal | The imaginary line drawn perpendicular to the reflecting surface at the point of incidence. |
| Angle of incidence (∠i) | The angle between the incident ray and the normal. |
| Angle of reflection (∠r) | The angle between the reflected ray and the normal. |
The Laws of Reflection
- First law: The incident ray, the reflected ray and the normal at the point of incidence all lie in the same plane.
- Second law: The angle of incidence is equal to the angle of reflection, that is ∠i = ∠r.
For any reflecting surface – plane or spherical – the two laws of reflection always hold: the three (incident ray, normal, reflected ray) are coplanar, and ∠i = ∠r.
Why Angles Are Measured from the Normal
The angles ∠i and ∠r are always measured with respect to the normal, never with respect to the mirror surface. This is a very common examination trick.
Regular Reflection vs Diffuse Reflection
| Feature | Regular (Specular) Reflection | Diffuse (Irregular) Reflection |
|---|---|---|
| Surface | Smooth, polished surface | Rough, uneven surface |
| Parallel rays | Reflect parallel to one another | Scatter in many directions |
| Image formed | Clear image | No clear image |
| Examples | Plane mirror, still water, polished metal | Paper, walls, bark of a tree |
Both regular and diffuse reflection obey the laws of reflection. In diffuse reflection each tiny ray still obeys ∠i = ∠r, but because the surface is uneven, the normals point in different directions and so the reflected rays scatter.
A page of printed paper reflects light diffusely, so light from a lamp reaches your eyes from any direction. A mirror, however, reflects regularly and only sends light in one direction.
3. Spherical Mirrors
A mirror whose reflecting surface is a part of a hollow sphere is called a spherical mirror.
Types of Spherical Mirrors
- Concave mirror: reflection takes place on the inner (curved-in) surface. It converges parallel light to a point, so it is called a converging mirror.
- Convex mirror: reflection takes place on the outer (curved-out) surface. It makes parallel light diverge, so it is called a diverging mirror.
Basic Terms for a Spherical Mirror
| Term | Meaning |
|---|---|
| Pole (P) | The centre of the mirror surface; the midpoint of the spherical shell. |
| Centre of curvature (C) | The centre of the hollow sphere of which the mirror is a part. |
| Radius of curvature (R) | The radius of the hollow sphere; the distance from the pole to the centre of curvature. |
| Principal axis | The straight line joining the pole and the centre of curvature. |
| Principal focus (F) | The point on the principal axis where rays parallel to the axis meet (or appear to meet) after reflection. |
| Focal length (f) | The distance between the pole and the principal focus. |
Relationship Between R and f
For a spherical mirror of small aperture, the focal length is half the radius of curvature:
R = 2f or f = R/2
The focus lies midway between the pole and the centre of curvature. For a concave mirror this focus is in front of the mirror; for a convex mirror it is behind the mirror (where the reflected rays appear to diverge from).
The relation R = 2f is strictly valid for mirrors of small aperture (small opening), where the mirror behaves almost like a parabola. In board-level questions it is always used directly.
Labelled Diagrams of Spherical Mirrors
Concave vs Convex Mirror at a Glance
| Feature | Concave Mirror | Convex Mirror |
|---|---|---|
| Reflecting surface | Inner (curved inwards) | Outer (bulging outwards) |
| Nature of light | Converges parallel rays | Diverges parallel rays |
| Focus | Real, in front of mirror | Virtual, behind mirror |
| Image by itself | Real or virtual | Always virtual, erect, diminished |
| Typical use | Shaving mirror, headlight, solar cooker | Rear-view mirror, security mirror |
4. Image Formation by a Concave Mirror
To locate the image of a point object placed in front of a concave mirror, we use two construction rays and find where they meet (or appear to meet) after reflection:
- A ray parallel to the principal axis reflects through the principal focus F.
- A ray passing through the centre of curvature C reflects back along the same path.
- A ray striking the pole P reflects making an equal angle (∠i = ∠r) on the other side of the principal axis.
Image for Different Object Positions
| Position of Object | Position of Image | Size of Image | Nature of Image | Orientation |
|---|---|---|---|---|
| At infinity | At the focus F | Highly diminished, point-sized | Real | Inverted |
| Beyond C | Between F and C | Diminished (smaller than object) | Real | Inverted |
| At C | At C | Same size as the object | Real | Inverted |
| Between C and F | Beyond C | Enlarged (larger than object) | Real | Inverted |
| At F | At infinity | Highly enlarged | Real | Inverted |
| Between F and P | Behind the mirror | Enlarged | Virtual | Erect |
Object at C → image at C, same size: a ray through C retraces its path and a parallel ray reflects through F; the two meet again at C. Object between C and F → image beyond C, enlarged: the positions simply swap compared with the "beyond C" case (image and object are conjugate).
Only concave mirrors can form real (inverted) images and magnified virtual images. The enlargeable upright image appears only when the object lies between the focus and the pole.
5. Image Formation by a Convex Mirror
Object at Infinity
Parallel rays falling on a convex mirror diverge after reflection and appear to come from the focus F. The image forms at the focus F, behind the mirror. It is virtual, erect and highly diminished (point-sized).
The image formed by a convex mirror is always virtual and erect – it is not real. For an object at infinity the virtual, point-sized, erect image appears at the focus F behind the mirror.
Object at a Finite Distance
For any finite object position in front of a convex mirror, the image is formed between the pole P and the focus F, behind the mirror, and is virtual, erect and diminished (smaller than the object).
General Characteristics of a Convex Mirror Image
- Always forms a virtual image (cannot be obtained on a screen).
- Always erect (upright).
- Always diminished (smaller than the object).
- Image lies behind the mirror, between the pole and the focus.
Because a convex mirror always gives a diminished, erect, virtual image, it shows a much wider field of view than a plane mirror of the same size – that is why it is used as a vehicle rear-view mirror (see section 8).
6. Mirror Formula
1/f = 1/v + 1/u
where u = object distance, v = image distance, f = focal length, all measured from the pole of the mirror along the principal axis.
This formula connects the object distance, the image distance and the focal length of any spherical mirror. It works for both concave and convex mirrors as long as the Cartesian sign convention is applied.
Cartesian Sign Convention for Mirrors
- The pole P of the mirror is taken as the origin.
- The principal axis is the X-axis; distances are measured from the pole along the axis.
- Distances measured in the direction of the incident light (from the mirror towards the right) are taken as positive.
- Distances measured against the direction of the incident light (to the left of the pole) are taken as negative.
