Gonit Vigyan

🌟 Understanding Light: An Introduction

Light enables us to see the world by reflecting off objects and reaching our eyes. It travels in a straight line, forming sharp shadows and creating fascinating phenomena like reflection, refraction, and rainbows.

This post explores how light interacts with spherical mirrors, bends through different media, and applies to real life. While it mostly moves in straight lines, effects like diffraction reveal its wave nature. Modern physics shows light behaves as both waves and particles, leading to quantum theory.

Notes on the Nature of Light

1️⃣ Diffraction of Light

• When an opaque object is very small, light bends around it instead of traveling in a straight line.
• This phenomenon is called diffraction.
• The ray model of light (straight-line propagation) fails in such cases.

2️⃣ Wave Nature of Light

• To explain diffraction, light is considered a wave.
• Wave theory helps understand phenomena like interference and diffraction.

3️⃣ Particle Nature of Light

• In the early 20th century, wave theory failed to explain light-matter interaction.
• Light sometimes behaves like a stream of particles (photons).

4️⃣ Quantum Theory of Light

• Light is neither purely a wave nor a particle.
• Quantum theory reconciles both properties.
• Light exhibits wave-particle duality, depending on the situation.

🔎 Conclusion: The nature of light is complex, evolving from ray optics to wave theory and finally to quantum mechanics.

Notes on Reflection of Light

1️⃣ Reflection and Its Laws

• A highly polished surface, like a mirror, reflects most of the light falling on it.
• The laws of reflection apply to all reflecting surfaces, including spherical mirrors:
  1. Angle of incidence = Angle of reflection.
  2. The incident ray, reflected ray, and normal lie in the same plane.

2️⃣ Image Formation by a Plane Mirror

• Always virtual and erect.
• Size of image = Size of object.
• Image is formed behind the mirror, at the same distance as the object is in front.
• Laterally inverted (left-right reversal).

3️⃣ Exploring Reflection on Curved Surfaces

Activity 10.1: Observing Reflection with a Spoon

• A shining spoon acts as a curved mirror.
• Observations when looking into the spoon:
  • The image can appear smaller or larger, depending on the curvature.
  • Moving the spoon changes the size and nature of the image.
  • Reversing the spoon alters the image characteristics.

4️⃣ Spherical Mirrors

• Curved mirrors with a reflecting surface forming part of a sphere are called spherical mirrors.
• They are the most commonly used type of curved mirrors.
• The reflection properties of spherical mirrors differ from plane mirrors and depend on the shape of the reflecting surface.

Notes on Spherical Mirrors

1️⃣ Types of Spherical Mirrors

• A spherical mirror has a curved reflecting surface forming part of a sphere.
• Two types of spherical mirrors:
  1. Concave Mirror – Reflecting surface curved inwards, facing the center of the sphere.
  2. Convex Mirror – Reflecting surface curved outwards.
• The shaded side in diagrams represents the non-reflecting side.

2️⃣ Important Terms in Spherical Mirrors

Term	Definition	Representation
Centre of Curvature (C)	The center of the sphere of which the mirror is a part. It is not on the mirror itself.	C
Radius of Curvature (R)	The radius of the sphere of which the mirror forms a part. It is the distance PC (Pole to Centre of Curvature).	R
Principal Axis	The imaginary straight line passing through the pole (P) and centre of curvature (C), normal to the mirror at P.	-
Pole (P)	The geometric center of the reflecting surface of the mirror.	P
Aperture (MN)	The diameter of the reflecting surface of the mirror.	MN
Principal Focus (F)	The point where parallel rays to the principal axis converge (concave mirror) or appear to diverge from (convex mirror).	F
Focal Length (f)	The distance between Pole (P) and Focus (F). It follows the relation R = 2f.	f

3️⃣ Activity: Finding the Focus of a Concave Mirror

⚠ Caution: Never look directly at the Sun or its reflection in a mirror.

🔹 Steps:

1. Hold a concave mirror facing the Sun.
2. Direct the reflected light onto a sheet of paper.
3. Move the paper back and forth until a sharp, bright spot appears.
4. Hold it in this position and observe.
5. The paper may catch fire due to the heat concentrated at the focus.

🔹 Inference:

• The bright spot is the image of the Sun at the focus of the mirror.
• The distance between this spot and the mirror is the focal length (f) of the concave mirror.

4️⃣ Ray Diagram Interpretation

• Concave Mirror: Parallel rays converge at a point on the principal axis, called the principal focus (F).
• Convex Mirror: Parallel rays appear to diverge from a point behind the mirror, which is the principal focus (F).

5️⃣ Relationship Between Radius of Curvature (R) and Focal Length (f)

• For small aperture spherical mirrors, the radius of curvature is twice the focal length: $$R = 2f$$
• This means the principal focus lies midway between the pole (P) and the centre of curvature (C).

6️⃣ Spoon as a Spherical Mirror Analogy

• Inner surface of a spoon = Concave Mirror (focuses light, forms real images).
• Outer surface of a spoon = Convex Mirror (diverges light, forms virtual images).

Notes on Image Formation by Spherical Mirrors

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1️⃣ Image Formation by Concave Mirrors

• The image formed by a concave mirror depends on the position of the object relative to P (Pole), F (Focus), and C (Centre of Curvature).
• The image can be real or virtual, enlarged or diminished, erect or inverted.

