Class 10 Light Notes Made Simple: Reflection, Refraction & Diagrams

Chapter 9 · CBSE · Class X

🔍

Diffraction of Light

Light Reflection of Light Refraction of Light Laws of Reflection Laws of Refraction Spherical Mirrors Concave Mirror Convex Mirror Mirror Formula Magnification Refraction through Glass Slab Lenses Convex Lens Concave Lens Lens Formula Power of a Lens Refractive Index Image Formation Ray Diagrams NCERT Class X Chapter 9

📘 Definition

💡 Concept

🔢 Formula

Important Formula

For single slit diffraction:

\[ a \sin \theta = n\lambda \]

a = width of slit
$\lambda$ = wavelength of light
$\theta$ = angle of diffraction
n = order of minima (n = 1,2,3,...)

📐 Derivation

Derivation (Conceptual)

Consider a slit of width $a$. According to Huygens’ principle, every point on the slit acts as a secondary source of wavelets.

Divide slit into two equal halves.
Light from corresponding points cancels due to destructive interference.
This condition leads to first minimum when:
\[ a \sin \theta = \lambda \]

Similarly, higher order minima occur at:

\[ a \sin \theta = n\lambda \]

🎨 SVG Diagram

Diffraction of Light through a Narrow Slit

✏️ Example

Why is diffraction not observed in everyday life easily?

Diffraction is significant only when obstacle size ≈ wavelength.

Compare wavelength of visible light (~10⁻⁷ m) with daily objects.

Since most objects are much larger than wavelength, diffraction is negligible.

What happens when slit width decreases?

Diffraction increases, spreading becomes more pronounced.

⚡ Exam Tip

❌ Common Mistakes

Confusing diffraction with refraction.
Writing wrong condition (using interference formula incorrectly).
Ignoring wavelength dependence.

📋 Case Study

A beam of monochromatic light passes through a narrow slit and spreads out. A student observes that decreasing slit width increases spreading.

Question: Explain the reason scientifically.

Answer: As slit width becomes comparable to wavelength, diffraction increases due to stronger wave interference effects.

🌟 Importance

🎬 Visualisation

diffraction of a light ray passing through a slit

Animation showing diffraction of a light ray passing through a slit

🔍

Reflection of Light

📘 Definition

🪞 Laws Of Reflection

The incident ray, reflected ray and normal lie in the same plane.
Angle of incidence equals angle of reflection:
\[ i = r \]

🎨 SVG Diagram

🗂️ Types / Category

Types of Reflection

Regular Reflection

Occurs when light reflects off a smooth, polished surface such as a plane mirror, where all reflected rays are parallel to each other. This produces a clear and well‑defined image, like the sharp reflection you see in a glass mirror or calm water surface.

Diffuse Reflection

Occurs when light reflects off a rough or uneven surface, causing the reflected rays to scatter in many directions. This is why we see non‑shiny surfaces clearly from different angles, such as paper, walls, or a waxy leaf, even though they do not form a clear mirror image.

✏️ Example

A ray strikes a mirror at 30°. Find angle of reflection.

Using law,

\[ r = i = 30^\circ \]

⚡ Exam Tip

🔍

Spherical Mirrors

📘 Definition

🗂️ Types / Category

Types of spherical mirror

Concave Mirror

A spherical mirror that bulges inward (curved surface facing the source of light); it converges parallel incident light rays towards a focal point in front of the mirror. Common examples include shaving mirrors, dentist’s head mirrors, and the reflecting mirrors used in Newtonian telescopes and solar concentrators.

Convex Mirror

A spherical mirror that bulges outward (curved surface facing away from the source of light); it diverges parallel incident light rays, forming a virtual, diminished, and erect image behind the mirror. Typical examples are the rear‑view mirrors in vehicles and security mirrors used in shops or at blind corners for a wider field of view.

🔢 Formula

Important Relations

Relation Between Radius of Curvature $'R\;'$ and focal length $'f\;'$

\[ f = \frac{R}{2} \]

Mirror Formula

\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]

Magnification

\[ m = \frac{h_i}{h_o} = \frac{-v}{u} \]

📌 Note

Sign Convention (Cartesian)

📐 Derivation

Derivation of Mirror Formula

Consider a spherical mirror of small aperture. Let an object AB be placed on the principal axis, and let its image A′B′ be formed after reflection. Using the geometry of the ray diagram, two pairs of triangles are obtained that are similar.

From the similar triangles formed near the pole of the mirror, we get a relation between the height of the object, the height of the image, and their distances from the pole:

\[ \frac{A'B'}{AB} = -\frac{v}{u} \]

By applying the second pair of similar triangles involving the focus F, we obtain another relation:

\[ \frac{A'B'}{AB} = -\frac{f-v}{f} \]

Equating the two expressions,

\[ -\frac{v}{u} = -\frac{f-v}{f} \]

On simplifying,

\[ \frac{f-v}{f} = \frac{v}{u} \]

\[ u(f-v)=fv \]

\[ uf-uv=fv \]

\[ uf=fv+uv \]

Dividing the whole equation by ufv, we get

\[ \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \]

Thus, the mirror formula is

\[ \boxed{\frac{1}{f}=\frac{1}{v}+\frac{1}{u}} \]

This relation holds for spherical mirrors when distances are measured according to the New Cartesian sign convention.

✏️ Example

Numerical Example

Object at 20 cm from concave mirror, f = -10 cm. Find image distance.

\[ \begin{aligned} \frac{1}{f} &= \frac{1}{v} + \frac{1}{u}\\\\ \frac{1}{-10} &= \frac{1}{v} + \frac{1}{-20}\\\\ \frac{1}{v} &= -\frac{1}{10} + \frac{1}{20} \\\\ &= -\frac{1}{20}\\\\ \Rightarrow v &= -20 \, \mathrm{cm} \end{aligned} \]

Image Distance is 20 cm before mirror

📌 Note

Applications

❌ Common Mistakes

Wrong sign convention
Confusing concave and convex formulas
Ignoring negative focal length for concave mirror

📋 Case Study

A driver uses a convex mirror. Explain why.

Answer: Convex mirrors provide wider field of view and always form erect images.

🔍

Image Formation by Spherical Mirrors

📘 Definition

📌 Note

Ray Rules

🎨 SVG Diagram

Ray Diagram: Real, Inverted, and Diminished Image

📊 Comparison Table

Concave Mirror Image Formation

Object Position	Image Position	Nature	Size
At infinity	At F	Real, inverted	Highly diminished
Beyond C	Between F and C	Real, inverted	Diminished
At C	At C	Real, inverted	Same size
Between C and F	Beyond C	Real, inverted	Magnified
At F	At infinity	Real, inverted	Highly enlarged
Between P and F	Behind mirror	Virtual, erect	Magnified

📊 Comparison Table

Convex Mirror Image Formation

Object Position	Image Position	Nature	Size
Anywhere	Between P and F	Virtual, erect	Diminished

💡 Concept

Conceptual Understanding

🔢 Formula

\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]

\[ m = \frac{-v}{u} \]

✏️ Example

Numerical Example

Object at 15 cm, f = -10 cm. Find image position.

\[ \begin{aligned} \frac{1}{-10} &= \frac{1}{v} + \frac{1}{-15}\\\\ \frac{1}{v} &= -\frac{1}{10} + \frac{1}{15} \\\\ &= -\frac{1}{30}\\\\ \Rightarrow v &= -30 \, cm \end{aligned} \]

Image postion 30 cm before mirror

⚡ Exam Tip

❌ Common Mistakes

Confusing real and virtual images.
Wrong image position (especially between C and F).
Forgetting convex mirror always gives virtual image.

📋 Case Study

A student places an object between F and P of a concave mirror.

Question: Describe the image formed.

Answer: Image is virtual, erect, and magnified behind the mirror.

🔍

Ray Diagrams for Image Formation by Spherical Mirrors

📘 Definition

🌟 Importance

Why Ray Diagrams Matter

📌 Note

Standard Rays Used in Construction

Parallel Ray: A ray parallel to the principal axis passes through the focus after reflection from a concave mirror, and appears to come from the focus after reflection from a convex mirror.
Focus Ray: A ray passing through the focus reflects parallel to the principal axis.
Centre Ray: A ray passing through the centre of curvature retraces its path after reflection.
Pole Ray: A ray incident at the pole reflects according to the law of reflection, with $ i = r $.

