Light – Reflection and Refraction-Notes
Physics - Notes
Diffraction
Diffraction is the bending or spreading out of light (or any wave) as it passes around an obstacle or through a narrow opening (like a slit).
- Diffraction shows that light behaves like a wave.
- When light waves encounter a small gap, edge, or object, they bend around it and spread into the region behind the obstacle.
- The effect is most noticeable when the size of the opening or obstacle is similar to the wavelength of light.
Reflection of Light
Reflection of light is the process where light rays bounce off the surface of an object rather than passing through it or being absorbed.
Law of Reflection:
- The incident ray, the reflected ray, and the normal (an imaginary line at right angles to the surface) all lie in the same plane.
- The angle of incidence is equal to the angle of reflection.
Spherical Mirror
A spherical mirror is a mirror whose reflecting surface is part of a sphere.
- Concave Mirror: The reflecting surface is curved inward, like the inside of a spoon.
- Convex Mirror: The reflecting surface is curved outward, like the back of a spoon.
- Pole (P): The centre of the mirror’s reflecting surface.
- Centre of curvature (C): The centre of the imaginary sphere of which the mirror is a part.
- Principal axis: A straight line passing through C and P.
- Radius of curvature (R): The distance between the pole (P) and centre of curvature (C).
- Focus (F): Point on the principal axis where rays parallel to the axis meet or appear to meet after reflection.
- Focal length (f): The distance between the pole (P) and focus (F).
- Aperture: The width of the reflecting surface of the mirror.
Concave Mirror (Converging Mirror): Reflects light inward to a real focal point.
Used in torches, car headlights, and shaving mirrors.
Convex Mirror (Diverging Mirror): Reflects light outward, rays appear to spread out
from a virtual focus. Used as rear-view mirrors in vehicles; cover a wider field of view.
Image Formation by Spherical Mirrors
Spherical mirrors (concave and convex) form images by reflecting light rays according to a set of rules:
Concave Mirror (Inward Curve)
| Object Position | Image Position | Nature | Size |
|---|---|---|---|
| At infinity | At focus (F) | Real, inverted | Highly small |
| Beyond centre (C) | Between F & C | Real, inverted | Smaller |
| At centre (C) | At centre (C) | Real, inverted | Same size |
| Between C and F | Beyond centre (C) | Real, inverted | Larger |
| At focus (F) | At infinity | Real, inverted | Highly large |
| Between pole (P) and focus (F) | Behind mirror | Virtual, erect | Larger |
Convex Mirror (Outward Curve)
| Object Position | Image Position | Nature | Size |
|---|---|---|---|
| Anywhere | Between P and F | Virtual, erect | Smaller |
Representation of Images Formed by Spherical Mirrors Using Ray Diagrams
Ray diagrams are a simple way to show how images are formed by spherical mirrors (concave and convex). These diagrams use drawn rays—lines with arrows—to trace the path of light reflected by the mirror, and shows the location, size, and nature of the image.
Behaviour Shown by Rays
- Parallel ray: Ray parallel to the principal axis; after reflection, passes through (concave) or appears to come from (convex) the focus.
- Ray through centre of curvature (C): Reflects back along itself.
- Ray through focus (F): Reflects parallel to principal axis.
- Ray towards pole (P): Reflects obeying the law of reflection.
Sign Convention for Reflection by Spherical Mirrors
While dealing with the reflection of light by spherical mirrors, we shall follow a set of sign conventions called the New Cartesian Sign Convention.
- The object is always placed to the left of the mirror. This implies that the light from the object falls on the mirror from the left-hand side.
- All distances parallel to the principal axis are measured from the pole of the mirror.
- All the distances measured to the right of the origin (along + x-axis) are taken as positive while those measured to the left of the origin (along – x-axis) are taken as negative.
- Distances measured perpendicular to and above the principal axis (along + y-axis) are taken as positive.
- Distances measured perpendicular to and below the principal axis (along –y-axis) are taken as negative.
Mirror Formula and Magnification
Mirror Formula
The mirror formula relates the focal length \(f\), object distance \(u\) and image
distance \(v\) for spherical mirrors:
\[\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\]
\(f\): focal length of the mirror
\(v\): distance from mirror to image
\(u\): distance from mirror to object
Magnification Formula
Magnification \(m\) is the ratio of the height of the image \(h'\) to the height of the object \(h\): \[m = \frac{h'}{h} = -\frac{v}{u}\]
- if \(|m| \ge 1\): Image is larger than object
- if \(|m| \le 1\): Image is smaller than object
- If \(m\) is negative: Image is inverted
- If \(m\) is positive: Image is upright
Examples
i) Find the position of the image.
ii) Is the image magnified?