- Heights measured upwards (above the axis) are positive; heights measured downwards are negative.
Simple Rules of Thumb (Mirrors)
| Quantity | Sign |
|---|---|
| Object distance u (real object in front) | Always negative |
| Focal length f of a concave mirror | Negative |
| Focal length f of a convex mirror | Positive |
| Image distance v (real image, in front) | Negative |
| Image distance v (virtual image, behind) | Positive |
| Height above the axis (erect image) | Positive |
| Height below the axis (inverted image) | Negative |
Always write the values of u and f with their signs, substitute into 1/v = 1/f − 1/u, and solve. Finally, interpret the sign of v: negative → real (front), positive → virtual (behind). Worked examples are given in section 22.
Practical note: the two construction rays used for drawing (parallel ray to F, ray through C retracing) come from the laws of reflection and confirm the mirror formula geometrically.
7. Magnification by Spherical Mirrors
Magnification is the ratio of the height of the image to the height of the object.
m = hi/ho = −v/u
where hi = height of the image, ho = height of the object, v = image distance, u = object distance.
Special Cases (Mirrors)
- m > 0 (positive) means the image is erect and virtual (as for convex mirrors and the object-between-F-and-P case of the concave mirror).
- m < 0 (negative) means the image is inverted and real.
- |m| = 1 → image same size as the object (object at C of a concave mirror).
- |m| > 1 → magnified; |m| < 1 → diminished.
The negative sign in m = −v/u is part of the formula for mirrors. If you forget it you will get the wrong orientation. (Note: for lenses this same formula has a different sign – see section 17.)
8. Uses of Mirrors
Uses of Concave Mirrors
| Use | Why a Concave Mirror |
|---|---|
| Shaving / make-up mirror | When the face is held between the pole and the focus, the concave mirror forms an enlarged, erect, virtual image, so details are seen clearly. |
| Dentist’s mirror | A small concave mirror gives a large, upright image of a tooth kept between F and P, helping the dentist inspect it closely. |
| Headlights / searchlights | The bulb is placed at the focus F; rays reflected by the concave mirror become a parallel beam that travels far. |
| Solar furnace / solar cooker | Parallel sun rays are concentrated onto the focus F, producing a strong heating point. |
Uses of Convex Mirrors
| Use | Why a Convex Mirror |
|---|---|
| Rear-view mirror in vehicles | A convex mirror always gives an erect, virtual, diminished image with a much wider field of view, so the driver can see more traffic behind. |
| Security / shop mirrors | Convex mirrors let a shopkeeper watch a large area (many aisles) at once. |
9. Refraction of Light
The bending of light when it travels from one transparent medium into another is called refraction of light.
Why Does Light Bend?
Light travels at different speeds in different media. When light goes from one transparent medium to another, its speed changes, which makes it change direction – this change of direction at the boundary is refraction.
| Term | Meaning |
|---|---|
| Optically denser medium | A medium in which light travels slower (e.g. water, glass). |
| Optically rarer medium | A medium in which light travels faster (e.g. air). |
Optical density is a measure of the speed of light in the medium, not of how heavy it is. A substance can be heavier (higher mass density) yet be optically rarer than another. For example, kerosene is lighter than water in mass but is optically denser (light is slower in kerosene).
Bending of Light
- From rarer to denser (air → glass): speed decreases and light bends towards the normal.
- From denser to rarer (glass → air): speed increases and light bends away from the normal.
A pencil partly dipped in water looks bent at the surface: the rays from the submerged part refract as they leave the water, so the eye sees the pencil displaced. Similarly, a pond looks shallower than it really is.
10. Laws of Refraction
- First law: The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
- Second law (Snell’s law): For a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant.
Snell’s law is written as:
sin i / sin r = constant (n)
The constant is the refractive index of the second medium with respect to the first medium. Snell’s law is true for every angle; it is the refraction analogue of ∠i = ∠r for reflection.
11. Refractive Index
The refractive index of a transparent medium is the ratio of the speed of light in vacuum to the speed of light in that medium:
n = c / v
where c = speed of light in vacuum (3 × 108 m s−1) and v = speed of light in the medium.
- Because light is fastest in vacuum, n is always > 1 for any real medium.
- Higher n → slower light in the medium → the medium is optically denser.
- Refractive index has no unit (it is a ratio).
Relative Refractive Index
When we say "refractive index of glass" we mean glass with respect to vacuum (or air), n = 1.5. If light goes from medium A into medium B, the relative refractive index of B with respect to A is:
nBA = vA / vB = nB / nA
It tells us how many times slower (or faster) light travels in B compared with A.
Refractive Indices of Some Common Media (Standard Values)
| Medium | Refractive index (n) |
|---|---|
| Air (approx. vacuum) | 1.00 |
| Water | 1.33 |
| Glass (crown) | 1.50 |
| Glass (flint) | 1.65 |
| Carbondisulphide | 1.63 |
| Diamond | 2.42 |
The values above are standard published reference values (typically at yellow light). They are commonly quoted in Class 10 textbooks; use them only in the numericals given here.
12. Refraction Through a Rectangular Glass Slab
When a ray of light passes through a rectangular (parallel-sided) glass slab, it is refracted twice: once entering (air → glass) and once leaving (glass → air).
| Term | Meaning |
|---|---|
| Incident ray | The ray falling on the slab from air. |
| Refracted ray | The ray inside the slab, bent towards the normal at the first surface. |
| Emergent ray | The ray leaving the slab at the second surface, bent away from the normal. |
| Lateral displacement | The perpendicular distance between the incident ray and the emergent ray. |
Since the two faces are parallel, the emergent ray is parallel to the incident ray, but it is shifted sideways. This sideways shift is the lateral displacement.
Why Lateral Displacement Happens
On entering the glass the ray bends towards the normal; on leaving, it bends away from the normal by the same amount (because the surfaces are parallel and the media are the same at both ends). So the direction inside the slab cancels back, leaving the emergent ray parallel to the incident ray but displaced sideways. The displacement is larger for thicker slabs and for rays entering at larger angles of incidence.