🎨 SVG Diagram

Stepwise Construction (Concave Mirror)<

The reflected rays meet in front of the mirror to form a real, inverted image.

Draw the principal axis and the concave mirror.
Mark the pole $P$, focus $F$, and centre of curvature $C$.
Draw one ray parallel to the principal axis; after reflection, it passes through the focus.
Draw a second ray through the focus; after reflection, it becomes parallel to the principal axis.
The intersection of the reflected rays gives the image position.

📌 Note

Convex Mirror Insight

💡 Concept

Conceptual Clarity

🔢 Formula

Connection with Mirror Formula

✏️ Example

Object placed beyond C in concave mirror. Where is image formed?

Use parallel ray + focus ray

Image forms between F and C, real and inverted.

⚡ Exam Tip

❌ Common Mistakes

Using incorrect ray rules.
Forgetting backward extension in convex mirror.
Not marking principal axis.
Drawing more than necessary rays (wastes time).

📋 Case Study

A student draws only one ray while constructing a ray diagram.

Question: Why is the diagram incorrect?

Answer: At least two rays are needed to determine exact image position.

🔍

Sign Convention for Reflection by Spherical Mirrors

📘 Definition

🎨 SVG Diagram

Coordinate System Representation

New Cartesian coordinate system for mirrors

Rules

Object is always placed on the left side of the mirror.
All distances are measured from the pole (P).
Distances to the right (+x direction) are positive.
Distances to the left (−x direction) are negative.
Heights above principal axis (+y) are positive.
Heights below principal axis (−y) are negative.

📊 Comparison Table

Quick Sign Reference Table

Quantity	Concave Mirror	Convex Mirror
Focal length (f)	Negative	Positive
Radius (R)	Negative	Positive
Object distance (u)	Always Negative	Always Negative
Image distance (v)	Negative (real), Positive (virtual)	Always Positive
Image height (hᵢ)	Negative (inverted), Positive (erect)	Positive

🔢 Formula

\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]

\[ m = \frac{-v}{u} \]

✏️ Example

🗒️ Quetion

Object at 15 cm in front of concave mirror, f = 10 cm. Find image distance.

Assign signs → $ u = -15 $, $ f = -10 $

\[ \begin{aligned} \frac{1}{-10} &= \frac{1}{v} + \frac{1}{-15}\\\\ \frac{1}{v} &= -\frac{1}{10} + \frac{1}{15} \\\\&= -\frac{1}{30}\\\\ \Rightarrow v &= -30 \, cm \end{aligned} \]

Image is real and formed in front of mirror 30 cm distant

💡 Concept

Conceptual Insight

⚡ Exam Tip

❌ Common Mistakes

Taking focal length of concave mirror as positive.
Ignoring negative sign of object distance.
Mixing lens and mirror sign conventions.

📋 Case Study

A student gets a positive image distance for a concave mirror in a numerical.

Question: What does this indicate?

Answer: Image is virtual and formed behind the mirror.

🔍

Mirror Formula and Magnification (Derivation, Numericals & Concepts)

📘 Definition

The mirror formula establishes a mathematical relationship between the object distance (u), image distance (v), and focal length (f) of a spherical mirror.

\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]

f → focal length of mirror
v → image distance
u → object distance

Magnification

Magnification describes how much larger or smaller the image is compared to the object.

\[ m = \frac{h_i}{h_o} = \frac{-v}{u} \]

$ m > 1 $ → image enlarged
$ 0 < m < 1 $ → image diminished
$ m < 0 $ → image inverted (real)
$ m > 0 $ → image erect (virtual)

📐 Derivation

Derivation of Mirror Formula

The mirror formula is derived using similar triangles formed in ray diagrams:

Consider object AB and image A'B'.
Using geometry of triangles:
\[ \frac{A'B'}{AB} = \frac{v}{u} \]
Using another pair of similar triangles involving focus:
\[ \frac{A'B'}{AB} = \frac{f - v}{f} \]

Equating and simplifying leads to:

\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]

Conceptual Diagram

Geometry for derivation of 1/f = 1/v + 1/u

✏️ Example

Numerical Example

An object is placed 20 cm in front of a concave mirror of focal length 15 cm. Find image position and magnification.

Assign signs → $ u = -20 $, $ f = -15 $

\[ \begin{aligned} \frac{1}{-15} &= \frac{1}{v} + \frac{1}{-20}\\\\ \frac{1}{v} &= -\frac{1}{15} + \frac{1}{20} \\\\&= -\frac{1}{60}\\\\ \Rightarrow v &= -60 \, \mathrm{cm}\\\\ \end{aligned} \]

\[ \begin{aligned} m &= \frac{-v}{u} \\\\ &= \frac{-(-60)}{-20} \\\\ &= -3 \end{aligned} \]

Image is real, inverted, and magnified. Image Position is 60 cm in front of Mirror with magnification 3 times

💡 Concept

Conceptual Insights

🗺️ Roadmap

Quick Problem-Solving Strategy

Draw rough ray diagram mentally.
Apply correct sign convention.
Use mirror formula.
Use magnification for verification.

⚡ Exam Tip

❌ Common Mistakes

Ignoring sign convention.
Wrong substitution of u and v.
Forgetting negative sign in magnification formula.

📋 Case Study

A student gets positive magnification for a concave mirror.

Question: What does this imply?

Answer: Image is virtual and erect (object placed between P and F).

🔍

Example-1

❓ Question

An object is placed at a distance of 4 cm in front of a concave mirror of radius of curvature 24 cm.

Find the position of the image.
State whether the image is magnified.

🗺️ Roadmap

Solution Roadmap

Apply mirror formula
Calculate image distance $ v $
Find magnification $ m $
Interpret nature of image

🧩 Solution

Given

Radius of curvature $ R = 24 \, cm $
Focal length $ f = \frac{R}{2} = 12 \, cm $
Object distance $ u = -4 \, cm $
For concave mirror → $ f = -12 \, cm $

\[ \begin{aligned} \frac{1}{f} &= \frac{1}{v} + \frac{1}{u} \\\\ \frac{1}{-12} &= \frac{1}{v} + \frac{1}{-4}\\\\ \frac{1}{v} &= -\frac{1}{12} + \frac{1}{4}\\\\ \frac{1}{v} &= \frac{1}{4} - \frac{1}{12} \\\\ &= \frac{3 - 1}{12} = \frac{2}{12} = \frac{1}{6}\\\\ v &= +6 \, cm \end{aligned} \]

Magnification

\[ \begin{aligned} m &= \frac{-v}{u} \\\\&= \frac{-6}{-4} \\\\&= 1.5 \end{aligned} \]

Image distance $ v = +6 \, cm $
Magnification $ m = 1.5 $

Interpretation

$ v $ is positive → image is virtual and formed behind the mirror
$ m > 1 $ → image is magnified
$ m > 0 $ → image is erect

💡 Concept

Concept Check

⚡ Exam Tip

❌ Common Mistakes

Taking $ f = +12 $ instead of $ -12 $
Ignoring sign of $ u $
Not interpreting final answer

🔍

Example-2

❓ Question

At what distance from a concave mirror of focal length 25 cm should an object be placed so that the image is of the same size as the object?