🔹Activity 10.3: Observing Image Formation with a Concave Mirror

🔹 Steps:

1. Find the approximate focal length of a concave mirror by forming the image of the Sun on paper.
2. Place the mirror on a marked line with points P, F, and C indicated at equal distances.
3. Keep a burning candle at different positions and observe the image.
4. Note the nature, size, and position of the image for each case.

🔹Summary of Image Formation by a Concave Mirror

Position of Object	Position of Image	Size of Image	Nature of Image
At infinity	At F	Highly diminished (point-sized)	Real and inverted
Beyond C	Between F and C	Diminished	Real and inverted
At C	At C	Same size	Real and inverted
Between C and F	Beyond C	Enlarged	Real and inverted
At F	At infinity	Highly enlarged	Real and inverted
Between P and F	Behind the mirror	Enlarged	Virtual and erect

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2️⃣ Image Formation by Convex Mirrors

• Convex mirrors always form virtual, erect, and diminished images.
• The image moves closer to the focus as the object moves away from the mirror.

🔹Activity 10.5: Observing Image Formation with a Convex Mirror

🔹 Steps:

1. Hold a convex mirror in one hand and a pencil in the other.
2. Observe whether the image is erect or inverted, diminished or enlarged.
3. Move the pencil away from the mirror and note changes in the image.

🔹Summary of Image Formation by a Convex Mirror

Position of Object	Position of Image	Size of Image	Nature of Image
At infinity	At F (behind mirror)	Highly diminished (point-sized)	Virtual and erect
Between infinity and P	Between P and F (behind mirror)	Diminished	Virtual and erect

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3️⃣ Representation of Image Formation Using Ray Diagrams

• Ray diagrams help in understanding image formation by spherical mirrors.
• At least two rays are needed to locate the image.

🔹Key Rays Used in Ray Diagrams:

1️⃣ A ray parallel to the principal axis → Passes through F (Concave) or appears to diverge from F (Convex).
2️⃣ A ray passing through F (Concave) or directed toward F (Convex) → Emerges parallel to the principal axis.
3️⃣ A ray passing through C (Concave) or directed toward C (Convex) → Reflects back along the same path.
4️⃣ A ray incident at the pole (P) → Reflects symmetrically following the laws of reflection.

📌 In all cases, the laws of reflection are followed:
Angle of incidence = Angle of reflection.

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4️⃣ Uses of Spherical Mirrors

🔹 Concave Mirrors:

• Used in torches, searchlights, and vehicle headlights for parallel beams of light.
• Used as shaving mirrors and dentist’s mirrors for magnified images.
• Used in solar furnaces to concentrate sunlight for heating.

🔹 Convex Mirrors:

• Used as rear-view mirrors in vehicles.
• Preferred because they provide a wider field of view and always give erect, diminished images.
• Used in security mirrors to monitor large areas.

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Q U E S T I O N S & A N S W E R S

1. Define the principal focus of a concave mirror.

Answer:
The principal focus (F) of a concave mirror is the point on the principal axis where light rays parallel to the principal axis converge after reflection.

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2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Answer:
The focal length (f) of a spherical mirror is given by the formula:

$$f = \frac{R}{2}$$

Given, R = 20 cm,

$$f = \frac{20}{2} = 10 \text{ cm}$$

So, the focal length is 10 cm.

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3. Name a mirror that can give an erect and enlarged image of an object.

Answer:
A concave mirror can produce an erect and enlarged image when the object is placed between the pole (P) and the focus (F) of the mirror.

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4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Answer:
A convex mirror is preferred as a rear-view mirror in vehicles because:

• It always gives an erect and diminished image, allowing drivers to see a wider field of view.
• It helps in monitoring more traffic behind the vehicle for safe driving.

10.2.3 Sign Convention for Reflection by Spherical Mirrors

🔹 New Cartesian Sign Convention

While dealing with spherical mirrors, we use a standard sign convention known as the New Cartesian Sign Convention.

📌 Rules of the Sign Convention:

1️⃣ The object is always placed to the left of the mirror.

• This means light always travels from left to right towards the mirror.

2️⃣ All distances are measured from the pole (P).

• The pole (P) is taken as the origin (0,0).
• The principal axis is considered as the x-axis of the coordinate system.

3️⃣ Distances along the principal axis (x-axis):

• Right of the pole (+x-axis) → Positive (+ve)
• Left of the pole (−x-axis) → Negative (−ve)

4️⃣ Distances perpendicular to the principal axis (y-axis):

• Above the principal axis (+y-axis) → Positive (+ve)
• Below the principal axis (−y-axis) → Negative (−ve)

📌 Application:
These sign conventions are used while applying the Mirror Formula and solving numerical problems related to spherical mirrors.

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10.2.4 Mirror Formula and Magnification

📌 Mirror Formula

The relationship between object distance (u), image distance (v), and focal length (f) for a spherical mirror is given by the Mirror Formula:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

• This formula is valid for all spherical mirrors (concave & convex) and for all object positions.
• New Cartesian Sign Convention must be followed when substituting values for u, v, and f.

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📌 Magnification

Magnification (m) tells how much larger or smaller the image is compared to the object. It is given by:

$$m = \frac{\text{Height of Image} (h')} {\text{Height of Object} (h)}$$

Or, in terms of image distance (v) and object distance (u):

$$m = -\frac{v}{u}$$

🔹 Interpretation of Magnification Values

• If $m > 1$ → The image is enlarged.
• If $m < 1$ → The image is diminished.
• If $m = 1$ → The image is of the same size as the object.
• If $m$ is positive → The image is virtual and erect.
• If $m$ is negative → The image is real and inverted.