13. Spherical Lenses
A lens is a transparent medium (usually glass) bounded by two curved surfaces (or one curved and one plane surface). A lens refracts light to form images.
| Feature | Convex Lens | Concave Lens |
|---|---|---|
| Middle vs edges | Thicker in the middle, thinner at the edges | Thinner in the middle, thicker at the edges |
| Action on parallel light | Converges to a point → converging lens | Diverges → diverging lens |
| Focal length sign | Positive | Negative |
| Image formed | Real (or virtual for close objects) | Always virtual, erect, diminished |
Basic Terms for a Lens
| Term | Meaning |
|---|---|
| Principal axis | The line joining the centres of curvature of the two surfaces of the lens. |
| Optical centre (O) | The central point of the lens through which a ray passes without deviation. |
| Principal focus (F) | The point on the principal axis where rays parallel to the axis converge (convex) or appear to diverge from (concave) after refraction. |
| Focal length (f) | The distance between the optical centre and the principal focus. A lens has two foci, F1 and F2, one on each side. |
14. Image Formation by a Convex Lens
Two construction rays locate the image for a convex lens:
- A ray parallel to the principal axis refracts through the principal focus F₂ on the other side.
- A ray passing through the optical centre O goes straight, without deviation.
| Position of Object | Position of Image | Size of Image | Nature | Orientation |
|---|---|---|---|---|
| At infinity | At focus F₂ | Highly diminished, point-sized | Real | Inverted |
| Beyond 2F₁ | Between F₂ and 2F₂ | Diminished | Real | Inverted |
| At 2F₁ | At 2F₂ | Same size as the object | Real | Inverted |
| Between F₁ and 2F₁ | Beyond 2F₂ | Enlarged | Real | Inverted |
| At F₁ | At infinity | Highly enlarged | Real | Inverted |
| Between O and F₁ | On the same side as the object | Enlarged | Virtual | Erect |
For a convex lens, objects far away give real, inverted images (camera and projector behaviour); objects very close (between O and F₁) give a virtual, erect, enlarged image (magnifying glass).
15. Image Formation by a Concave Lens
Object at Infinity
Parallel rays diverge after passing through a concave lens and appear to come from the focus F₁ (on the object side). The image is virtual, erect, highly diminished and point-sized, at the focus.
Object at a Finite Distance
For any finite object position, the image is formed between the focus F₁ and the optical centre O, on the same side as the object, and is virtual, erect and diminished (smaller than the object).
16. Lens Formula
1/f = 1/v − 1/u
where u = object distance, v = image distance, f = focal length, all measured from the optical centre O.
Mirror formula: 1/f = 1/v + 1/u (plus sign)
Lens formula: 1/f = 1/v − 1/u (minus sign)
The mirror formula uses a plus; the lens formula uses a minus in front of 1/u. Getting the sign wrong is the most common error in this chapter.
Cartesian Sign Convention for Lenses
- The optical centre O of the lens is taken as the origin.
- Distances measured in the direction of the incident light (towards the right) are positive.
- Distances measured against the incident light (towards the left) are negative.
- The focal length of a convex lens is positive; that of a concave lens is negative.
- Heights above the axis are positive, below the axis negative.
17. Magnification by a Lens
m = hi/ho = v/u
Unlike mirrors, the lens magnification is v/u (no minus sign).
Interpreting Sign and Magnitude (Lenses)
- m > 0 → erect, virtual image (e.g. magnifying glass, concave lens).
- m < 0 → inverted, real image (most convex lens images).
- |m| > 1 enlarged, |m| = 1 same size, |m| < 1 diminished.
Mirror: m = −v/u. Lens: m = v/u. The extra negative sign belongs to mirrors only.
18. Power of a Lens
The power of a lens is the reciprocal of its focal length measured in metres:
P = 1 / f (f in metres)
The SI unit of power is the dioptre (D): 1 D = 1 m−1.
- Positive power (P > 0) → convex (converging) lens.
- Negative power (P < 0) → concave (diverging) lens.
- Higher power → shorter focal length → stronger lens (bends light more sharply).
Focal length must be converted to metres before finding power:
1 cm = 0.01 m (so f = 40 cm means f = 0.40 m). Forgetting this conversion is a classic error.
For example, a lens of focal length +50 cm has power P = 1/0.50 = +2 D (convex); a lens of focal length −25 cm has power P = 1/(−0.25) = −4 D (concave).
19. Important Formulas (Summary)
| Topic | Formula | Notes |
|---|---|---|
| Focal length of spherical mirror | R = 2f | R and f measured from the pole; sign depends on the mirror. |
| Mirror formula | 1/f = 1/v + 1/u | All distances from the pole; u is always negative. |
| Mirror magnification | m = hi/ho = −v/u | Negative m → inverted real image. |
| Refractive index | n = c/v | c = 3 × 108 m s−1; n has no unit. |
| Snell’s law | sin i / sin r = constant | Constant equals n of the second medium w.r.t. the first. |
| Lens formula | 1/f = 1/v − 1/u | All distances from the optical centre. |
| Lens magnification | m = hi/ho = v/u | No negative sign for lenses. |
| Power of a lens | P = 1/f (f in metres) | Unit dioptre (D); positive convex, negative concave. |
Write Given (with signs) → Formula → Substitution → Calculation → Final Answer with units. Then interpret the signs of v and m words: real/virtual, erect/inverted, magnified/diminished.
20. Cartesian Sign Convention
All distances are measured from the pole P of a mirror or the optical centre O of a lens, with the incident light taken as moving from left to right.
- Distances measured in the direction of the incident light are positive.
- Distances measured opposite to the incident light are negative.
- Distances measured upwards (above the principal axis) are positive.
- Distances measured downwards (below the principal axis) are negative.
For a real object, u is always negative. Heights measured upward (object height and erect image height) are positive; inverted-image heights are negative.
| Quantity | Concave Mirror | Convex Mirror | Convex Lens | Concave Lens |
|---|---|---|---|---|
| Focal length f | Negative | Positive | Positive | Negative |
| Object distance u (real object) | Negative | Negative | Negative | Negative |
| Image distance v | Negative (real) / positive (virtual) | Positive (always virtual) | Positive (real) / negative (virtual) | Negative (always virtual) |
| Power P | − | + | + | − |
The sign convention is applied automatically when the mirror formula 1/f = 1/v + 1/u or the lens formula 1/f = 1/v − 1/u is used (see Figures 7 and 16 in sections 6 and 16).