💡 Concept

🗺️ Roadmap

Solution Roadmap

Use magnification formula
Substitute into mirror formula
Solve for object distance

🧩 Solution

Given

Focal length $ f = -25 \, cm $ (concave mirror)
Magnification $ m = -1 $

\[ \begin{aligned} m &= \frac{-v}{u}\\\\ -1 &= \frac{-v}{u}\\\\ \Rightarrow v &= u \end{aligned} \]

\[ \begin{aligned} \frac{1}{f} &= \frac{1}{v} + \frac{1}{u}\\\\ \frac{1}{-25} &= \frac{1}{u} + \frac{1}{u}\\\\ \frac{1}{-25} &= \frac{2}{u}\\\\ \Rightarrow u &= -50 \, cm \end{aligned} \]

Object distance $ u = -50 \, cm $
Distance from mirror = 50 cm in front of mirror

Interpretation

Image is formed at $ v = -50 \, cm $
Image is real and inverted
Object is placed at centre of curvature (C = 2f)

💡 Concept

Concept Check

⚡ Exam Tip

❌ Common Mistakes

Using $ m = +1 $ instead of $ -1 $
Forgetting negative sign of focal length
Writing only 50 cm without direction

🔍

Example-3

❓ Question

An object 5 cm high is placed at a distance of 60 cm in front of a concave mirror of focal length 10 cm. Find:

Position of image
Size (height) of image

🗺️ Roadmap

Solution Roadmap

Apply mirror formula to find $ v $
Use magnification to find image height
Interpret sign for nature of image

🧩 Solution

GIven

Object height $ h_o = 5 \, cm $
Object distance $ u = -60 \, cm $
Focal length $ f = -10 \, cm $

\[ \begin{aligned} \frac{1}{f} &= \frac{1}{v} + \frac{1}{u}\\\\ \frac{1}{-10} &= \frac{1}{v} + \frac{1}{-60}\\\\ \frac{1}{v} &= -\frac{1}{10} + \frac{1}{60}\\\\ \frac{1}{v} &= \frac{-6 + 1}{60} \\\\ &= -\frac{5}{60} \\\\ &= -\frac{1}{12}\\\\ v &= -12 \, cm \end{aligned} \]

MAgnification

\[ \begin{aligned} m = \frac{-v}{u} &= \frac{-(-12)}{-60} \\\\ &= -\frac{12}{60} \\\\ &= -\frac{1}{5}\\\\ m &= \frac{h_i}{h_o}\\\\ \Rightarrow -\frac{1}{5} &= \frac{h_i}{5}\\\\ h_i &= -1 \, cm \end{aligned} \]

Image distance $ v = -12 \, cm $
Image height $ h_i = -1 \, cm $

Interpretation

Negative $ v $ → image is real and formed in front of mirror
Negative $ h_i $ → image is inverted
$ |m| = \frac{1}{5} $ → image is diminished
Image forms between F and C

💡 Concept

Concept Link

⚡ Exam Tip

❌ Common Mistakes

Using $ m = \frac{v}{u} $ instead of correct formula
Ignoring negative sign of image height
Not stating nature of image

🔍

example-4

❓ Question

A point light source is placed at a distance of 40 cm in front of a convex mirror of focal length 40 cm. Find the position of the image.

🗺️ Roadmap

Solution Roadmap

Apply mirror formula
Solve for image distance $ v $
Interpret sign of $ v $

🧩 Solution

\[ \begin{aligned} \frac{1}{f} &= \frac{1}{v} + \frac{1}{u}\\\\ \frac{1}{40} &= \frac{1}{v} + \frac{1}{-40}\\\\ \frac{1}{40} &= \frac{1}{v} - \frac{1}{40}\\\\ \frac{1}{v} &= \frac{1}{40} + \frac{1}{40} \\\\ &= \frac{2}{40} \\\\ &= \frac{1}{20}\\\\ v &= +20 \, \mathrm{cm} \end{aligned} \]

Image distance $ v = +20 \, cm $

Interpretation

Positive $ v $ → image is formed behind the mirror
Image is virtual and erect
Image is diminished

💡 Concept

Concept Link

⚡ Exam Tip

❌ Common Mistakes

Taking $ f $ as negative for convex mirror
Forgetting sign of $ u $
Not interpreting the final answer

🔍

Example-5

🗒️ Quetion

An object of height 4 cm is placed 30 cm in front of a concave mirror of focal length 15 cm.

a) Find the position of the image
b) Find the height of the image

🗺️ Roadmap

Solution ROadmap

Apply mirror formula to find $ v $
Use magnification to find $ h_i $
Interpret signs

🧩 Solution

\[ \begin{aligned} \frac{1}{f} &= \frac{1}{v} + \frac{1}{u}\\\\ \frac{1}{-15} &= \frac{1}{v} + \frac{1}{-30}\\\\ -\frac{1}{15} &= \frac{1}{v} - \frac{1}{30}\\\\ \frac{1}{v} &= -\frac{1}{15} + \frac{1}{30}\\\\ &= \frac{-2 + 1}{30}\\\\ &= -\frac{1}{30}\\\\ v &= -30 \, cm \end{aligned} \]

Magnification

\[ \begin{aligned} m &= \frac{-v}{u} &= \frac{-(-30)}{-30} \\\\ &= -1\\\\ m &= \frac{h_i}{h_o}\\\\ \Rightarrow -1 &= \frac{h_i}{4}\\\\ h_i &= -4 \, cm\\\\ \end{aligned} \]

Image distance $ v = -30 \, cm $
Image height $ h_i = -4 \, cm $

Interpretation

Negative $ v $ → image is real (in front of mirror)
Negative $ h_i $ → image is inverted
$ |m| = 1 $ → image is same size
Object is at centre of curvature (C = 2f)

💡 Concept

Concept Link

⚡ Exam Tip

❌ Common Mistakes

Writing $ h' = +4 \, cm $ instead of $ -4 \, cm $
Ignoring sign of magnification
Not identifying centre of curvature case

🔍

Example-6

❓ Question

A concave mirror forms a virtual image twice the size of the object when the object is placed 5 cm from it. Find:

a) Focal length of the mirror
b) Position of the image

🗺️ Roadmap

Solution Raodmap

Use magnification formula to find $ v $
Substitute in mirror formula to find $ f $

🧩 Solution

\[ \begin{aligned} m &= \frac{-v}{u}\\\\ 2 &= \frac{-v}{-5}\\\\ \Rightarrow 2 &= \frac{v}{5}\\\\ \Rightarrow v &= 10 \, \mathrm{cm}\\\\ \frac{1}{f} &= \frac{1}{v} + \frac{1}{u}\\\\ \frac{1}{f} &= \frac{1}{10} + \frac{1}{-5}\\\\ &= \frac{1}{10} - \frac{2}{10}\\\\ &= -\frac{1}{10}\\\\ f &= -10 \, \mathrm{cm} \end{aligned} \]

Focal length $ f = -10 \, cm $
Image distance $ v = +10 \, cm $

Interpretation

Positive $ v $ → image formed behind the mirror
Image is virtual and erect
Magnified ($ m = 2 $)
Object is placed between P and F

💡 Concept

Concept Link

⚡ Exam Tip

❌ Common Mistakes

Using $ m = \frac{v}{u} $ instead of correct formula
Taking $ u = +5 $ instead of $ -5 $
Missing interpretation of virtual image

🔍

Refraction of Light

📘 Definition

🤔 Did You Know?

Why Does Refraction Occur?

Speed of light changes in different media.
Light slows down in denser medium and bends towards the normal.
Light speeds up in rarer medium and bends away from the normal.

📍 Key Point

Laws of Refraction (Snell’s Law)

Incident ray, refracted ray and normal lie in the same plane.
\[ \frac{\sin i}{\sin r} = \text{constant} \]

🧠 Remember

Refractive Index

\[ n = \frac{\sin i}{\sin r} \]

Also defined as:

\[ n = \frac{c}{v} \]

$ c $ = speed of light in vacuum
$ v $ = speed of light in medium

📍 Key Point

Optical Density

Medium with higher refractive index → optically denser
Medium with lower refractive index → optically rarer

🎨 SVG Diagram

Ray Diagram

Light bending toward the normal as it enters a denser medium.

⭐ Special Case

🧪 Experiment

Refraction Through Glass Slab

Emergent ray is parallel to incident ray
Lateral displacement occurs

Corrected Geometry:

● EO: Incident ray hits at O.
● OO': Refracted ray bends toward normal inside glass.
● O'H: Emergent ray shifts away from the original path (dashed line).
● d (Lateral Displacement): Perpendicular distance O'L between the path extension and emergent ray.