📌 Important:

• The height of the object (h) is always positive since the object is placed above the principal axis.
• The height of the image (h′) is:
  • Positive for virtual images.
  • Negative for real images.

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🔹 Summary of Key Formulas

Formula	Description
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$	Mirror Formula (relates object distance, image distance, and focal length).
$m = \frac{h'}{h}$	Magnification Formula (ratio of image height to object height).
$m = -\frac{v}{u}$	Magnification in terms of distances (negative sign indicates real image).

Q U E S T I O N S & A N S W E R S

1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.

Answer:
The focal length (f) of a spherical mirror is related to the radius of curvature (R) by the formula:

$$f = \frac{R}{2}$$

Given, R = 32 cm,

$$f = \frac{32}{2} = 16 \text{ cm}$$

Since it is a convex mirror, the focal length is positive:

$$f = +16 \text{ cm}$$

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2. A concave mirror produces a three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?

Answer:
Given:

• Object distance (u) = –10 cm (negative because the object is in front of the concave mirror).
• Magnification (m) = –3 (negative sign indicates a real and inverted image).
• Image distance (v) = ?

Using the magnification formula:

$$m = \frac{-v}{u}$$

Substituting values:

$$-3 = \frac{-v}{-10}$$ $$v = -30 \text{ cm}$$

Thus, the image is formed 30 cm in front of the concave mirror.

Solved Examples on Spherical Mirrors

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1️⃣ A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position, nature, and size of the image.

Solution:

🔹 Given Data:

• Radius of curvature (R) = +3.00 m (Convex mirror → R is positive).
• Object distance (u) = –5.00 m (Negative as the object is in front).
• Focal length (f): $$f = \frac{R}{2} = \frac{3.00}{2} = +1.50 \text{ m}$$
• Image distance (v) = ?
• Height of the image (h') = ?

🔹 Using the Mirror Formula:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

Substituting values:

$$\frac{1}{v} = \frac{1}{1.50} + \frac{1}{5.00}$$ $$\frac{1}{v} = \frac{5.00}{7.50}$$ $$v = \frac{7.50}{6.50} = +1.15 \text{ m}$$

🔹 Interpretation:

• The image is 1.15 m behind the mirror.
• The positive v confirms that the image is virtual.

🔹 Magnification Calculation:

$$m = \frac{-v}{u} = \frac{1.15}{5.00} = +0.23$$

🔹 Final Answer:
✅ Position: 1.15 m behind the mirror.
✅ Nature: Virtual and erect.
✅ Size: Diminished (23% of the object size).

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2️⃣ An object, 4.0 cm in size, is placed at 25.0 cm in front of a concave mirror of focal length 15.0 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and the size of the image.

Solution:

🔹 Given Data:

• Object size (h) = +4.0 cm
• Object distance (u) = –25.0 cm
• Focal length (f) = –15.0 cm (Concave mirror → f is negative)
• Image distance (v) = ?
• Image size (h') = ?

🔹 Using the Mirror Formula:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

Substituting values:

$$\frac{1}{v} = \frac{1}{-15.0} - \frac{1}{-25.0}$$ $$\frac{1}{v} = \frac{-5.0}{75.0} + \frac{3.0}{75.0} = \frac{-2.0}{75.0}$$ $$v = -37.5 \text{ cm}$$

🔹 Interpretation:

• The negative v means the image is real and formed in front of the mirror.
• The screen should be placed 37.5 cm from the mirror to get a sharp image.

🔹 Magnification Calculation:

$$m = \frac{-v}{u} = \frac{-37.5}{-25.0} = 1.5$$ $$h' = mh = (1.5)(4.0) = 6.0 \text{ cm}$$

🔹 Final Answer:
✅ Position: 37.5 cm in front of the mirror.
✅ Nature: Real and inverted.
✅ Size: Enlarged (6.0 cm, 1.5× the object size).

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📌 Key Takeaways:
✔ Convex mirrors always form virtual, erect, and diminished images.
✔ Concave mirrors can form real, inverted, and magnified images, depending on the object position.

📖 10.3 Refraction of Light

🔹 What is Refraction?

• Light travels in straight lines in a single medium.
• When light enters from one transparent medium to another, it changes direction.
• This bending of light is called refraction.
• Refraction occurs due to a change in the speed of light as it moves between different media.

🔬 Daily Life Examples of Refraction

✅ The bottom of a pond appears raised.
✅ A pencil in water looks bent at the surface.
✅ A lemon in water appears larger than its actual size.
✅ Letters appear raised when viewed through a glass slab.

📌 Why does this happen?

• Light rays change direction when moving between different materials.
• Different liquids (water, kerosene, turpentine) cause different extents of refraction.

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🔹 10.3.1 Refraction Through a Rectangular Glass Slab

📝 Key Observations from Experiments

1️⃣ When light enters from air to glass, it bends towards the normal.
2️⃣ When light exits from glass to air, it bends away from the normal.
3️⃣ The emergent ray is parallel to the incident ray but is slightly shifted sideways.

📌 Conclusion:

• Refraction occurs at both air-glass and glass-air interfaces.
• The bending of light depends on the material and its optical properties.