21. Important Diagrams (Reference)
All diagrams in these notes are drawn cleanly with the construction rays explained. The following table lists the diagrams you must be able to draw and label.
| # | Diagram | Key labels |
|---|---|---|
| Fig. 1 | Laws of reflection (plane mirror) | Incident ray, normal, reflected ray, ∠i, ∠r, point of incidence |
| Fig. 2 | Concave mirror terminology | Pole P, centre of curvature C, focus F, principal axis, R |
| Fig. 3 | Convex mirror terminology | P, C, F (behind mirror), principal axis |
| Fig. 4 | Concave mirror – object beyond C | C, F, P, image position/half-arrow |
| Fig. 5 | Concave mirror – object between F and P | C, F, P, virtual image (dashed) |
| Fig. 6 | Convex mirror ray diagram | P, F, virtual image (dashed) |
| Fig. 7 | Cartesian sign convention for mirrors | Pole as origin, +x / −x directions |
| Fig. 8 | Refraction air → glass | Normal, ∠i, ∠r, rarer/denser labels |
| Fig. 9 | Refraction glass → air | Normal, ∠i, ∠r |
| Fig. 10 | Refraction through a glass slab | Incident, refracted, emergent rays; lateral displacement |
| Fig. 11 | Convex lens terminology | O, F₁, F₂, 2F₁, 2F₂, principal axis |
| Fig. 12 | Concave lens terminology | O, F₁, F₂, principal axis |
| Fig. 13 | Convex lens – object beyond 2F₁ | O, F₁, F₂, 2F₁, 2F₂ |
| Fig. 14 | Convex lens as magnifier | O, F₁, virtual image (dashed) |
| Fig. 15 | Concave lens ray diagram | O, F₁, F₂, virtual image (dashed) |
| Fig. 16 | Cartesian sign convention for lenses | Optical centre as origin, +x / −x |
Use a ruler, draw the principal axis and the mirror/lens first, then the construction rays with arrowheads, mark P/C/F (or O/F₁/F₂/2F) with dots and letters, and finish with the object and image as vertical half-arrows. Virtual images are drawn with dashed half-arrows.
22. Solved Numerical Examples
All examples below are original, CBSE exam-style questions. None is claimed as a CBSE previous-year question.
A. Basic Conceptual
State the two laws of reflection of light.
Show Solution
Answer: (1) The incident ray, the reflected ray and the normal at the point of incidence lie in the same plane. (2) The angle of incidence equals the angle of reflection: ∠i = ∠r.
Why do parallel rays scattered by a rough surface not form a clear image, even though reflection still obeys the laws?
Show Solution
Reasoning: On a rough surface each tiny patch has a different normal. Each ray still respects ∠i = ∠r locally, but reflected rays point in many directions (diffuse reflection), so a clear image cannot form.
Final Answer: Diffusion of rays because of differently oriented normals.
A ray falls normally on a plane mirror. What are the angle of incidence and the angle of reflection?
Show Solution
Solution: Falling normally means the ray is along the normal, so the angle with the normal is 0°.
Final Answer: ∠i = 0° and ∠r = 0° → the ray reflects back along its own path.
The radius of curvature of a spherical mirror is 30 cm. Find its focal length and state the type of mirror if the focus is real.
Show Solution
Given: R = 30 cm.
Formula: f = R/2.
Calculation: f = 30/2 = 15 cm.
Final Answer: f = 15 cm. A real focus lies in front of the mirror, so the mirror is concave.
Is the relation R = 2f exact for all mirrors? Give a one-line reason.
Show Solution
Answer: It is exact only for spherical mirrors of small aperture, where the spherical surface behaves like a parabola and rays near the axis meet at one focus.
B. Mirror Numericals
An object is placed 30 cm in front of a concave mirror of focal length 15 cm. Find the image position and nature.
Show Solution
Given: u = −30 cm, f = −15 cm.
Formula: 1/f = 1/v + 1/u.
Substitution: 1/v = 1/(−15) − 1/(−30) = −1/15 + 1/30 = −1/30.
Calculation: v = −30 cm.
Final Answer: Image at 30 cm in front of the mirror (object at C): real, inverted, same size.
An object is placed 60 cm in front of a concave mirror of focal length 20 cm. Locate the image.
Show Solution
Given: u = −60 cm, f = −20 cm.
Formula: 1/v = 1/f − 1/u = −1/20 + 1/60 = (−3 + 1)/60 = −2/60 = −1/30.
Calculation: v = −30 cm.
Final Answer: Real, inverted, diminished image 30 cm in front of the mirror (between F and C).
An object is placed 10 cm in front of a concave mirror of focal length 20 cm. Describe the image.
Show Solution
Given: u = −10 cm, f = −20 cm (object lies between F and the pole).
Formula: 1/v = 1/f − 1/u = −1/20 + 1/10 = (−1 + 2)/20 = 1/20.
Calculation: v = +20 cm (positive → behind the mirror).
Final Answer: Virtual, erect, enlarged image 20 cm behind the mirror.
A convex mirror has a focal length of 20 cm. An object is 60 cm in front of it. Find the image distance and state the image nature.
Show Solution
Given: u = −60 cm, f = +20 cm.
Formula: 1/v = 1/f − 1/u = 1/20 + 1/60 = (3 + 1)/60 = 4/60 = 1/15.
Calculation: v = +15 cm.
Final Answer: Virtual, erect, diminished image 15 cm behind the mirror (between P and F).
A concave mirror of focal length 30 cm forms a real image three times the size of the object. Find the object position.
Show Solution
Given: f = −30 cm. Real and 3× means m = −3.
Formulas: m = −v/u = −3 → v = 3u. Substitute into 1/f = 1/v + 1/u.
Substitution: −1/30 = 1/(3u) + 1/u = (1/3 + 1)/u = 4/(3u) → u = −40 cm.
Final Answer: Object at 40 cm in front of the mirror; image v = −120 cm.
An object in front of a spherical mirror gives a virtual erect image twice the object size when placed 30 cm away. Find the mirror's focal length and identify it.
Show Solution
Given: u = −30 cm, m = +2.
Formula: m = −v/u → 2 = −v/(−30) → v = +60 cm.
Mirror formula: 1/f = 1/v + 1/u = 1/60 − 1/30 = −1/60 → f = −60 cm.
Final Answer: f = −60 cm, R = −120 cm; focal length negative → concave mirror.
A convex mirror produces a virtual image half the size of an object placed 30 cm in front. Find the focal length and radius.
Show Solution
Given: u = −30 cm, m = +1/2.
Formula: m = −v/u → v = −mu = +15 cm.
Mirror formula: 1/f = 1/15 − 1/30 = 1/30 → f = +30 cm.
Final Answer: f = +30 cm, R = 2f = +60 cm.
An object is 20 cm from a concave mirror and its real image forms 60 cm from the mirror on the same side. Find f and R.
Show Solution
Given: u = −20 cm, v = −60 cm (real, in front).