🔢 Formula

\[ n_1 \sin i = n_2 \sin r \]

✏️ Example

Numerical Example

If $ i = 30^\circ $, $ r = 20^\circ $, find refractive index.

\[ \begin{aligned} n &= \frac{\sin 30^\circ}{\sin 20^\circ}\\ &\approx \frac{0.5}{0.342} \approx 1.46 \end{aligned} \]

Refractive index $\approx 1.46$

Conceptual Insight

Greater bending → higher refractive index
Speed and refractive index are inversely related

🗒️ Everyday Applications

Everyday Applications

Coin appears raised in water
Pencil appears bent in water
Lenses in spectacles and cameras
Mirage and rainbow formation

⚡ Exam Tip

❌ Common Mistakes

Confusing denser and rarer medium
Writing wrong formula
Ignoring units in speed-based formula

🔍

Refractive Index

📘 Definition

Refractive index is a measure of how much light slows down in a medium compared to another medium. It determines the bending of light during refraction.

Relation with Snell’s Law

\[ n_{21} = \frac{\sin i}{\sin r} \]

This gives refractive index of medium 2 with respect to medium 1.

Relative Refractive Index

\[ n_{21} = \frac{v_1}{v_2} \]

$ v_1 $ = speed of light in medium 1
$ v_2 $ = speed of light in medium 2

Similarly:

\[ n_{12} = \frac{v_2}{v_1} \]

🗒️ Important

\[ n_{12} = \frac{1}{n_{21}} \]

📌 Note

Absolute Refractive Index

When refractive index is measured with respect to vacuum (or air), it is called absolute refractive index.

\[ n = \frac{c}{v} \]

$ c = 3 \times 10^8 \, m/s $ (speed of light in vacuum)
$ v $ = speed of light in medium

👁️ Observation

Optical Density & Refractive Index

📊 Comparison Table

Common Refractive Indices

Medium	Refractive Index
Air	~1.0003
Water	1.33
Glass	1.5
Diamond	2.42

✏️ Example

Speed of light in a medium is $ 2 \times 10^8 \, m/s $. Find refractive index.

\[ n = \frac{c}{v} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5 \]

Refractive index = 1.5

Higher refractive index → more bending of light
Speed decreases as refractive index increases
Light bends towards normal in denser medium

⚡ Exam Tip

❌ Common Mistakes

Confusing relative and absolute refractive index
Writing $ n = v/c $ (wrong)
Ignoring units in speed-based questions

🔍

Optical Density (Concept, Relation & Applications)

📘 Definition

🗒️ Important Clarification

Important Clarification

Optical density is not related to mass density
A medium can be optically denser but physically lighter
It depends on interaction of light with medium, not mass

📌 Note

Relation with Refractive Index

📊 Comparison Table

Comparison of Media

Property	Optically Denser Medium	Optically Rarer Medium
Refractive Index	Higher	Lower
Speed of Light	Lower	Higher
Refraction	Towards normal	Away from normal

💡 Concept

Conceptual Understanding

✏️ Example

Between water (n = 1.33) and glass (n = 1.5), which is optically denser?

Glass is optically denser because it has higher refractive index.

⚡ Exam Tip

❌ Common Mistakes

Confusing optical density with mass density
Assuming heavier object is optically denser
Ignoring relation with refractive index

🔍

Refraction by Spherical Lenses

📘 Definition

🗂️ Types / Category

Types of Lenses

Convex Lens (Converging)

Thicker at the centre and thinner at the edges; it converges parallel rays to a principal focus. It can form real and inverted images or virtual and enlarged images depending on the object position.

Concave Lens (Diverging)

Thinner at the centre and thicker at the edges; it diverges parallel rays outward as if they came from a principal focus. It always forms a virtual, erect, and diminished image.

🎨 SVG Diagram

Basic Structure

Pure lens profiles: Converging (Thick middle) vs Diverging (Thin middle).

🧠 Remember

Important Terms

🗒️ Important

Ray Rules

Ray parallel to principal axis → passes through focus (convex) or appears from focus (concave)
Ray through focus → emerges parallel to axis
Ray through optical centre → undeviated

🔢 Formula

Lens Formula

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

🔢 Formula

Magnification

\[ m = \frac{h_i}{h_o} = \frac{v}{u} \]

📌 Note

Sign Convention

✏️ Example

Object at 20 cm, convex lens $ f = 10 \, cm $. Find image distance.

\[ \begin{aligned} \frac{1}{10} &= \frac{1}{v} - \frac{1}{-20}\\\\ \frac{1}{10} &= \frac{1}{v} + \frac{1}{20}\\\\ \Rightarrow \frac{1}{v} &= \frac{1}{10} - \frac{1}{20} \\\\ &= \frac{1}{20}\\\\ v &= 20 \, \mathrm{cm} \end{aligned} \]

Image distance = 20 cm

💡 Concept

Conceptual Insights

🛠️ Application

Convex lens → magnifying glass, camera
Concave lens → spectacles for myopia

⚡ Exam Tip

❌ Common Mistakes

Using mirror formula for lens
Wrong sign of focal length
Forgetting $ v/u $ relation in magnification

🔍

Image Formation by Lenses (Ray Diagrams + Formula + Analysis)

📘 Definition

🗒️ Standard Rays for Construction

Standard Rays for Construction

Ray parallel to principal axis → passes through focus (convex) or appears from focus (concave)
Ray through focus → emerges parallel to axis
Ray through optical centre → undeviated

🎨 SVG Diagram

Ray Diagram (Convex Lens)

Real, Inverted Image formed by Convex Lens

📊 Comparison Table

Convex Lens (Converging)

Object Position	Image Position	Size	Nature
At infinity	At F₂	Highly diminished	Real, inverted
Beyond 2F₁	Between F₂ & 2F₂	Diminished	Real, inverted
At 2F₁	At 2F₂	Same size	Real, inverted
Between F₁ & 2F₁	Beyond 2F₂	Magnified	Real, inverted
Between F₁ & O	Same side	Magnified	Virtual, erect

📊 Comparison Table

Concave Lens (Diverging)

Object Position	Image Position	Size	Nature
At infinity	At F₁	Highly diminished	Virtual, erect
Anywhere	Between F₁ & O	Diminished	Virtual, erect

🔢 Formula

Lens Formula & Magnification

\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\]

\[m = \frac{v}{u}\]

💡 Concept

Conceptual Insights

✏️ Example

Numerical Example

Object at 15 cm from convex lens, $ f = 10 \, cm $. Fid distance of Image

\[ \begin{aligned} \frac{1}{10} &= \frac{1}{v} - \frac{1}{-15}\\\\ \frac{1}{10} &= \frac{1}{v} + \frac{1}{15}\\\\ \Rightarrow \frac{1}{v} &= \frac{1}{10} - \frac{1}{15} = \frac{1}{30}\\\\ v &= 30 \, \mathrm{cm} \end{aligned} \]

Distance of Image = 30 cm

⚡ Exam Tip

❌ Common Mistakes

Confusing F₁ and F₂ sides
Using mirror formula instead of lens formula
Wrong sign of focal length

🔍

Image Formation in Lenses Using Ray Diagrams (Stepwise Construction)

📘 Definition

🌟 Importance

📌 Note

Ray Rule

Parallel Ray: After refraction, passes through focus (convex) or appears from focus (concave).
Ray through Focus: Emerges parallel to principal axis.
Ray through Optical Centre: Passes undeviated.

🎨 SVG Diagram

Ray Diagram (Convex Lens)

CONVEX CONVERGENCE

Step by step Construction

Draw principal axis and lens.
Mark optical centre (O) and focal points (F₁, F₂).
Place object at given position
Draw any two standard rays.
Locate intersection of refracted rays → image position.
If rays diverge, extend them backward to locate virtual image.

📊 Comparison Table

Convex vs Concave Lens Behaviour

Feature	Convex Lens	Concave Lens
Shape	Thicker at center, thinner at edges	Thinner at center, thicker at edges
Curvature	Bulging outward (convex surfaces)	Curved inward (concave surfaces)
Action on light rays	Converges parallel rays to a focal point	Diverges parallel rays outward
Type of lens	Converging lens	Diverging lens
Focal length (sign)	Positive (f > 0)	Negative (f < 0)
Image type (in general)	Can form both real and virtual images	Always forms virtual images
Image orientation (typical)	Can be real and inverted or virtual and erect	Always erect
Image size (typical)	Can be magnified, same size, or diminished	Always diminished (smaller than object)
Common uses	Magnifying glass, cameras, telescopes, projectors, correcting hyperopia	Correcting myopia, peepholes, some laser‑beam expanders

💡 Concept

Conceptual Understanding

🔢 Formula

Connection with Lens Formula

✏️ Example

Object placed between F and 2F of convex lens. Where is image formed?

