Light – Concept Booster | Class 10 Science CBSE

Q: Think of the mirror’s pole (or the lens’s optical center) as the origin $(0,0)$ on a standard X-Y graph. 1. The object is always placed on the left side of the mirror/lens. Therefore, object distance ($u$) is ALWAYS NEGATIVE. 2. All distances measured to the right are Positive (+). Distances to the left are Negative (-). 3. Heights measured upwards (above principal axis) are Positive (+). Heights downwards are Negative (-). 4. Focal Length Hack: Convex = Positive $f$. Concave = Negative $f$. (This applies to BOTH mirrors and lenses!).

Given: Object distance $u = -10 \text{ cm}$ (Always negative) Focal length $f = +15 \text{ cm}$ (Convex mirror is positive) Mirror Formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ $\frac{1}{v} + \frac{1}{-10} = \frac{1}{15}$ $\frac{1}{v} = \frac{1}{15} + \frac{1}{10} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6}$ $v = +6 \text{ cm}$. Since $v$ is positive, the image forms behind the mirror. Thus, it is Virtual and Erect. (Magnification $m = -v/u = -6/-10 = +0.6$, so it is diminished).

Q: Convex mirrors are used as rear-view mirrors in vehicles. Why? Because they always produce an erect, diminished image, which gives the driver a much wider field of view to see the traffic behind them. Concave mirrors are used in car headlights and searchlights. Why? Because if you place a light bulb exactly at the focus ($F$) of a concave mirror, the reflected rays will travel completely parallel to each other, creating a powerful, focused beam of light that cuts through the darkness.

Using the absolute refractive index formula: $n_m = \frac{c}{v}$ $1.50 = \frac{3 \times 10^8}{v}$ $v = \frac{3 \times 10^8}{1.50} = 2 \times 10^8 \text{ m/s}$.

Q: An object is placed at a distance of $10 \text{ cm}$ from a convex mirror of focal length $15 \text{ cm}$. Find the position and nature of the image. [3 marks]

Given Power $P = +1.5 \text{ D}$. Since $P = \frac{1}{f \text{ (in m)}}$, we have $f = \frac{1}{P}$. $f = \frac{1}{1.5} = \frac{10}{15} = \frac{2}{3} \text{ m} \approx +0.67 \text{ m}$ (or $+66.7 \text{ cm}$). Because the focal length and power are Positive, it must be a Convex Lens, which is a Converging lens.

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Home Concept Boosters CBSE CBSE Class 10 Science Light

Table of Contents

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How to Use This Page

Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page completely covers Light: Reflection and Refraction for CBSE Class 10 Science, opening the door to the fascinating world of optics.

Key Concepts

Class 10 · Science · Physics

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Optics Basics

How light bounces and bends

Class 10 · Ch 9

Laws of Reflection Rule

Applicable to all reflecting surfaces (plane or curved):
1. The angle of incidence is always equal to the angle of reflection.
2. The incident ray, the normal at the point of incidence, and the reflected ray all lie in the same plane.

$$\angle i = \angle r$$

Spherical Mirrors Definition

Mirrors whose reflecting surfaces are spherical.
Concave Mirror: Reflecting surface curves inwards. It is a converging mirror (focuses light to a point).
Convex Mirror: Reflecting surface curves outwards. It is a diverging mirror.

Focal Length and Radius of Curvature Formula

The center of the spherical shell is the Center of Curvature ($C$). The midpoint of the mirror is the Pole ($P$). The Principal Focus ($F$) lies exactly halfway between $P$ and $C$ for mirrors of small apertures.

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

Mirror Formula & Magnification Formula

Relates Object Distance ($u$), Image Distance ($v$), and Focal Length ($f$). Magnification ($m$) tells us how large or small the image is compared to the object.

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \quad | \quad m = \frac{h’}{h} = -\frac{v}{u}$$

Refraction & Snell’s Law Formula

The bending of light as it passes from one transparent medium to another due to a change in speed. Snell’s Law states that the ratio of the sine of angle of incidence to the sine of angle of refraction is a constant.

$$\frac{\sin i}{\sin r} = \text{constant} = n_{21}$$

Absolute Refractive Index ($n$) Formula

The ratio of the speed of light in a vacuum ($c$) to the speed of light in the medium ($v$). Higher $n$ means the medium is more “optically dense” and light travels slower.

$$n_m = \frac{\text{Speed of light in vacuum (c)}}{\text{Speed of light in medium (v)}} = \frac{c}{v}$$

Spherical Lenses Definition

A transparent material bound by two surfaces (at least one spherical).
Convex Lens: Thicker in the middle, thinner at edges. Converging lens.
Concave Lens: Thinner in the middle, thicker at edges. Diverging lens.

Lens Formula & Magnification Formula

Similar to the mirror formula but with a crucial sign difference.

$$\frac{1}{v} – \frac{1}{u} = \frac{1}{f} \quad | \quad m = \frac{h’}{h} = +\frac{v}{u}$$

Power of a Lens ($P$) Formula

The degree of convergence or divergence of light rays achieved by a lens. It is the reciprocal of its focal length expressed in meters. The SI unit is the Diopter ($\text{D}$).

$$P = \frac{1}{f \text{ (in meters)}} \quad | \quad 1 \text{ D} = 1 \text{ m}^{-1}$$

Concept Deep Dive

The New Cartesian Sign Convention

The secret to flawless numericals

Core Rule

Think of the mirror’s pole (or the lens’s optical center) as the origin $(0,0)$ on a standard X-Y graph.