Radius of Curvature = 24cm
Therefore; focal length = \(\frac{24}{2}\) = 12cm
Object distance \(u\) = -4cm
$$\text{Magnification }m=1\\m=\frac{v}{u}$$ $$\begin{aligned}\Rightarrow \frac{v}{u}&=1\\ v&=u\\\\\frac{1}{f}&=\frac{1}{v}+\frac{1}{u}\\\frac{1}{f}&=\frac{1}{u}+\frac{1}{u}\\ \frac{1}{f}&=\frac{2}{u}\\ u&=2f\\&=2\times 25\\&=50\ cm\end{aligned}$$
1) the position
2) size of Image
$$ \begin{align} h &= 5\,\text{cm} \\ u &= -60\,\text{cm} \\ f &= -10\,\text{cm} \\ \\ \color{blue} \frac{1}{f} &= \color{blue} \frac{1}{v} + \frac{1}{u} \\ -\frac{1}{10} &= \frac{1}{v} - \frac{1}{60} \\ \frac{1}{v} &= \frac{1}{60} - \frac{1}{10} \\ &= \frac{1 - 6}{60} \\ &= -\frac{5}{60} \\ \frac{1}{v} &= -\frac{1}{12} \\ v &= -12\,\text{cm} \\ \\ \text{Magnification } \color{blue} m &= \color{blue} \frac{v}{u} \\ \end{align} $$ $$\text{Putting the values of } u \text{ and } v:$$ $$ \begin{align} m &= \frac{-12}{-60} \\ m&= \frac{1}{5} \tag{1}\\ m &= \frac{h'}{h} \\ h &= 5\,\text{ (Given)} \\ m&=\frac{h'}{5}\tag{2}\end{align} $$ $$\text{Equating equation (1) and (2):}$$ $$ \begin{align} \frac{1}{5} &= \frac{h'}{5} \\ h' &= \frac{5 \times 1}{5} \\ &= 1 \end{align} $$
a) Where will the image form?
b) What will be the length of image?
a) The focal length of the mirror
b) Position of the image
Refraction of Light
Refraction occurs when light travels from one transparent medium to another (such as from air to water or glass). In doing so, the speed of light changes, causing the light to bend at the boundary between the two media. If light enters a denser medium (e.g., from air into glass), it slows down and bends towards the normal (an imaginary line perpendicular to the surface). If it enters a rarer medium (e.g., from glass to air), it speeds up and bends away from the normal.
Laws of Refraction
- The incident ray (incoming ray), the refracted ray (bent ray), and the normal all lie in the same plane.
- The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media. This is known as Snell's Law: \[\frac{\sin i}{\sin r}=\text{constant}\] where \(i\) is the angle of incidence and \(r\) is the angle of refraction.
- This constant is called the refractive index of the second medium with respect to the first.
Everyday Examples
- A coin appears raised when placed under water in a glass.
- Objects look bent or broken at the water surface.
- Formation of rainbows.
- Lenses in eyeglasses, cameras, and microscopes use refraction to focus light.
Refractive Index
The refractive index of medium 2 with respect to medium 1 is given by the ratio of the speed of light in medium 1 and the speed of light in medium 2. This is usually represented by the symbol \(n_{21}\). \[\scriptsize \mathrm n_{21}=\frac{\text{Speed of light in medium 1}}{\text{Speed of light in medium 2}}\\\\=\frac{v_1}{v_2}\] By the same argument, the refractive index of medium 1 with respect to medium 2 is represented as \(n_{12}\). It is given by \[\scriptsize \mathrm n_{12}=\frac{\text{Speed of light in medium 2}}{\text{Speed of light in medium 1}}\\\\=\frac{v_2}{v_1}\] If medium 1 is vacuum or air, then the refractive index of medium 2 is considered with respect to vacuum. This is called the absolute refractive index of the medium. It is simply represented as \(n_2\). If c is the speed of light in air and v is the speed of light in the medium, then the refractive index of the medium \(n_m\) is given by \[\scriptsize \mathrm n_m=\frac{\text{Speed of light in air}}{\text{Speed of light in the medium}}\\\\=\frac{c}{v}\]
Optical Density
Optical density refers to the ability of a medium to refract light. It does not mean the
mass density. In comparing two media, the one with the larger refractive index is called optically
denser, and the other with a lower refractive index is optically rarer. The speed of light is higher
in an optically rarer medium than in an optically denser medium.
Optical Density is a measure
of how well a material transmits light.
Refraction by Spherical Lenses
A spherical lens is a piece of transparent material (like glass or plastic) bounded by at least one spherical surface. There are two types:
- Convex lens (converging lens): Thicker in the middle and thinner at the edges, it converges (brings together) rays of light.
- Concave lens (diverging lens): Thinner in the middle and thicker at the edges, it diverges (spreads out) rays of light.
Important Points
- The line passing through the centers of both spherical surfaces is called the principal axis.
- The center of the lens is the optical centre.
- A ray passing through the optical centre continues undeviated.
- A ray parallel to the principal axis passes through the principal focus after refraction by a convex lens (or appears to come from the focus for a concave lens).
- The distance from the optical centre to the principal focus is called the focal length.
Image Formation by Lenses
When light rays pass through a lens, they bend (refract) and form images whose nature (real or virtual), position, and size depend on the type of lens (convex or concave) and the position of the object:
Convex Lens (Converging Lens)
- Ray Directions:
- Ray parallel to the principal axis passes through the principal focus on the other side.
- Ray passing through the optical centre goes straight without deviation.