Beyond 2F, real, inverted, magnified.

⚡ Exam Tip

❌ Common Mistakes

Forgetting backward extension for concave lens
Using incorrect ray rules
Not marking focal points

📋 Case Study

A student draws only one ray for a convex lens diagram.

Question: Why is it incorrect?

Answer: At least two rays are required to locate image accurately.

🔍

Sign Convention for Spherical Lenses (New Cartesian System)

📘 Definition

📜 Rules

Cartesian Sign Convention

Academic Standards: ISO Optical Notation

Rules

All distances are measured from optical centre (O).
Distances in direction of incident light (left → right) are positive.
Distances opposite to incident light are negative.
Heights above principal axis are positive; below are negative.
Object is always placed on left → $ u $ is always negative.

📌 Note

Quick Sign Reference Table

Quantity	Sign	Condition
Object distance (u)	Negative	Always (object on left)
Image distance (v)	Positive	Real image (right side)
Image distance (v)	Negative	Virtual image (left side)
Focal length (f)	Positive	Convex lens
Focal length (f)	Negative	Concave lens
Image height (hᵢ)	Positive	Erect image
Image height (hᵢ)	Negative	Inverted image

🔢 Formula

Use in Lens Formula

✏️ Example

Numerical Example

Object at 20 cm from convex lens of focal length 10 cm.

Assign signs: $ u = -20 $, $ f = +10 $

\[ \begin{aligned} \frac{1}{10} &= \frac{1}{v} - \frac{1}{-20}\\ \frac{1}{10} = \frac{1}{v} + \frac{1}{20}\\ \Rightarrow \frac{1}{v} &= \frac{1}{20}\\ \Rightarrow v &= +20 \, \mathrm{cm} \end{aligned} \]

💡 Concept

Conceptual Insights

⚡ Exam Tip

❌ Common Mistakes

Taking $ f $ negative for convex lens
Forgetting $ u $ is always negative
Confusing sign of magnification

📋 Case Study

A student obtains negative image distance for convex lens.

Question: What does it indicate?

Answer: Image is virtual and formed on same side as object.

🔍

Lens Formula and Magnification (Derivation, Numericals & Concepts)

📘 Definition

The lens formula gives the relationship between object distance (u), image distance (v), and focal length (f) of a spherical lens.

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

f → focal length
v → image distance
u → object distance

Magnification

Magnification tells how much the image is enlarged or diminished compared to the object.

\[ m = \frac{h_i}{h_o} = \frac{v}{u} \]

$ m > 1 $ → enlarged image
$ 0 < m < 1 $ → diminished image
$ m > 0 $ → erect image
$ m < 0 $ → inverted image

📐 Derivation

Derivation of Lens Formula

GEOMETRIC OPTICS: LENS FORMULA DERIVATION

Derivation of Lens Formula

Using ray‑diagram geometry and similar triangles formed by the object, image, and focal points, we can derive the lens formula. Consider a real object $AB$ and its real image $A'B'$ formed by a convex lens, with heights $h_o$ and $h_i$ respectively, object distance $u$, image distance $v$, and focal length $f$.

From similar triangles $\triangle ABO \sim \triangle A'B'O$ (ray through the optical center): $$\frac{h_i}{h_o} = -\frac{v}{u}$$
From similar triangles $\triangle ODF \sim \triangle A'B'F$ (ray parallel to the principal axis passing through the focal point, where $OD = h_o$): $$\frac{h_i}{h_o} = \frac{v - f}{f}$$

Equating the two expressions for $\frac{h_i}{h_o}$ yields: $$-\frac{v}{u} = \frac{v - f}{f}$$

On cross‑multiplying and simplifying: $$ \begin{aligned} -vf &= u(v - f) \\ -vf &= uv - uf \\ uf - vf &= uv \\ \end{aligned} $$

Dividing throughout by $uvf$: $$ \begin{aligned} \frac{uf}{uvf} - \frac{vf}{uvf} &= \frac{uv}{uvf} \\\\ \frac{1}{v} - \frac{1}{u} &= \frac{1}{f} \end{aligned} $$

Final Lens Formula: $$\boxed{\bbox[2pt]{\frac{1}{f} = \frac{1}{v} - \frac{1}{u}}}$$

✏️ Example

An object is placed 15 cm from a convex lens of focal length 10 cm. Find image position and magnification.

Assign signs → $ u = -15 $, $ f = +10 $

\[ \begin{aligned} m &= \frac{v}{u} \\\\&= \frac{30}{-15} \\\\&= -2 \end{aligned} \]

$ \text{Image is real, inverted, and magnified.} $

💡 Concept

Conceptual Insights

📌 Note

Quick Problem-Solving Strategy

⚡ Exam Tip

❌ Common Mistakes

Using wrong formula sign
Ignoring negative value of u
Not interpreting magnification

📋 Case Study

A student gets positive magnification for convex lens.

Question: What does it indicate?

Answer: Image is virtual and erect.

🔍

Power of a Lens (Formula, Unit, Combination & Numericals)

📘 Definition

🔢 Formula

\[ P = \frac{1}{f} \]

$ P $ = power of lens (dioptre, D)
$ f $ = focal length (in metres)

Unit of Power

SI unit: dioptre (D)
$ 1 \, D = 1 \, m^{-1} $
Always convert focal length into metres before calculation

Sign Convention

Convex lens → $ f > 0 $ → $ P > 0 $
Concave lens → $ f < 0 $ → $ P < 0 $

Important Conversion

If focal length is given in cm:

\[ P = \frac{100}{f(\text{in cm})} \]

📌 Note

Combination of Lenses

✏️ Example

Example

Find power of lens of focal length 50 cm.

\[ f = 50 \, \mathrm{cm} = 0.5 \, \mathrm{m} \]

\[ P = \frac{1}{0.5} = 2 \, \mathrm{D} \]

Example

Two lenses of powers +2 D and −1 D are combined. Find total power.

\[P = 2 + (-1) = 1 \, \mathrm{D}\]

💡 Concept

Conceptual Insights

🤔 Did You Know?

Quick Shortcuts

25 cm → 4 D
50 cm → 2 D
100 cm → 1 D

🛠️ Application

Used in spectacles to correct vision defects
Used in cameras and microscopes

⚡ Exam Tip

❌ Common Mistakes

Not converting cm to metres
Wrong sign of focal length
Forgetting unit (D)

📋 Case Study

A lens has power −2 D. Identify the type of lens and its focal length.

Answer: Concave lens, $ f = -0.5 \, m $

🔍

Example-7

❓ Question

A convex lens of focal length 10 cm is placed at a distance of 12 cm from a wall. At what distance from the lens should an object be placed so that a real image is formed on the wall?

🗺️ Roadmap

Solution Roadmap

Apply lens formula
Substitute values with correct signs
Solve for $ u $

🧩 Solution

Given

Focal length $ f = +10 \, cm $
Image distance $ v = +12 \, cm $ (real image on wall)
Object distance $ u = ? $

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

\[ \frac{1}{10} = \frac{1}{12} - \frac{1}{u} \]

\[ \frac{1}{u} = \frac{1}{12} - \frac{1}{10} \]

\[ \frac{1}{u} = \frac{5 - 6}{60} = -\frac{1}{60} \]

\[ u = -60 \, cm \]

Object distance $ u = -60 \, cm $
Object should be placed 60 cm in front of the lens

📌

Interpretation

Negative $ u $ → object is on left side (as expected)
Since image is real → object is placed beyond focal length

💡 Concept

Concept Link

⚡ Exam Tip

❌ Common Mistakes

Taking $ v $ negative for real image
Not applying correct formula sign
Writing only 60 cm without sign or direction

🔍

Example-8

❓ Question

An object of height 7 cm is placed at a distance of 12 cm from a convex lens of focal length 8 cm. Find the position, nature, and height of the image.