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🔹 Laws of Refraction (Snell’s Law)

1️⃣ The incident ray, refracted ray, and normal at the point of incidence lie in the same plane.
2️⃣ The ratio of the sine of the angle of incidence (i) to the sine of the angle of refraction (r) is a constant:

$$\frac{\sin i}{\sin r} = \text{constant}$$

• This constant is called the refractive index of the second medium relative to the first.

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🔹 10.3.2 Refractive Index (n)

💡 The refractive index is a measure of how much light slows down when entering a medium.

🔹 Formula:

$$n_{21} = \frac{\text{Speed of light in medium 1}}{\text{Speed of light in medium 2}}$$ $$n_m = \frac{c}{v}$$

Where:

• $n_{21}$ = refractive index of medium 2 relative to medium 1.
• $c$ = speed of light in air/vacuum (3 × 10⁸ m/s).
• $v$ = speed of light in the medium.

📌 Key Points:

• If $n > 1$ → Light slows down in the medium.
• If $n < 1$ → Light speeds up in the medium.
• A higher refractive index means the medium is optically denser.

🔹 Refractive Index Between Two Media:

$$n_{12} = \frac{\text{Speed of light in medium 2}}{\text{Speed of light in medium 1}}$$ $$n_{21} = \frac{\text{Speed of light in medium 1}}{\text{Speed of light in medium 2}}$$ $$n_{12} \times n_{21} = 1$$

📌 Important Property:

• The product of refractive indices of two media relative to each other is always 1.

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🔹 Optical Density vs Mass Density

📌 Important Difference:

• Optical density depends on how much light slows down in a medium.
• Mass density is the mass per unit volume.
  🔹 Example:
• Kerosene has a higher refractive index than water, meaning it is optically denser but less massive.

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📊 Table: Absolute Refractive Index of Some Materials

Material 🌎	Refractive Index (n) 📊
Air 🌬️	1.0003
Ice ❄️	1.31
Water 💧	1.33
Alcohol 🍶	1.36
Kerosene 🛢️	1.44
Fused Quartz 🪨	1.46
Turpentine Oil 🧴	1.47
Benzene 🏺	1.50
Crown Glass 🏡	1.52
Canada Balsam 🌲	1.53
Rock Salt 🧂	1.54
Carbon Disulfide ⚗️	1.63
Dense Flint Glass 🏺	1.65
Ruby 💎	1.71
Sapphire 🔹	1.77
Diamond 💎✨	2.42

📌 Observations:

• Diamond has the highest refractive index (2.42), making it highly optically dense.
• Water (1.33) and air (1.0003) are optically rarer.

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🔹 Key Takeaways

✔ Refraction occurs when light changes direction while passing from one medium to another.
✔ Laws of refraction govern how light bends.
✔ Refractive index determines how much light slows down in a medium.
✔ Optically denser mediums have a higher refractive index but may not have higher mass density.
✔ $n_{12} \times n_{21} = 1$ is always true for two media.

Q U E S T I O N S & A N S W E R S

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1️⃣ A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?

Answer:

• The light ray bends towards the normal when it enters water from air.
• This happens because water is optically denser than air.
• As light moves from a rarer medium (air) to a denser medium (water), its speed decreases, causing it to bend towards the normal.

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2️⃣ Light enters from air to glass having a refractive index of 1.50. What is the speed of light in the glass?

Answer:

• Given:
  • Refractive index of glass, $n_g = 1.50$
  • Speed of light in vacuum, $c = 3 \times 10^8$ m/s
• Using the formula: $$n = \frac{c}{v}$$ $$v = \frac{c}{n}$$ $$v = \frac{3 \times 10^8}{1.50}$$ $$v = 2 \times 10^8 \text{ m/s}$$

✅ Speed of light in glass = 2 × 10⁸ m/s.

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3️⃣ Find out, from Table 10.3, the medium having the highest optical density. Also, find the medium with the lowest optical density.

Answer:

• The highest optical density is in Diamond (Refractive Index = 2.42).
• The lowest optical density is in Air (Refractive Index = 1.0003).

📌 Conclusion: A medium with a higher refractive index is optically denser, while a medium with a lower refractive index is optically rarer.

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4️⃣ You are given kerosene, turpentine, and water. In which of these does light travel fastest? Use the information given in Table 10.3.

Answer:

• The speed of light in a medium is inversely proportional to its refractive index.
• From Table 10.3:
  • Refractive index of water = 1.33
  • Refractive index of kerosene = 1.44
  • Refractive index of turpentine = 1.47
• Since water has the lowest refractive index, light travels fastest in water.

✅ Light travels fastest in water, followed by kerosene, and slowest in turpentine.

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5️⃣ The refractive index of diamond is 2.42. What is the meaning of this statement?

Answer:

• The refractive index of diamond = 2.42 means that light slows down 2.42 times when it enters diamond from vacuum.
• It also means that the speed of light in diamond is: $$v = \frac{c}{n} = \frac{3 \times 10^8}{2.42}$$ $$v \approx 1.24 \times 10^8 \text{ m/s}$$

✅ This means light travels 2.42 times slower in diamond than in vacuum.

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📌 Key Takeaways:
✔ Light bends towards the normal when entering a denser medium and away from the normal when entering a rarer medium.
✔ Speed of light in a medium is given by $v = \frac{c}{n}$.
✔ Diamond has the highest optical density, while air has the lowest.
✔ Light travels fastest in water among water, kerosene, and turpentine.
✔ A refractive index of 2.42 means light slows down significantly in diamond.