1. The object is always placed on the left side of the mirror/lens. Therefore, object distance ($u$) is ALWAYS NEGATIVE.
2. All distances measured to the right are Positive (+). Distances to the left are Negative (-).
3. Heights measured upwards (above principal axis) are Positive (+). Heights downwards are Negative (-).
4. Focal Length Hack: Convex = Positive $f$. Concave = Negative $f$. (This applies to BOTH mirrors and lenses!).

Rear-view Mirrors vs. Headlight Reflectors

Why the shape matters

Real-world Physics

Convex mirrors are used as rear-view mirrors in vehicles. Why? Because they always produce an erect, diminished image, which gives the driver a much wider field of view to see the traffic behind them.

Concave mirrors are used in car headlights and searchlights. Why? Because if you place a light bulb exactly at the focus ($F$) of a concave mirror, the reflected rays will travel completely parallel to each other, creating a powerful, focused beam of light that cuts through the darkness.

Compare & Contrast

✗ Real Image

Formed when light rays actually intersect after reflection/refraction.
Can be captured on a screen (like a movie projector).
Always Inverted (upside down) relative to the object.
Formed by Concave mirrors and Convex lenses (most of the time).

✓ Virtual Image

Formed when light rays appear to diverge from a point; they do not actually intersect.
Cannot be captured on a screen (like your reflection in a bathroom mirror).
Always Erect (right-side up) relative to the object.
Formed by Plane mirrors, Convex mirrors, and Concave lenses.

Common Mistakes to Avoid

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Mistake 1

Using Centimeters for Lens Power: The formula $P = 1/f$ strictly requires focal length to be in Meters. If a question says $f = 20 \text{ cm}$, you cannot do $P = 1/20 = 0.05 \text{ D}$. You must convert: $f = 0.2 \text{ m}$, so $P = 1/0.2 = +5 \text{ D}$.

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Mistake 2

Mixing up Magnification Signs:
Mirror Magnification: $m = -v/u$
Lens Magnification: $m = +v/u$
If you drop the negative sign for mirrors, your final height calculation will be upside down!

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Mistake 3

Drawing Ray Diagrams without Arrows: A line without an arrow is just a geometry line, not a light ray. Examiners will heavily penalize or give zero marks for ray diagrams that do not clearly show the direction of incident and reflected/refracted rays.

Exam Tips

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Tip 1

Decoding Magnification ($m$):
– If $m$ is Negative, the image is Real & Inverted.
– If $m$ is Positive, the image is Virtual & Erect.
– If $|m| > 1$, it is Magnified. If $|m| < 1$, it is Diminished. If $|m| = 1$, it's the exact same size.

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Tip 2

The “Where is the Image?” Trick for Convex Lenses/Concave Mirrors: As the object moves closer to the lens/mirror (from Infinity to F), the real image moves farther away and gets larger. The turning point is $C$ (or $2F$): Object at $2F$ means Image at $2F$, same size. If the object crosses $F$, the image suddenly becomes Virtual and Erect!

Expected Exam Questions

Board Pattern Questions

Class 10 · Science · CBSE Exam

Class 10 · Physics

An object is placed at a distance of $10 \text{ cm}$ from a convex mirror of focal length $15 \text{ cm}$. Find the position and nature of the image. [3 marks]

Answer $v = +6 \text{ cm}$. Virtual, Erect, and Diminished. 📝

Explanation

Given:
Object distance $u = -10 \text{ cm}$ (Always negative)
Focal length $f = +15 \text{ cm}$ (Convex mirror is positive)

Mirror Formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} + \frac{1}{-10} = \frac{1}{15}$
$\frac{1}{v} = \frac{1}{15} + \frac{1}{10} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6}$
$v = +6 \text{ cm}$.

Since $v$ is positive, the image forms behind the mirror. Thus, it is Virtual and Erect. (Magnification $m = -v/u = -6/-10 = +0.6$, so it is diminished).

Light enters from air to glass having a refractive index of $1.50$. What is the speed of light in the glass? The speed of light in vacuum is $3 \times 10^8 \text{ m/s}$. [2 marks]

Answer $v = 2 \times 10^8 \text{ m/s}$ 📝

Explanation

Using the absolute refractive index formula: $n_m = \frac{c}{v}$
$1.50 = \frac{3 \times 10^8}{v}$
$v = \frac{3 \times 10^8}{1.50} = 2 \times 10^8 \text{ m/s}$.

A doctor prescribes a corrective lens of power $+1.5 \text{ D}$. Find the focal length of the lens. Is the prescribed lens diverging or converging? [2 marks]

Answer $f = +0.67 \text{ m}$ ($+66.7 \text{ cm}$). Converging lens. 📝

Explanation

Given Power $P = +1.5 \text{ D}$.
Since $P = \frac{1}{f \text{ (in m)}}$, we have $f = \frac{1}{P}$.
$f = \frac{1}{1.5} = \frac{10}{15} = \frac{2}{3} \text{ m} \approx +0.67 \text{ m}$ (or $+66.7 \text{ cm}$).
Because the focal length and power are Positive, it must be a Convex Lens, which is a Converging lens.

Concept Map

Light Reflection & Refraction connects to →

Optics

Human Eye (Natural convex lens)

Ray Optics (Class 12 deep dive)

Wave Nature of Light (Advanced Physics)

Light – Concept Booster | Class 10 Science CBSE

Key Concepts

Optics Basics