🗺️ Roadmap

Solution Roadmap

Apply lens formula to find $ v $
Use magnification to find $ h_i $
Interpret signs

🧩 Solution

Given

Object height $ h_o = 7 \, cm $
Object distance $ u = -12 \, cm $
Focal length $ f = +8 \, cm $

\[ \begin{aligned} \frac{1}{f} &= \frac{1}{v} - \frac{1}{u}\\\\ \frac{1}{8} &= \frac{1}{v} - \left(\frac{1}{-12}\right)\\\\ \frac{1}{8} &= \frac{1}{v} + \frac{1}{12}\\\\ \frac{1}{v} &= \frac{1}{8} - \frac{1}{12}\\\\ &= \frac{3 - 2}{24}\\\\ &= \frac{1}{24}\\\\ v &= +24 \, cm \end{aligned} \]

Magnification

\[ \begin{aligned} m &= \frac{v}{u} \\ &= \frac{24}{-12} \\ &= -2\\\\ m &= \frac{h_i}{h_o}\\ \Rightarrow -2 &= \frac{h_i}{7}\\ \Rightarrow h_i &= -14 \end{aligned} \]

Image distance $ v = +24 \, cm $
Image height $ h_i = -14 \, cm $

📌

Note

Positive $ v $ → image formed on right side (real image)
Negative $ h_i $ → image is inverted
$ |m| = 2 $ → image is magnified (twice the size)
Image is formed beyond 2F

💡 Concept

Concept Link

⚡ Exam Tip

❌ Common Mistakes

Writing $ h_i = +14 \, cm $ instead of $ -14 \, cm $
Ignoring sign of magnification
Not interpreting final result

🔍

Example-9

❓ Question

An object 2 cm tall is placed on the principal axis of a convex (converging) lens of focal length 8 cm. Find the position, nature, and size of the image when the object is

(i) 12 cm from the lens
(ii) 6 cm from the lens

🧩 Solution

Given

Object height $ h_o = 2 \, cm $
Focal length $ f = +8 \, cm $

(i) Object at 12 cm

$ u = -12 \, cm $

\[ \begin{aligned} \frac{1}{f} &= \frac{1}{v} - \frac{1}{u}\\\\ \frac{1}{8} &= \frac{1}{v} + \frac{1}{12}\\\\ \frac{1}{v} &= \frac{1}{8} - \frac{1}{12} = \frac{1}{24}\\\\ v &= +24 \, \mathrm{cm} \end{aligned} \]

Magnification

\[ \begin{aligned} m &= \frac{v}{u} \\&= \frac{24}{-12} \\&= -2 \end{aligned} \]

\[ \begin{aligned} h_i &= m \times h_o \\&= (-2)(2) \\&= -4 \, \mathrm{cm} \end{aligned} \]

Conclusion

Image at $ 24 \, cm $ (right side)
Real and inverted
Magnified (2 times)

(ii) Object at 6 cm

$ u = -6 \, cm $

\[ \begin{aligned} \frac{1}{8} &= \frac{1}{v} + \frac{1}{6}\\ \frac{1}{v} &= \frac{1}{8} - \frac{1}{6} \\ &= -\frac{1}{24}\\ v &= -24 \, \mathrm{cm} \end{aligned} \]

Magnification

\[ \begin{aligned} m &= \frac{v}{u} \\&= \frac{-24}{-6} \\&= 4 \end{aligned} \]

\[ \begin{aligned} h_i &= 4 \times 2 \\&= 8 \, \mathrm{cm} \end{aligned} \]

Conclusion

📌

Note

Image at $ 24 \, cm $ (same side as object)
Virtual and erect
Highly magnified (4 times)

💡 Concept

Concept Link

⚡ Exam Tip

❌ Common Mistakes

Wrong substitution in magnification
Forgetting sign of $ u $
Writing “imaginary” instead of “virtual”

🔍

Example-10

❓ Question

An object 3 cm high is placed 24 cm away from a convex lens of focal length 8 cm.

(a) Find the position, height, and nature of the image.
(b) If the object is moved to 3 cm from the lens, find the new position, height, and nature of the image.
(c) Which case represents a magnifying glass?

🧩 Solution

Given

Object height $ h_o = 3 \, cm $
Focal length $ f = +8 \, cm $

(a) Object at 24 cm

$ u = -24 \, cm $

\[ \begin{aligned} \frac{1}{8} &= \frac{1}{v} + \frac{1}{24}\\ \frac{1}{v} &= \frac{1}{8} - \frac{1}{24} \\ &= \frac{2}{24} \\ &= \frac{1}{12}\\ \Rightarrow v &= +12 \, \mathrm{cm} \end{aligned} \]

\[ \begin{aligned} m &= \frac{v}{u} \\ &= \frac{12}{-24} \\ &= -\frac{1}{2} \end{aligned} \]

\[ h_i = -\frac{1}{2} \times 3 = -1.5 \, \mathrm{cm} \]

Conclusion:

Image distance $ v = +12 \, cm $
Image height $ h_i = -1.5 \, cm $
Real, inverted, diminished

(b) Object at 3 cm

$ u = -3 \, cm $

\[ \begin{aligned} \frac{1}{8} &= \frac{1}{v} + \frac{1}{3}\\ \frac{1}{v} &= \frac{1}{8} - \frac{1}{3} \\ &= -\frac{5}{24}\\ \Rightarrow v &= -4.8 \, \mathrm{cm} \end{aligned} \]

\[ \begin{aligned} m &= \frac{v}{u} \\ &= \frac{-4.8}{-3} \\ &= 1.6 \end{aligned} \]

\[ \begin{aligned} h_i &= 1.6 \times 3 \\&= 4.8 \, \mathrm{cm} \end{aligned} \]

Conclusion:

Image distance $ v = -4.8 \, cm $
Image height $ h_i = +4.8 \, cm $
Virtual, erect, magnified

(c) Magnifying Glass

Case (b) represents the working of a magnifying glass because:

Object is placed within focal length
Image formed is virtual, erect, and magnified

💡 Concept

Concept Link

⚡ Exam Tip

❌ Common Mistakes

Missing conclusion about nature of image
Confusing real vs virtual using sign of $ v $
Not identifying magnifying glass condition

🔍

Example-11

❓ Question

Find the position, nature, and magnification of the images formed by a convex lens of focal length 0.20 m when the object is placed at:

(i) 0.50 m
(ii) 0.25 m

🧩 Solution

Given

Focal length $ f = +0.20 \, m = +20 \, cm $

(i) Object at 0.50 m

$ u = -50 \, cm $

\[ \begin{aligned} \frac{1}{20} &= \frac{1}{v} + \frac{1}{50}\\ \frac{1}{v} &= \frac{1}{20} - \frac{1}{50}\\ &= \frac{5 - 2}{100}\\ &= \frac{3}{100}\\ v &= \frac{100}{3} \\ &\approx 33.3 \, \mathrm{cm} \end{aligned} \]

\[ \begin{aligned} m &= \frac{v}{u} \\ &= \frac{33.3}{-50} \\ &\approx -0.67 \end{aligned} \]

Conclusion:

Image distance $ v \approx +33.3 \, cm $
Real and inverted
Diminished ($ |m| < 1 $)

(ii) Object at 0.25 m

$ u = -25 \, cm $

\[ \begin{aligned} \frac{1}{20} &= \frac{1}{v} + \frac{1}{25}\\ \frac{1}{v} &= \frac{1}{20} - \frac{1}{25}\\ &= \frac{5 - 4}{100}\\ &= \frac{1}{100}\\ v &= 100 \, cm \end{aligned} \]

\[ \begin{aligned} m &= \frac{v}{u} \\ &= \frac{100}{-25} \\ &= -4 \end{aligned} \]

Conclusion:

Image distance $ v = +100 \, cm $
Real and inverted
Highly magnified ($ |m| = 4 $)

💡 Concept

Concept Link

⚡ Exam Tip

❌ Common Mistakes

Forgetting sign of $ u $
Not interpreting magnification
Mixing units (m and cm)

🔍

Important Points

💡 Concept

Core Concepts

⚖️ Laws

Laws Summary

Reflection: $ i = r $
Refraction (Snell’s Law): $ n = \frac{\sin i}{\sin r} $
Incident ray, normal, and refracted/reflected ray lie in same plane.