📖 10.3.3 Refraction by Spherical Lenses

🔹 What is a Lens?

A lens is a transparent object that bends light by refraction. It is bounded by two surfaces, at least one of which is spherical. Lenses are widely used in spectacles, magnifying glasses, cameras, microscopes, and telescopes.

📌 Key Properties of Lenses:
✔ A lens has two spherical surfaces.
✔ It bends light rays due to refraction.
✔ It can be convex (converging) or concave (diverging).

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🔹 Types of Lenses

1️⃣ Convex Lens (Converging Lens) 🔍

✅ Definition:

• A convex lens is thicker in the middle and thinner at the edges.
• It converges (brings together) parallel light rays to a single point after refraction.
• Hence, it is also called a converging lens.

✅ Uses:
✔ Used in magnifying glasses, cameras, microscopes, and telescopes.
✔ Used in spectacles to correct farsightedness (hypermetropia).

✅ Diagram:
📌 See Figure 10.12 (a) for the converging action of a convex lens.

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2️⃣ Concave Lens (Diverging Lens) 🔍

✅ Definition:

• A concave lens is thicker at the edges and thinner in the middle.
• It diverges (spreads out) parallel light rays after refraction.
• Hence, it is also called a diverging lens.

✅ Uses:
✔ Used in spectacles to correct nearsightedness (myopia).
✔ Used in peepholes (door viewers) and lasers.

✅ Diagram:
📌 See Figure 10.12 (b) for the diverging action of a concave lens.

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🔹 Important Terms Related to Lenses

Term 📌	Definition 🔍
Centre of Curvature (C₁, C₂)	The center of the spheres from which the lens surfaces are formed.
Principal Axis	The imaginary straight line passing through both centres of curvature.
Optical Centre (O)	The central point of the lens; a ray passing through O does not deviate.
Principal Focus (F₁, F₂)	The point where parallel rays converge (convex lens) or appear to diverge (concave lens).
Focal Length (f)	The distance between the optical centre (O) and the principal focus (F).
Aperture	The diameter of the circular outline of the lens.

📌 Key Concept:

• A lens has two principal foci (F₁ and F₂) because light can pass through it from both sides.

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🔹 Activity: Burning Paper with a Convex Lens 🔥

🔬 Activity 10.11: Finding the Focal Length of a Convex Lens

⚠ Caution: Never look at the Sun directly through a lens!

✅ Steps:
1️⃣ Hold a convex lens and direct it towards the Sun.
2️⃣ Focus the light from the Sun onto a sheet of paper.
3️⃣ Adjust the lens until you get a sharp bright spot on the paper.
4️⃣ Observe what happens to the paper.

✅ Observation:

• The paper starts to burn and produces smoke.
• This happens because the lens converges the Sun’s rays at a single point, generating heat.
• The distance from the lens to the burning spot is the approximate focal length of the convex lens.

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🔹 Refraction of Light by Lenses

📌 What happens when parallel rays of light pass through a lens?

1️⃣ Refraction Through a Convex Lens (Figure 10.12 a)

• Light rays parallel to the principal axis converge at a point called the principal focus (F) after refraction.

2️⃣ Refraction Through a Concave Lens (Figure 10.12 b)

• Light rays parallel to the principal axis appear to diverge from a point on the principal axis called the principal focus (F).

📌 Key Concept:

• A convex lens has a real principal focus.
• A concave lens has a virtual principal focus.

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🔹 Principal Focus & Focal Length

📌 Key Facts:

• Convex lens: Parallel rays converge at F.
• Concave lens: Parallel rays appear to diverge from F.
• Focal length (f) is the distance between O and F.
• A lens has two focal points (F₁, F₂) since light can pass through both sides.

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🔹 Summary of Key Concepts

✔ Convex lenses converge light, forming real and inverted images (except when the object is very close).
✔ Concave lenses diverge light, forming virtual, upright, and diminished images.
✔ The principal focus (F) is where parallel light rays converge (convex) or appear to diverge (concave).
✔ The focal length (f) is the distance between the optical centre (O) and the principal focus (F).
✔ The centre of curvature (C₁, C₂) is the centre of the spheres forming the lens surfaces.

📖 10.3.4 Image Formation by Lenses

🔹 How Do Lenses Form Images?

Lenses form images by refracting light. The nature, position, and size of the image depend on:
✔ The type of lens (convex or concave).
✔ The position of the object relative to the lens.

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🔹 Image Formation by a Convex Lens 🔍

📝 Activity 10.12: Studying Image Formation by a Convex Lens

✅ Steps:
1️⃣ Find the approximate focal length of the convex lens using Activity 10.11.
2️⃣ Draw five parallel lines on a table with equal spacing = focal length of the lens.
3️⃣ Place the lens on the central line so that its optical centre (O) lies on it.
4️⃣ Mark F₁, 2F₁, F₂, and 2F₂ on the table.
5️⃣ Place a burning candle at different positions and observe the image on a screen.
6️⃣ Record observations for different object positions.