Formula: 1/f = 1/v + 1/u = −1/60 − 1/20 = (−1 − 3)/60 = −4/60 = −1/15.
Calculation: f = −15 cm; R = 2f = −30 cm.
Final Answer: f = 15 cm (focal length magnitude), the mirror is concave.
C. Magnification Problems
A 4 cm tall object gives an inverted image 12 cm tall. Find the magnification and its meaning.
Show Solution
Given: ho = +4 cm, hi = −12 cm (inverted).
Formula: m = hi/ho = −12/4 = −3.
Final Answer: m = −3: image is 3× larger, inverted and real.
What does a magnification of +1/4 tell you about an image? Give real and virtual features.
Show Solution
Answer: m = +1/4 means the image is erect, virtual, one-fourth the size of the object, and (for a mirror) formed behind the mirror – a convex-mirror situation.
An image forms at 100 cm from a concave mirror while the object is 20 cm from it. Find the magnification.
Show Solution
Given: u = −20 cm, v = −100 cm (real image in front).
Formula: m = −v/u = −(−100)/(−20) = −5.
Final Answer: m = −5 → inverted, real, 5× enlarged.
A mirror gives a virtual erect image twice the size of a 5 cm object. Find the image height and the sign of m.
Show Solution
Given: ho = +5 cm, m = +2.
Formula: hi = m × ho = 2 × 5 = +10 cm.
Final Answer: hi = +10 cm (erect, virtual image); m positive.
A convex mirror forms an image one-third the object size when the object is 45 cm away. Find the image distance and the object's image height for a 9 cm object.
Show Solution
Given: u = −45 cm, m = +1/3, ho = +9 cm.
Formula: m = −v/u → v = −mu = −(1/3)(−45) = +15 cm. hi = m ho = 3 cm.
Final Answer: v = +15 cm, hi = +3 cm (erect, diminished).
D. Refraction and Refractive Index
Light travels at 2 × 108 m s−1 in a glass. Find the refractive index of the glass (c = 3 × 108 m s−1).
Show Solution
Given: v = 2 × 108 m s−1, c = 3 × 108 m s−1.
Formula: n = c/v.
Calculation: n = 3/2 = 1.5.
Final Answer: n = 1.5.
The refractive index of water is 1.33. Find the speed of light in water.
Show Solution
Given: n = 1.33, c = 3 × 108 m s−1.
Formula: v = c/n.
Calculation: v = (3 × 108)/1.33 = 2.26 × 108 m s−1.
Final Answer: v ≈ 2.26 × 108 m s−1.
Diamond has a refractive index of 2.42. Compute the speed of light in diamond.
Show Solution
Formula: v = c/n = (3 × 108)/2.42 = 1.24 × 108 m s−1.
Final Answer: v ≈ 1.24 × 108 m s−1 – diamond is optically much denser than glass.
For a light ray entering glass, the angle of incidence is 60° and the angle of refraction is 30°. Find the refractive index of glass with respect to air (sin 60° = 0.866, sin 30° = 0.5).
Show Solution
Formula (Snell): n = sin i / sin r.
Calculation: n = 0.866 / 0.5 = 1.732.
Final Answer: n = 1.73 = √3.
nglass = 1.5 and nwater = 1.33. Find the refractive index of glass relative to water.
Show Solution
Formula: ng w.r.t. w = ng/nw.
Calculation: 1.5/1.33 = 1.13.
Interpretation: light is about 1.13× slower in glass than in water, so it bends towards the normal on entering glass from water.
E. Lens Numericals
An object is 30 cm from a convex lens of focal length 20 cm. Find the image position and nature.
Show Solution
Given: u = −30 cm, f = +20 cm.
Formula: 1/v = 1/f + 1/u = 1/20 − 1/30 = (3 − 2)/60 = 1/60.
Calculation: v = +60 cm; m = v/u = 60/(−30) = −2.
Final Answer: Real, inverted, enlarged image 60 cm on the other side of the lens.
An object is 30 cm from a convex lens of focal length 10 cm. Locate the image.
Show Solution
Formula: 1/v = 1/10 − 1/30 = (3 − 1)/30 = 2/30 = 1/15 → v = +15 cm.
Magnification: m = 15/(−30) = −1/2.
Final Answer: Real, inverted, diminished image 15 cm on the other side.
An object is held 10 cm from a convex lens of focal length 15 cm. Describe the image.
Show Solution
Given: u = −10 cm, f = +15 cm (object between O and F₁).
Formula: 1/v = 1/15 − 1/10 = (2 − 3)/30 = −1/30 → v = −30 cm.
Magnification: m = v/u = (−30)/(−10) = +3.
Final Answer: Virtual, erect, 3× enlarged image, 30 cm on the same side as the object.
An object is 30 cm from a concave lens of focal length 15 cm. Find the image position and magnification.
Show Solution
Given: u = −30 cm, f = −15 cm.
Formula: 1/v = 1/f + 1/u = −1/15 − 1/30 = (−2 − 1)/30 = −3/30 = −1/10 → v = −10 cm.
Magnification: m = v/u = (−10)/(−30) = +1/3.
Final Answer: Virtual, erect, one-third size, 10 cm on the object side.
A lens forms a virtual image five times the object size when the object is 10 cm away. Find the focal length and lens type.
Show Solution
Given: u = −10 cm, m = +5 → v = mu = −50 cm.
Lens formula: 1/f = 1/v − 1/u = −1/50 + 1/10 = 4/50 = 2/25 → f = +12.5 cm.
Final Answer: f = 12.5 cm; positive focal length → convex lens used as a magnifying glass.
F. Power of a Lens
Find the power of a lens of focal length 40 cm. State its type.
Show Solution
Given: f = 40 cm = 0.40 m (conversion: 1 cm = 0.01 m).
Formula: P = 1/f.
Calculation: P = 1/0.40 = +2.5 D.
Final Answer: P = +2.5 D; positive power → convex (converging) lens.
A lens has a power of −4 D. Find its focal length in metres and centimetres, and identify the lens.
Show Solution
Formula: f = 1/P.
Calculation: f = 1/(−4) = −0.25 m = −25 cm.
Final Answer: f = −25 cm; negative power → concave (diverging) lens.
23. Application-Based Examples
Original – CBSE Exam-Style
Why is a convex mirror preferred as the rear-view mirror in vehicles instead of a plane or concave mirror?
Show Solution
Physics: A convex mirror gives a virtual, erect, diminished image and a much wider field of view than a plane mirror of the same size. A concave mirror would show an inverted image in most positions.