🔢 Formula

Formula Sheet

Mirror Formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
Lens Formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
Magnification (Mirror): $ m = -\frac{v}{u} $
Magnification (Lens): $ m = \frac{v}{u} $
Refractive Index: $ n = \frac{c}{v} $
Power of Lens: $ P = \frac{1}{f} $ (f in metre)

🔗 Relations

Key Relations

For spherical mirrors: $ f = \frac{R}{2} $
Higher refractive index → lower speed of light
Power and focal length are inversely related

📜 Rules

Direction Rules

📜 Rules

Denser medium → light bends towards normal
Rarer medium → light bends away from normal
Normal incidence → no bending

🤔 Did You Know?

Sign Convention Quick View

Object distance $ u $ → always negative
Real image → $ v > 0 $
Virtual image → $ v < 0 $
Convex lens → $ f > 0 $, Concave lens → $ f < 0 $

📍 Key Point

High-Yield Exam Points

Concave lens always forms virtual, erect image
Convex mirror always forms virtual, erect image
Magnifying glass → object within focal length
Real image can be obtained on screen; virtual cannot

NCERT Class X · Chapter 9 · Physics

Light — Reflection & Refraction

A comprehensive AI-powered learning engine with interactive solvers, concept builders, and practice questions

📚 Core Concepts

Organised by sub-topic. Click any concept card to expand.

Reflection of Light

Nature of Light & Reflection Laws ▼

Light travels in straight lines (rectilinear propagation). When light hits a smooth, polished surface it bounces back — this is reflection.

⚡

Laws of Reflection:
① The angle of incidence (∠i) equals the angle of reflection (∠r). ② The incident ray, reflected ray, and the normal at the point of incidence all lie in the same plane.

These laws hold for all types of reflecting surfaces — plane, concave, and convex.

∠i = ∠r Angle of incidence = Angle of reflection

Spherical Mirrors — Concave & Convex ▼

Spherical mirrors are parts of a hollow sphere with a reflecting surface inside (concave) or outside (convex).

Feature	Concave Mirror	Convex Mirror
Also called	Converging mirror	Diverging mirror
Reflecting surface	Inner (curved inward)	Outer (curved outward)
Focal point	Real, in front of mirror	Virtual, behind mirror
Image of distant object	Real, inverted, at F	Virtual, erect, diminished
Common uses	Shaving mirror, dentist's, headlights	Rear-view mirrors, security mirrors

🔑

Key Terms: Pole (P) — centre of mirror surface. Centre of Curvature (C) — centre of the sphere. Radius of Curvature (R) — radius of the sphere. Principal Axis — line through P and C. Principal Focus (F) — point where rays parallel to principal axis converge (concave) or appear to diverge from (convex).

f = R / 2 Focal length = Half of radius of curvature

Image Formation by Concave Mirror (All Cases) ▼

Object Position	Image Position	Nature & Size
At infinity	At F	Real, inverted, highly diminished
Beyond C	Between F and C	Real, inverted, diminished
At C	At C	Real, inverted, same size
Between C and F	Beyond C	Real, inverted, magnified
At F	At infinity	Real, inverted, highly magnified
Between F and P	Behind mirror	Virtual, erect, magnified

Mirror Formula & Magnification ▼

1/v + 1/u = 1/f Mirror Formula | v = image distance, u = object distance, f = focal length

📐

Sign Convention (New Cartesian): All distances measured from pole P. Distances along the direction of incident light → positive. Against incident light → negative. Heights above principal axis → positive. Below → negative.

m = hₙ / hₒ = −v / u Magnification | h' = image height, h = object height

Interpreting magnification:

✔
m > 0 (positive): image is virtual and erect
✔
m < 0 (negative): image is real and inverted
✔
|m| > 1: image is magnified (larger than object)
✔
|m| < 1: image is diminished (smaller than object)

Refraction of Light

What is Refraction? Laws of Refraction ▼

When light passes from one transparent medium to another, it changes its direction of travel. This bending is called refraction and happens because light travels at different speeds in different media.

⚡

Laws of Refraction (Snell's Law):
① The incident ray, refracted ray, and normal lie in the same plane.
② sin(i) / sin(r) = constant (for a given pair of media and colour of light).

n₁ · sin(i) = n₂ · sin(r) Snell's Law | n₁, n₂ = refractive indices; i = angle of incidence; r = angle of refraction

When light goes denser → rarer medium: it bends away from normal (r > i).
When light goes rarer → denser medium: it bends toward normal (r < i).

Refractive Index — Absolute & Relative ▼

n = c / v Absolute refractive index | c = speed of light in vacuum ≈ 3 × 10⁸ m/s, v = speed in medium

ₙ₁n₂ = n₂ / n₁ = sin(i) / sin(r) Relative refractive index of medium 2 with respect to medium 1

Medium	Refractive Index (approx.)
Vacuum / Air	1.00
Water	1.33
Crown Glass	1.52
Dense Glass	1.65
Diamond	2.42

A higher refractive index means the medium is optically denser and light travels slower in it.

Refraction Through a Glass Slab ▼

When light passes through a rectangular glass slab, it refracts at both surfaces. The emergent ray is parallel to the incident ray but laterally displaced (shifted sideways).

💡

Key fact: At the first surface, light bends toward the normal. At the second surface (emerging from denser to rarer), it bends away from the normal by the same angle. So the net deviation is zero but there's a lateral shift that depends on the thickness of the slab and the angle of incidence.

Spherical Lenses — Convex & Concave ▼

A lens is a transparent optical element bounded by two curved (or one curved, one plane) surfaces.

Feature	Convex (Converging) Lens	Concave (Diverging) Lens
Centre	Thicker at centre	Thinner at centre
Effect on rays	Converges parallel rays to F	Diverges rays; appear from virtual F
Focal length	Positive (+)	Negative (−)
Power	Positive	Negative
Uses	Reading glasses, camera, magnifier	Myopia correction

🔑

Key Terms for Lenses: Optical Centre (O) — centre of lens. Principal Focus (F) — convex lens has real F on both sides; concave has virtual F. Focal Length (f) — distance O to F. Every lens has two foci (F₁ and F₂).

Image Formation by Convex Lens (All Cases) ▼

Object Position	Image Position	Nature & Size
At infinity	At F₂	Real, inverted, highly diminished
Beyond 2F₁	Between F₂ and 2F₂	Real, inverted, diminished
At 2F₁	At 2F₂	Real, inverted, same size
Between F₁ and 2F₁	Beyond 2F₂	Real, inverted, magnified
At F₁	At infinity	Real, inverted, highly magnified
Between O and F₁	Same side as object	Virtual, erect, magnified

For a concave lens, image is always virtual, erect, and diminished, regardless of object position.

Lens Formula, Magnification & Power ▼

1/v − 1/u = 1/f Lens Formula | Same sign convention as mirrors (from optical centre)

m = hₙ / hₒ = v / u Magnification for lenses (note: no negative sign unlike mirrors)

P = 1 / f (in metres) | Unit: Dioptre (D) Power of a lens | f must be in metres

➕

Combination of lenses: When lenses are placed in contact, total power = P₁ + P₂ + P₃ + … and effective focal length: 1/f = 1/f₁ + 1/f₂.

⚗️ All Formulas at a Glance

Every formula from Chapter 9 with variables explained and conditions noted.

Mirrors

Mirror Formula

1/f = 1/v + 1/u f = focal length | v = image distance from pole | u = object distance from pole ⚠ All distances measured from pole P with sign convention.

Relationship: Focal Length & Radius

f = R / 2 ⟺ R = 2f

Linear Magnification (Mirrors)

m = hₙ / hₒ = −v / u h' = height of image | h = height of object

Concave Mirror

f is negative. u is always negative. v is negative for real image, positive for virtual image.

Convex Mirror

f is positive. u is always negative. v is always positive (virtual image always).