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📊 Table 10.4: Image Formation by a Convex Lens

Object Position	Image Position	Relative Size	Nature
At Infinity	At F₂	Highly diminished (point-sized)	Real & Inverted
Beyond 2F₁	Between F₂ and 2F₂	Diminished	Real & Inverted
At 2F₁	At 2F₂	Same Size	Real & Inverted
Between F₁ and 2F₁	Beyond 2F₂	Enlarged	Real & Inverted
At F₁	At Infinity	Infinitely large	Real & Inverted
Between F₁ and O	On the same side of the lens as the object	Enlarged	Virtual & Erect

📌 Key Observations:
✔ When the object is beyond 2F₁, the image is diminished, real, and inverted.
✔ When the object is between F₁ and 2F₁, the image is magnified, real, and inverted.
✔ When the object is at F₁, the image is formed at infinity.
✔ When the object is between F₁ and O, the image is magnified, virtual, and erect.

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🔹 Image Formation by a Concave Lens 🔍

📝 Activity 10.13: Studying Image Formation by a Concave Lens

✅ Steps:
1️⃣ Place a concave lens on a lens stand.
2️⃣ Position a burning candle on one side of the lens.
3️⃣ Look through the lens and try to capture the image on a screen.
4️⃣ Observe the nature, size, and position of the image.
5️⃣ Move the candle farther away and observe how the image size changes.

📌 Conclusion:
✔ A concave lens always forms a virtual, erect, and diminished image regardless of object position.

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📊 Table 10.5: Image Formation by a Concave Lens

Object Position	Image Position	Relative Size	Nature
At Infinity	At F₁	Highly diminished (point-sized)	Virtual & Erect
Between Infinity and O	Between F₁ and O	Diminished	Virtual & Erect

📌 Key Observations:
✔ A concave lens always forms a virtual, erect, and diminished image.
✔ The image is always on the same side as the object.

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📖 10.3.5 Image Formation in Lenses Using Ray Diagrams

Ray diagrams help us understand how lenses form images by refraction.

✏ Rules for Drawing Ray Diagrams

1️⃣ Ray parallel to principal axis → After refraction:

• Convex lens: Passes through F₂.
• Concave lens: Appears to diverge from F₁.

2️⃣ Ray passing through (or directed toward) focus (F₁) → After refraction:

• Convex lens: Emerges parallel to the principal axis.
• Concave lens: Emerges parallel to the principal axis.

3️⃣ Ray passing through the optical centre (O) → Passes undeviated.

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📊 Image Formation in Convex and Concave Lenses

✅ Figure 10.16 shows ray diagrams for different object positions in a convex lens.
✅ Figure 10.17 shows ray diagrams for different object positions in a concave lens.

📌 Key Points from Ray Diagrams:
✔ Convex Lens: Can form real or virtual images depending on object position.
✔ Concave Lens: Always forms a virtual, erect, and diminished image.

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🔹 Summary of Key Concepts

✔ Convex lenses form real, inverted images, except when the object is between F₁ and O, where the image is virtual and erect.
✔ Concave lenses always form virtual, erect, and diminished images.
✔ Ray diagrams help visualize how lenses form images.
✔ The position and size of the image depend on the object’s position.

📖 10.3.6 Sign Convention for Spherical Lenses

🔹 Sign Convention for Lenses

The sign convention for spherical lenses is similar to that for spherical mirrors, but all measurements are taken from the optical centre (O) of the lens.

📌 Rules of Sign Convention:

✔ The object is always placed to the left of the lens.
✔ Distances measured in the direction of incident light (right side) are positive.
✔ Distances measured against the direction of incident light (left side) are negative.
✔ Distances measured above the principal axis are positive.
✔ Distances measured below the principal axis are negative.
✔ Focal length (f) of a convex lens is positive (+f).
✔ Focal length (f) of a concave lens is negative (-f).

📌 Key Point:
Always use proper signs while solving numerical problems related to lenses.

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📖 10.3.7 Lens Formula and Magnification

🔹 Lens Formula 📏

The relationship between object distance (u), image distance (v), and focal length (f) for spherical lenses is given by:

$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$

📌 Key Points:
✔ This formula is valid for both convex and concave lenses.
✔ Use sign convention while substituting values.

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🔹 Magnification Formula 🔍

Magnification (m) is the ratio of the height of the image (h') to the height of the object (h).

$$m = \frac{h'}{h}$$

Magnification is also related to object distance (u) and image distance (v):

$$m = \frac{v}{u}$$

📌 Key Points:
✔ m > 1 → Image is enlarged.
✔ m < 1 → Image is diminished.
✔ m is positive → Image is virtual and erect.
✔ m is negative → Image is real and inverted.

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🔹 Summary of Key Concepts

✔ Lens formula relates object distance (u), image distance (v), and focal length (f).
✔ Convex lenses have positive focal length, while concave lenses have negative focal length.
✔ Magnification (m) tells us the size and nature of the image.
✔ Use proper signs while solving problems.

📖 Example Problems on Lenses

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🔹 Example 10.3

❓ Question:

A concave lens has a focal length of 15 cm. At what distance should the object be placed so that it forms an image at 10 cm from the lens? Also, find the magnification produced by the lens.

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📝 Given Data:

• Focal length (f) = -15 cm (Concave lens → Negative focal length)
• Image distance (v) = -10 cm (Virtual image on the same side as the object)
• Object distance (u) = ?
• Magnification (m) = ?