Final Answer: Wide upright field of view keeps more traffic visible, helping the driver judge vehicles behind safely.
Why does a shaving mirror enlarge your face, and where should your face be relative to the mirror?
Show Solution
Physics: A concave mirror magnifies when the object (your face) is between the pole and the focus; the image is virtual, erect and enlarged.
Final Answer: Hold the face between F and P of the concave mirror to see an enlarged, upright image.
A dentist uses a small concave mirror with a long focal length. Why is this helpful for examining a tooth?
Show Solution
Physics: Held close, the tooth lies between the focus and the pole, giving a magnified erect image; the longer focal length gives a comfortable working distance and a larger |m| for close objects.
Final Answer: It gives an enlarged, upright image, so small cavities are visible clearly.
In a car headlight the bulb is placed at the focus of a concave reflector. What happens to the reflected rays, and why is that useful?
Show Solution
Physics: Rays from the focus become a parallel beam on reflection (the ray diagram reversed).
Final Answer: The reflected light stays a parallel beam, so it travels far ahead without spreading, illuminating the road well.
Explain the working of a solar furnace that concentrates sunlight.
Show Solution
Physics: Parallel rays from the Sun fall on a large concave reflector and converge at the focus.
Final Answer: Sunlight is concentrated on the focus, producing a strong, hot point useful for heating or cooking.
Why is a convex lens with a short focal length used as a magnifying glass?
Show Solution
Physics: When the object is between O and F₁, a convex lens gives a virtual, erect, enlarged image. A short f means the object can be kept close yet still produce a large magnified image.
Final Answer: The lens magnifies because the object lies between the optical centre and the focus.
A long-sighted person is prescribed a convex lens of suitable power. How does the lens help him read nearby text?
Show Solution
Physics: A convex (converging) lens bends the light more so that the eye forms a sharp image of a nearby object on the retina. (The concave lens helps a short-sighted person by diverging light slightly.)
Final Answer: The convex lens converges the rays that would otherwise focus beyond the retina, giving a clear near image.
Why does a pencil appear bent when partly dipped in water, and what does the same principle predict for light emerging from a glass slab?
Show Solution
Physics: Rays from the submerged part refract while leaving water (denser to rarer), so they bend away from the normal and the eye sees the pencil displaced. In a glass slab the emergent ray is parallel to the incident ray, just shifted sideways.
Final Answer: Refraction at the water surface bends the rays, bending the apparent pencil; a glass slab shifts the emergent ray laterally but keeps it parallel.
24. Competency-Based and Reasoning Questions
Original – CBSE Exam-Style
A student needs a mirror that always forms an erect, diminished, virtual image to observe a wide room. Choose the correct mirror and justify.
Show Solution
Reasoning: A convex mirror always gives an erect, diminished, virtual image with a wide field of view.
Final Answer: Convex mirror – it is the only mirror that satisfies all three conditions for every object position.
An object is placed (a) at C and (b) between F and P of a concave mirror. Predict the image nature in each case and state which can be obtained on a screen.
Show Solution
Solution: (a) At C: real, inverted, same size – can be caught on a screen. (b) Between F and P: virtual, erect, enlarged, behind the mirror – cannot be caught on a screen.
In a ray diagram for a concave mirror, a ray parallel to the principal axis passes through F after reflection. Which physical law guarantees this?
Show Solution
Reasoning: The two laws of reflection (specifically ∠i = ∠r with the normal to the spherical surface). For a small-aperture spherical mirror all such reflected rays cross at the focus F.
Why is the object distance u always taken as negative in the sign convention for both mirrors and lenses?
Show Solution
Reasoning: The origin is at the pole/optical centre and the incident light travels from left to right. A real object is on the left, i.e. against the direction of incident light, so u is negative.
A student is asked to find the image distance for a mirror. When should he use the mirror formula and when the lens formula?
Show Solution
Reasoning: The mirror formula 1/f = 1/v + 1/u applies when light reflects (mirrors); the lens formula 1/f = 1/v − 1/u applies when light refracts (lenses).
Final Answer: Use mirror formula for any spherical mirror, lens formula for any lens – never interchange them.
A lens has power +2 D. (a) Identify the lens. (b) Find its focal length in cm. (c) Where must an object be for this lens to act as a magnifier?
Show Solution
Solution: (a) Positive power → convex lens. (b) f = 1/2 = 0.5 m = 50 cm. (c) Object between the optical centre and F (less than 50 cm from the lens).
25. Case-Based Learning
Original – CBSE Exam-Style
Scenario: A driver fits a rear-view mirror in his car and notices that the vehicles behind appear smaller but upright.
(a) Which mirror is fitted? (b) Why do vehicles appear smaller yet upright? (c) Why does this help the driver? (d) How would the appearance change if a plane mirror of the same size were used?
Show Solution
Concept: Curvature of a convex mirror and field of view.
Solution: (a) Convex mirror. (b) It always forms a virtual, erect, diminished image. (c) The widened field of view lets the driver see more traffic without turning. (d) A plane mirror would give a same-size, erect image but a much narrower field of view.
Scenario: A student holds a coin close to a lens and sees a magnified, upright image; when the coin is moved far away, the image becomes inverted and smaller on the other side.
(a) Which lens is it? (b) Why did the image change from upright to inverted? (c) Where was the coin when the image was upright? (d) If f = 10 cm, and the coin was 6 cm away, where was the image?
Show Solution
Concept: Convex lens image formation for different object positions.
Solution: (a) Convex lens. (b) Close objects (between O and F₁) give virtual erect images; objects beyond F₁ give real inverted images. (c) Between the optical centre and F₁. (d) u = −6 cm, f = +10 cm: 1/v = 1/10 − 1/6 = (3 − 5)/30 = −2/30 → v = −15 cm (virtual, erect, enlarged, same side).
Scenario: In a laboratory, a laser beam enters a rectangular glass slab at 45° and leaves it shifted sideways but parallel to its original path.
(a) Name the phenomenon. (b) Why is the emergent ray parallel to the incident ray? (c) What is the sideways shift called? (d) What would happen if the two faces were not parallel?
Show Solution
Concept: Double refraction through parallel faces.
Solution: (a) Refraction. (b) Bending towards the normal on entry is cancelled by the equal bending away from the normal at the parallel second face. (c) Lateral displacement. (d) The emergent ray would not be parallel to the incident ray (e.g. a prism deviates the ray).
Scenario: A concave reflector in a solar cooker concentrates sunlight onto a single bright point where food can be cooked.