Refraction & Refractive Index

Snell's Law

n₁ · sin(θ₁) = n₂ · sin(θ₂) n₁, n₂ = refractive indices of medium 1 & 2 | θ₁ = angle of incidence | θ₂ = angle of refraction

Absolute Refractive Index

n = c / v = (speed of light in vacuum) / (speed in medium) c = 3 × 10⁸ m/s | n is always ≥ 1 for a physical medium

Relative Refractive Index

₁n₂ = n₂ / n₁ = v₁ / v₂ = sin(i) / sin(r)

Lenses

Lens Formula

1/v − 1/u = 1/f Note the minus sign (−1/u) — different from mirror formula!

Magnification (Lenses)

m = hₙ / hₒ = v / u No negative sign here — unlike mirrors

Power of a Lens

P = 1 / f(m) [Unit: Dioptre, D] Convert focal length to metres before applying. Convex: P > 0. Concave: P < 0.

Combination of Lenses

P_total = P₁ + P₂ + P₃ + … Also: 1/f_net = 1/f₁ + 1/f₂ + …

Convex Lens

f is positive. u negative (object on left). v positive = real image (right side); v negative = virtual image (left side).

Concave Lens

f is negative. Image always virtual, erect, diminished. v always negative (same side as object).

📐 Sign Convention — Quick Reference

For Mirrors

−ve u (object always in front)
−ve f for concave mirror
+ve f for convex mirror
−ve v for real image (in front)
+ve v for virtual image (behind)

For Lenses

−ve u (object always on left)
+ve f for convex lens
−ve f for concave lens
+ve v for real image (right side)
−ve v for virtual image (left side)

🧮 Step-by-Step AI Solver

Enter known values for Mirror, Lens, or Refraction problems. The engine solves and explains every step.

🪞 Mirror Solver

Enter any two of the three values. Leave one blank to solve for it.

Object Distance u (cm) — enter with sign

Image Distance v (cm)

Focal Length f (cm)

Object Height h (cm) (optional, for magnification)

📝 Concept-Building Questions

Original questions with full step-by-step solutions, organised by concept. Not replicated from textbook.

🎯 Knowledge Quiz

Test your conceptual understanding. Feedback and explanation after each answer.

Score

Answered

—

Total

0% complete

🔬 Interactive Modules

Visual, hands-on learning tools — ray diagrams, refraction visualiser, and more.

🪞 Concave Mirror — Live Ray Diagram

Drag the object slider to change its position. See how the image changes in real time.

Object Distance: 35 cm

Focal Length: 20 cm

💡 Tips, Tricks & Common Mistakes

Curated from common student errors and exam patterns. Study these before your exam.

✨ Tricks & Study Tips

🎯

Mirror vs Lens Formula Memory Trick: Mirror → 1/v + 1/u = 1/f (both positive sides, add). Lens → 1/v − 1/u = 1/f (subtract u side). The minus in lens formula trips students up most often.

🌊

Denser means slower, slower means bends toward normal: When light enters a denser medium (higher n), it slows down and bends toward the normal. Rarer → faster → bends away. Visualise it as a car wheel hitting mud — the wheel in mud slows and turns the car.

🔢

Power sign shortcut: Convex lens → always positive P. Concave lens → always negative P. If two lenses are combined and the total P is positive → system is converging.

📏

Magnification sign tells you everything: Positive m = virtual + erect. Negative m = real + inverted. |m| > 1 = enlarged. |m| < 1 = diminished. You can determine the nature of image just from m without drawing any ray diagram.

🪞

Convex mirror — always use it for rear-view: Because it gives a wider field of view and always forms a virtual, erect, diminished image regardless of object position — making it safe and consistent for drivers.

💎

Diamond sparkles because of high n (2.42): A higher refractive index leads to a smaller critical angle — enabling total internal reflection. Diamonds are cut to exploit this, causing light to bounce internally multiple times, creating brilliance.

📐

Quick cross-check: After solving a mirror/lens problem, always verify the sign and nature of the image matches the case you'd expect (e.g., object between F and mirror/lens should give virtual image). This catches errors in seconds.

🔭

Ray diagram drawing rule of three: Any two of these three special rays uniquely locate the image — (1) parallel to principal axis → passes through F after refraction, (2) through F → emerges parallel, (3) through optical centre → passes undeviated (lens) or reflects at equal angle (mirror).

⚠ Common Mistakes to Avoid

❌

Forgetting the negative sign on u: Object distance u is always negative in mirror formula (object always in front). Students often plug in +30 instead of −30 and get completely wrong answers.

❌

Using mirror formula for lenses (or vice versa): Mirror: 1/v + 1/u = 1/f. Lens: 1/v − 1/u = 1/f. These are different — confusing them is one of the most common exam errors.

❌

Focal length in cm instead of m for Power: Power formula P = 1/f requires f in metres. A 20 cm lens has f = 0.20 m → P = 5 D. Using 20 gives P = 0.05 D — completely wrong.

❌

Confusing radius of curvature with focal length: R = 2f. Students sometimes use R directly in mirror formula instead of f. Always halve R to get f before substituting.

❌

Wrong formula for magnification in lenses: Lens magnification m = v/u (no negative sign). Mirror magnification m = −v/u (with negative sign). Applying mirror formula to a lens or vice versa gives a wrong sign and wrong nature of image.

❌

Stating Snell's Law incompletely: Students often write only "sin i / sin r = constant" but forget to state the coplanarity condition (incident ray, refracted ray, and normal in same plane). Full Snell's Law has two parts.

❌

Declaring concave mirror always forms real image: Wrong! When object is placed between the focus F and pole P, the concave mirror forms a virtual, erect, magnified image — behind the mirror.

🃏 Flashcard Revision

Click a card to reveal the answer. Navigate through all cards for quick revision.

of —

Category: —

click to reveal answer ↓

📚

ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025

Sharing this chapter

Class 10 Light Notes Made Simple: Reflection, Refraction & Diagrams

Class 10 Light Notes Made Simple: Reflection, Refraction & Diagrams — Complete Notes & Solutions · academia-aeternum.com

Chapter 9, "Light – Reflection and Refraction," introduces students to the fascinating world of light and its behavior when it encounters different surfaces and media. The chapter begins by explaining the nature of light and the concept of rectilinear propagation. It covers the laws of reflection, types of mirrors (plane, concave, convex), and the formation of images by spherical mirrors using ray diagrams. Students learn about important terms like pole, principal axis, centre of curvature,…

🎓 Class 10 📐 Science 📖 NCERT ✅ Free Access 🏆 CBSE · JEE

Share on

X / Twitter

Facebook

Instagram

Post to Instagram

Threads

Quora

Discord

academia-aeternum.com/class-10/science/light-reflection-and-refraction/notes/ Copy link

💡

Exam tip: Sharing chapter notes with your study group creates a reinforcement loop. Teaching a concept is the fastest path to mastering it.

Light – Reflection and Refraction

Chapter Snapshot

Why This Chapter Matters for Exams

Key Concept Highlights

Important Formulae & Reactions

What You Will Learn

Navigate to Chapter Resources

🏆 Exam Strategy & Preparation Tips

Important Formula

Derivation (Conceptual)

Important Relations

Sign Convention (Cartesian)

Derivation of Mirror Formula

Applications

Ray Rules

Concave Mirror Image Formation

Convex Mirror Image Formation

Standard Rays Used in Construction

Stepwise Construction (Concave Mirror)<

Connection with Mirror Formula

Coordinate System Representation

Rules

Quick Sign Reference Table

Magnification

Derivation of Mirror Formula

Conceptual Diagram

Given

Magnification

Given

Interpretation

GIven

MAgnification

Interpretation

Interpretation

Magnification

Interpretation

Interpretation

Laws of Refraction (Snell’s Law)

Refractive Index

Optical Density

Refraction Through Glass Slab

Optical Density & Refractive Index

Relation with Refractive Index

Comparison of Media

Magnification

Derivation of Lens Formula

Unit of Power

Sign Convention

Important Conversion

Combination of Lenses

Given

Given

Magnification

Given

(i) Object at 12 cm

Magnification

Conclusion

(ii) Object at 6 cm

Magnification

Conclusion

Given

(a) Object at 24 cm

(b) Object at 3 cm

(c) Magnifying Glass

Given

(i) Object at 0.50 m

(ii) Object at 0.25 m

Recent posts

Light - Reflection and Refraction — Learning Resources

Let's Connect