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🔢 Solution:

Using the Lens Formula:

$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ $$\frac{1}{u} = \frac{1}{v} - \frac{1}{f}$$ $$\frac{1}{u} = \frac{1}{-10} - \frac{1}{-15}$$ $$\frac{1}{u} = -\frac{1}{10} + \frac{1}{15}$$ $$\frac{1}{u} = \frac{-3 + 2}{30} = \frac{-1}{30}$$ $$u = -30 cm$$

📌 The object must be placed 30 cm from the lens.

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🔢 Finding Magnification:

$$m = \frac{v}{u} = \frac{-10}{-30} = +\frac{1}{3} = +0.33$$

📌 The positive sign indicates that the image is virtual and erect.
📌 The image is one-third the size of the object.

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🔹 Example 10.4

❓ Question:

A 2.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position, and size of the image. Also, find its magnification.

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📝 Given Data:

• Height of object (h) = +2.0 cm
• Focal length (f) = +10 cm (Convex lens → Positive focal length)
• Object distance (u) = -15 cm (Object placed to the left of the lens)
• Image distance (v) = ?
• Image height (h') = ?

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🔢 Solution:

Using the Lens Formula:

$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ $$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$$ $$\frac{1}{v} = \frac{1}{10} + \frac{1}{-15}$$ $$\frac{1}{v} = \frac{3 - 2}{30} = \frac{1}{30}$$ $$v = +30 cm$$

📌 The positive sign of v indicates that the image is on the other side of the lens (real image).
📌 The image is formed at a distance of 30 cm from the lens.

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🔢 Finding Magnification:

$$m = \frac{v}{u} = \frac{+30}{-15} = -2$$ $$h' = h \times m = 2.0 \times (-2) = -4.0 cm$$

📌 The negative sign of h' indicates that the image is inverted and real.
📌 The image is 4 cm tall, twice the size of the object.

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📌 Key Takeaways:

✔ Concave lenses always form virtual, erect, and diminished images.
✔ Convex lenses can form real and inverted images or virtual and erect images, depending on object placement.
✔ Use correct sign conventions to avoid calculation errors.

📖 10.3.8 Power of a Lens

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❓ What is the Power of a Lens?

The power of a lens (P) describes its ability to converge or diverge light. It is defined as the reciprocal of the focal length (f) of the lens.

🔢 Formula for Power of a Lens:

$$P = \frac{1}{f}$$

where,

• P = Power of the lens (in dioptres, D)
• f = Focal length of the lens (in metres)

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🔹 SI Unit of Power 🔍

• The SI unit of power is Dioptre (D).
• 1 Dioptre (1D) = Power of a lens with focal length 1 metre.
• If f is in metres, P is in dioptres.

📌 Sign Convention for Power

✔ Convex lens (converging lens) → Positive Power (+P)
✔ Concave lens (diverging lens) → Negative Power (-P)

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🔹 Examples of Lens Power Calculation

1️⃣ A convex lens with focal length +0.50 m has power:

$$P = \frac{1}{0.50} = +2.0 D$$

📌 Since P is positive, the lens is convex.

2️⃣ A concave lens with focal length -0.40 m has power:

$$P = \frac{1}{-0.40} = -2.5 D$$

📌 Since P is negative, the lens is concave.

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🔹 Combination of Lenses 🔗

When multiple lenses are placed in contact, the net power (P) is the sum of the individual powers:

$$P = P_1 + P_2 + P_3 + \dots$$

📌 Example:

If two lenses of power +2.0 D and +0.25 D are placed together, the total power is:

$$P = 2.0 D + 0.25 D = +2.25 D$$

📌 This is equivalent to a single lens of power +2.25 D.

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🔹 Importance of Power in Optics 👓

✔ Used by opticians to prescribe corrective lenses.
✔ Helps design optical instruments like microscopes, telescopes, and cameras.
✔ Used in eye testing by combining different lenses to find the required correction.

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🔹 Summary of Key Concepts

Lens Type	Power (P)	Focal Length (f)	Nature of Lens
Convex Lens	Positive (+P)	Positive (+f)	Converging
Concave Lens	Negative (-P)	Negative (-f)	Diverging

📌 Power makes lens calculations easier for opticians and optical instrument design.

📖 Q U E S T I O N S & A N S W E R S

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❓ Q1: Define 1 dioptre of power of a lens.

Answer:
📌 1 dioptre (1D) is the power of a lens whose focal length is 1 metre.
📌 It is given by the formula:

$$P = \frac{1}{f}$$

where P is the power in dioptres (D) and f is the focal length in metres.
📌 A convex lens has positive power (+D), and a concave lens has negative power (-D).

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❓ Q2: A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.

Answer:
📌 Given Data:

• Image distance (v) = +50 cm = +0.50 m
• Since the image is real, inverted, and same size as the object, the object must be placed at 2F (twice the focal length, 2f).
• So, Object distance (u) = -50 cm = -0.50 m
• Focal length (f) = ?
• Power (P) = ?

📌 Step 1: Finding Focal Length (f)

Since the object is at 2F, we use the formula:

$$u = 2f, \quad v = 2f$$ $$f = \frac{u}{2} = \frac{50}{2} = 25 cm = 0.25 m$$

📌 Step 2: Finding Power (P)

$$P = \frac{1}{f} = \frac{1}{0.25} = +4 D$$

✅ Final Answer:

• The needle is placed 50 cm (0.50 m) in front of the lens.
• The power of the lens is +4 dioptres (D).

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❓ Q3: Find the power of a concave lens of focal length 2 m.