(a) What happens to the parallel sun rays? (b) Where is the bright point located? (c) Name the mirror behaviour used here. (d) Safety: why should food not block all the light?
Show Solution
Concept: Convergence of parallel rays by a concave mirror.
Solution: (a) They reflect and converge. (b) At the principal focus F. (c) Converging (focusing) of parallel light by a concave mirror. (d) The whole concentrated energy lands at F; leaving the focus clear keeps the effect safe and efficient.
Scenario: A long-sighted student needs +1.5 D spectacles.
(a) Identify the lens in the spectacles. (b) Find its focal length in cm. (c) Introduce conversion: 1 cm = 0.01 m. (d) What kind of image must this lens form so the student can read a book held close?
Show Solution
Concept: Power and convergence of a convex lens.
Solution: (a) Convex (converging) lens. (b) f = 1/P = 1/1.5 = 0.667 m = 66.7 cm. (c) A focal length of 0.667 m equals 66.7 cm since 1 cm = 0.01 m. (d) The lens must converge the rays so that the eye sees a sharp, erect (upright) image of the nearby text.
26. Important Comparison Tables
Concave Mirror vs Convex Mirror
| Feature | Concave Mirror | Convex Mirror |
|---|---|---|
| Reflecting surface | Inner (curved in) | Outer (bulging out) |
| Focus | Real, in front | Virtual, behind |
| Focal length sign | Negative | Positive |
| Image | Real (mostly) or virtual | Always virtual, erect, diminished |
| Use | Shaving mirror, headlight, solar cooker | Rear-view and security mirrors |
Concave Lens vs Convex Lens
| Feature | Convex Lens | Concave Lens |
|---|---|---|
| Shape | Thicker in the middle | Thinner in the middle |
| Action | Converges light | Diverges light |
| Focal length | Positive | Negative |
| Image | Real or virtual (magnifier) | Always virtual, erect, diminished |
| Use | Camera, magnifying glass, projector | Correcting short-sightedness |
Reflection vs Refraction
| Feature | Reflection | Refraction |
|---|---|---|
| What happens | Light bounces back in the same medium | Light bends into a new medium |
| Speed of light | Unchanged | Changes |
| Law | ∠i = ∠r | sin i / sin r = constant |
| Requires | A reflecting surface | Two transparent media |
Real Image vs Virtual Image
| Feature | Real Image | Virtual Image |
|---|---|---|
| Formed by converging rays | Yes (rays actually meet) | No (rays only appear to meet) |
| Obtained on a screen | Yes | No |
| Orientation | Usually inverted | Always erect |
| Example | Image of a candle by concave mirror / convex lens | Plane-mirror image, concave-lens image |
Converging Lens vs Diverging Lens
| Feature | Converging (Convex) Lens | Diverging (Concave) Lens |
|---|---|---|
| Parallel beam | Meets at the focus | Spreads out (appears from focus) |
| Power | Positive | Negative |
| Magnification sign | Negative (real) or positive (virtual) | Always positive (virtual erect) |
Regular Reflection vs Diffuse Reflection
| Feature | Regular Reflection | Diffuse Reflection |
|---|---|---|
| Surface | Smooth, polished | Rough |
| Reflected rays | Stay parallel | Scatter |
| Image | Clear | None |
| Example | Mirror, still water | Paper, chalkboard |
Optically Denser vs Optically Rarer Medium
| Feature | Optically Denser | Optically Rarer |
|---|---|---|
| Speed of light | Smaller | Greater |
| Refractive index | Higher | Lower |
| On entering from rarer/denser | Bends towards the normal | Bends away from the normal |
| Not related to | Mass/heavy or light feeling of the substance | |
27. Common Mistakes Students Make
| Common Mistake | Correction |
|---|---|
| Using the wrong formula for mirrors. | Mirror formula has a plus: 1/f = 1/v + 1/u. |
| Using the wrong formula for lenses. | Lens formula has a minus: 1/f = 1/v − 1/u. |
| Forgetting the Cartesian sign convention. | u is always negative; write f and v with their correct signs before substituting. |
| Mixing cm and m in power calculations. | Convert focal length to metres first: 1 cm = 0.01 m, so 40 cm = 0.40 m. |
| Forgetting the negative sign in mirror magnification. | mirror: m = −v/u; lens: m = v/u (no minus). |
| Confusing focal length and radius of curvature. | f = R/2 and R = 2f; f is half of R. |
| Confusing convex and concave lenses. | Convex = thicker middle = converging (+); concave = thinner middle = diverging (−). |
| Confusing optical density with mass density. | Optical density depends on the speed of light in the medium, not on weight. |
| Misidentifying real and virtual images. | Real images can be obtained on a screen (rays meet); virtual images cannot (rays only appear to meet). |
| Drawing incorrect ray-diagram labels. | Always label P/C/F (mirror) or O/F₁/F₂ (lens), the arrows, and dashed arrows for virtual images. |
28. Exam Tips
- Solve optics numericals stepwise: Given (with signs) → Formula → Substitution → Calculation → Final Answer with unit → one line of interpretation.
- Never skip the sign convention: mirror vs lens formula differ by one sign; u is always negative for a real object.
- Draw ray diagrams with construction: parallel ray to focus, ray through C retracing / through O passing straight; add arrowheads and labels.
- Identify image characteristics from the object position: use the tables in sections 4 and 14 – position determines real/virtual, size, nature and orientation.
- Apply signs to mirror/lens formulas: real v and concave f are negative for mirrors; convex lens f is positive; power sign tells the lens type.
- For power questions: always give f in metres, and state the unit dioptre (D) in the answer.
- For reasoning questions: quote the physical principle (laws of reflection, Snell’s law, optical density, sign convention) and then apply it to the given situation.
29. Quick Revision (5–10 minutes)
- Reflection: bouncing back of light; regular on smooth, diffuse on rough surfaces.
- Laws of reflection: coplanar (incident, normal, reflected); ∠i = ∠r.
- Spherical mirror terms: P (pole), C (centre of curvature), F (focus), R = 2f.
- Concave mirror: converging; real images except object between F and P (virtual erect enlarged); f negative.
- Convex mirror: diverging; always virtual, erect, diminished; f positive.
- Mirror formula: 1/f = 1/v + 1/u.
- Mirror magnification: m = hi/ho = −v/u.
- Refraction: bending of light due to a change of speed between media.
- Snell’s law: sin i / sin r = constant.
- Refractive index: n = c/v; towards normal rarer→denser, away from normal denser→rarer.
- Glass slab: emergent ray parallel to incident ray, with lateral displacement.
- Convex lens: converging; real inverted images except object between O and F₁ (virtual erect enlarged); f positive.
- Concave lens: diverging; always virtual, erect, diminished; f negative.
- Lens formula: 1/f = 1/v − 1/u.
- Lens magnification: m = v/u (no minus).
- Power of lens: P = 1/f (f in metres), unit dioptre (D); positive = convex, negative = concave.
- Sign convention: origin at pole (mirror) or optical centre (lens); distances in the direction of incident light are positive.
30. Self-Practice Questions
Original – CBSE Exam-Style (answers and hints are in section 31)
MCQs
- The angle of incidence of a ray on a plane mirror is 30°. The angle of reflection is: (a) 30° (b) 60° (c) 90° (d) 0°
- Which mirror always forms an erect, diminished, virtual image? (a) Concave mirror (b) Convex mirror (c) Plane mirror (d) Any spherical mirror
- The radius of curvature of a concave mirror is 20 cm. Its focal length is: (a) +20 cm (b) −20 cm (c) +10 cm (d) −10 cm
- When light travels from air into glass, it: (a) bends towards the normal (b) bends away from the normal (c) does not bend (d) reflects completely
- Which statement is NOT true for a real image? (a) It can be caught on a screen (b) It is generally inverted (c) It is formed by rays actually meeting (d) It is always virtual
- The refractive index of a medium is given by: (a) n = v/c (b) n = c/v (c) n = c × v (d) n = c + v
- The power of a lens of focal length 0.5 m is: (a) 0.5 D (b) 2 D (c) 5 D (d) 20 D
- For a convex lens, a negative magnification means the image is: (a) virtual and erect (b) real and inverted (c) virtual and inverted (d) real and erect
- The lens formula is: (a) 1/f = 1/v + 1/u (b) 1/f = 1/v − 1/u (c) f = 1/v + 1/u (d) 1/f = v + u
- A concave mirror is used in a: (a) rear-view mirror (b) shaving mirror (c) security mirror (d) shop mirror
Very Short Answer Questions
- Define the focal length of a spherical mirror.
- State the SI unit of power of a lens.
- Under what object position does a concave mirror form a virtual, erect, magnified image?
- What is meant by lateral displacement of light in a glass slab?
- A ray strikes a plane mirror at 40° to its surface. What is the angle of incidence?
Short Answer Questions
- State the two laws of reflection of light.
- Why is a convex mirror preferred as a rear-view mirror in vehicles?
- Differentiate between a real image and a virtual image.
- Why does a ray of light bend towards the normal when it enters a denser medium?
- Write the mirror formula and the lens formula, and state which sign differs between them.
Numerical Problems
- An object is placed 30 cm in front of a concave mirror of focal length 20 cm. Find the image distance and describe the image.
- An object is placed 30 cm in front of a convex mirror of focal length 15 cm. Find the image distance and magnification.
- Light travels in a medium at 2 × 108 m s−1. Find the refractive index of the medium (c = 3 × 108 m s−1).
- An object is placed 8 cm from a convex lens of focal length 12 cm. Find the image distance and magnification.
- Find the focal length (in cm) of a lens of power +5 D, and identify the type of lens.
Reasoning / Application Questions
- Why can a concave mirror form a real image, while a convex mirror can never form a real image?
- How would you distinguish a convex lens from a concave lens without touching the surface?
- An object is placed at infinity in front of a convex lens. Where is the image formed, and what is its size?
Case-Based Questions
- Case: A torch is fitted with a concave reflector and the bulb is placed at the focus. Predict what the reflected beam looks like at a distance, and justify using a ray diagram idea.
- Case: A student passes a laser through a rectangular glass slab and observes the emergent ray shifted sideways. (a) Is the emergent ray parallel to the incident ray? (b) What would happen if the slab were replaced by a triangular prism?
31. Answer Key / Hints
Attempt section 30 fully before reading these answers and hints.
| Q.No. | Answer / Hint |
|---|---|
| 1 | (a) ∠r = ∠i = 30°. |
| 2 | (b) Convex mirror. |
| 3 | (d) f = R/2 = 10 cm, and for a concave mirror f is −10 cm. |
| 4 | (a) bends towards the normal. |
| 5 | (d) a real image is never virtual. |
| 6 | (b) n = c/v. |
| 7 | (b) P = 1/0.5 = 2 D. |
| 8 | (b) real and inverted. |
| 9 | (b) 1/f = 1/v − 1/u. |
| 10 | (b) shaving mirror. |
| 11 | The distance between the pole and the principal focus of the mirror. |
| 12 | Dioptre (D). |
| 13 | When the object lies between the focus F and the pole P. |
| 14 | The perpendicular distance between the incident ray and the emergent ray in a glass slab. |
| 15 | 50° (90° − 40°), angles are measured from the normal. |
| 16 | (1) Same plane: incident, normal, reflected. (2) ∠i = ∠r. |
| 17 | Gives a wide, upright, diminished view of traffic behind. |
| 18 | Real: rays meet, catchable on a screen, usually inverted. Virtual: rays appear to meet, not catchable, erect. |
| 19 | Speed decreases on entering the denser medium, so the ray bends towards the normal. |
| 20 | Mirror: 1/f = 1/v + 1/u; Lens: 1/f = 1/v − 1/u (the sign of 1/u differs). |
| 21 | u = −30 cm, f = −20 cm → v = −60 cm; real, inverted, enlarged image in front of the mirror. |
| 22 | u = −30, f = +15 → v = +10 cm; m = +1/3 (virtual, erect, diminished). |
| 23 | n = c/v = 3/2 = 1.5. |
| 24 | u = −8 cm, f = +12 cm → v = −24 cm; m = +3 (virtual, erect, enlarged magnifier). |
| 25 | f = 1/P = 1/5 = 0.2 m = 20 cm; convex lens. |
| 26 | A concave mirror actually converges rays at F and beyond, forming real images; a convex mirror always diverges rays, so its image is always virtual. |
| 27 | (Convex thickens at the centre; concave thins at the centre; also hold over a page – convex enlarges, concave shrinks the print.) |
| 28 | At the focus F₂; highly diminished, real, point-sized image. |
| 29 | A parallel beam emerges; the bulb at the focus of a concave mirror gives parallel rays that travel far (reverse of the parallel-to-focus rule). |
| 30 | (a) Yes, parallel but laterally displaced. (b) In a prism the faces are inclined, so the emergent ray is deviated, not parallel to the incident ray. |