Answer:
📌 Given Data:

• Focal length (f) = -2 m (Negative sign because it's a concave lens)
• Power (P) = ?

Using the formula:

$$P = \frac{1}{f}$$ $$P = \frac{1}{-2} = -0.50 D$$

✅ Final Answer:
The power of the concave lens is -0.50 dioptres (D).

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🔹 Summary of Key Concepts:

✔ 1 dioptre (D) = power of a lens with focal length 1 metre.
✔ Convex lens → Positive power (+P), Concave lens → Negative power (-P).
✔ Use correct sign conventions for focal length while calculating power.

📖 E X E R C I S E S

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❓ Q1: Which one of the following materials cannot be used to make a lens?

(a) Water
(b) Glass
(c) Plastic
(d) Clay

Answer:
✅ (d) Clay cannot be used to make a lens because it is opaque and does not allow light to pass through, which is essential for a lens to function.

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❓ Q2: The image formed by a concave mirror is observed to be virtual, erect, and larger than the object. Where should be the position of the object?

(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus

Answer:
✅ (d) Between the pole of the mirror and its principal focus
📌 A concave mirror forms a virtual, erect, and magnified image when the object is placed between the pole (P) and the principal focus (F).

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❓ Q3: Where should an object be placed in front of a convex lens to get a real image of the same size as the object?

(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus

Answer:
✅ (b) At twice the focal length
📌 A convex lens forms a real, inverted, and same-sized image when the object is placed at twice the focal length (2F).

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❓ Q4: A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be:

(a) both concave.
(b) both convex.
(c) the mirror is concave and the lens is convex.
(d) the mirror is convex, but the lens is concave.

Answer:
✅ (a) Both concave.
📌 A negative focal length indicates that both the spherical mirror is concave and the lens is concave.

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❓ Q5: No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be:

(a) only plane.
(b) only concave.
(c) only convex.
(d) either plane or convex.

Answer:
✅ (d) Either plane or convex.
📌 A plane mirror always forms an erect image of the same size.
📌 A convex mirror also always forms a virtual, erect, and diminished image.

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❓ Q6: Which of the following lenses would you prefer to use while reading small letters in a dictionary?

(a) A convex lens of focal length 50 cm.
(b) A concave lens of focal length 50 cm.
(c) A convex lens of focal length 5 cm.
(d) A concave lens of focal length 5 cm.

Answer:
✅ (c) A convex lens of focal length 5 cm.
📌 A convex lens acts as a magnifying glass and is used for reading small letters. A smaller focal length (5 cm) provides greater magnification.

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❓ Q7: We wish to obtain an erect image of an object using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram.

Answer:
📌 To obtain an erect image using a concave mirror, the object must be placed between the pole (P) and the focus (F).
📌 The image will be virtual, erect, and magnified.

✅ Object distance range: Less than 15 cm (i.e., between P and F).
✅ Nature of image: Virtual and erect.
✅ Size of image: Larger than the object (magnified).

📌 Ray Diagram: (Draw a concave mirror with the object between P and F, showing a virtual, magnified image).

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❓ Q8: Name the type of mirror used in the following situations.

(a) Headlights of a car.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.

Answer:
✅ (a) Concave Mirror → Produces a powerful, focused beam of light.
✅ (b) Convex Mirror → Provides a wider field of view and always forms an erect image.
✅ (c) Concave Mirror → Concentrates sunlight at a single point for high temperature.

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❓ Q9: One-half of a convex lens is covered with black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

Answer:
✅ Yes, the convex lens will still produce a complete image.
📌 However, the brightness of the image will reduce because fewer rays will pass through the uncovered part of the lens.

🔬 Experiment:

• Take a convex lens and cover half of it with black paper.
• Focus light from an object onto a screen.
• Observe that a full image is still formed, but dimmer.

📌 Conclusion: Every part of the lens refracts light and contributes to image formation.

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❓ Q10: An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size, and nature of the image formed.

Answer:
📌 Given Data:

• Object height (h) = 5 cm
• Object distance (u) = -25 cm
• Focal length (f) = +10 cm
• Image distance (v) = ?

Using the Lens Formula:

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$ $$\frac{1}{v} = \frac{1}{10} + \frac{1}{25}$$ $$\frac{1}{v} = \frac{5+2}{50} = \frac{7}{50}$$ $$v = + 50/7 = +7.14 cm$$

✅ Image Position: 7.14 cm behind the lens
✅ Image Nature: Real and inverted
✅ Image Size: Diminished

📌 Ray Diagram: (Draw a convex lens with object at 25 cm, image at 7.14 cm, and show the inverted image).

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❓ Q16: Find the focal length of a lens of power –2.0 D. What type of lens is this?

Answer:
📌 Given:

• Power (P) = -2.0 D
• Focal length (f) = ?

Using the formula:

$$P = \frac{1}{f}$$ $$f = \frac{1}{P} = \frac{1}{-2.0} = -0.50 m$$

✅ The focal length is -0.50 m (or -50 cm).
✅ Since focal length is negative, the lens is concave.

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❓ Q17: A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Answer:
📌 Given:

• Power (P) = +1.5 D
• Focal length (f) = ?

Using the formula:

$$f = \frac{1}{P} = \frac{1}{1.5} = 0.67 m = 67 cm$$

✅ The focal length is +0.67 m (or 67 cm).
✅ Since the power is positive, the lens is convex (converging).

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