- Ray passing through (or appearing to pass through) the principal focus emerges parallel to the principal axis.
| Object Position | Image Position | Size of Image | Nature |
|---|---|---|---|
| At infinity | At focus (F₂) | Highly diminished, point-sized | Real & inverted |
| Beyond 2F₁ | Between F₂ and 2F₂ | Diminished | Real & inverted |
| At 2F₁ | At 2F₂ | Same size | Real & inverted |
| Between F₁ and 2F₁ | Beyond 2F₂ | Enlarged | Real & inverted |
| At F₁ | At infinity | Infinitely large | Real & inverted |
| Between F₁ and O | Same side as object | Enlarged | Virtual & erect |
Concave Lens (Diverging Lens)
- Ray Directions:
- Ray parallel to the principal axis appears to diverge from the principal focus.
- Ray directed towards the focal point emerges parallel to the principal axis.
- Ray through optical centre passes straight.
| Object Position | Image Position | Size of Image | Nature |
|---|---|---|---|
| At infinity | At focus (F₁) | Highly diminished | Virtual & erect |
| Between infinity and O | Between F₁ and O (same side) | Diminished | Virtual & erect |
Image Formation in Lenses Using Ray Diagrams
Image formation in lenses using ray diagrams is a visual way to show how lenses (convex or concave) create images by bending (refracting) light rays.
- Ray parallel to principal axis: After passing through the lens, it either passes through the focus (convex lens) or appears to diverge from the focus (concave lens).
- Ray passing through the optical centre: This continues in the same direction without getting refracted.
For convex lenses, rays converge to form real, inverted images (except when the object is close—then the image is virtual and erect).
For concave lenses, rays diverge so they appear to come from a virtual, erect image.
Sign Convention for Spherical Lenses
To solve lens formula and magnification problems, we follow a standardised sign convention (New Cartesian Sign Convention):
- Principal Axis: The principal axis is taken as the reference, and all distances are measured from the optical centre (O) of the lens.
- Directions: Distances measured in the direction of incident light (usually from left to right) is taken as positive. Distances measured opposite to the direction of incident light is taken as negative.
- Vertical Distances: Distances measured upwards from the principal axis (above axis) is positive. Distances measured downwards (below the axis) are negative.
- Object Position: The object is always placed to the left of the lens (by convention), so the object distance \((u)\) is always negative.
- Focal Length \((f)\): For a convex lens (converging lens), the focal length is positive. For a concave lens (diverging lens), the focal length is negative.
- Image Distance \((v)\): If the image forms on the opposite side of the light’s entry (real image, on right of lens), \(v\) is positive. If the image forms on the same side as the light’s entry (virtual image, on left of lens), \(v\) is negative.
| Quantity | Positive (+) Side | Negative (–) Side |
|---|---|---|
| Object distance \((u)\) | - | Always negative |
| Image distance \((v)\) | Right of lens (real) | Left of lens (virtual) |
| Focal length \((f)\) | Convex lens | Concave lens |
| Heights \((h, h')\) | Above principal axis | Below principal axis |
Lens Formula and Magnification
The relationship between the object distance \((u)\), image distance \((v)\), and focal
length \((f)\) of a spherical lens is given by the lens formula:
\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\]
where:
\(f\) = focal length of the lens
\(v\) = image distance from the optical centre
\(u\) = object distance from the optical centre
Magnification Formula: Magnification \((m)\) produced by a lens is the ratio of the height of the image \((h')\) to the height of the object \((h)\): \[m = \frac{h'}{h} = \frac{v}{u}\]
Power of a Lens
The power of a lens is a measure of how strongly the lens converges (convex) or diverges
(concave) light rays. It is defined as the reciprocal of the focal length:
\[P = \frac{1}{f}\]
where:
\(P\) = power of the lens (in dioptres, D)
\(f\) = focal length of the lens (in metres)
Unit: The SI unit of lens power is dioptre (D).
1 dioptre = 1/metre
Sign Convention:
- Power of a convex lens (focal length positive) is positive.
- Power of a concave lens (focal length negative) is negative.
Examples
Using Lens Formula $$\begin{aligned}\dfrac{1}{f}&=\dfrac{1}{v}-\dfrac{1}{u}\\\\ \dfrac{1}{10}&=\dfrac{1}{12}-\dfrac{1}{u}\\\\ \dfrac{1}{u}&=\dfrac{1}{12}-\dfrac{1}{10}\\\\ &=\dfrac{5-6}{60}\\\\ &=-\dfrac{1}{60}\\\\ \dfrac{1}{u}&=-\dfrac{1}{60}\\\\ u&=-60cm\end{aligned}$$
Solution:
Height of object \((h)\) = 7cm
Object distance \((u)\) =-12cm
Focal length \((f)\) = 8 cm
Using Lens Formula: $$\begin{aligned}\dfrac{1}{f}&=\dfrac{1}{v}-\dfrac{1}{u}\\\\ \dfrac{1}{8}&=\dfrac{1}{v}-\left( \dfrac{1}{-12}\right) \\\\ \dfrac{1}{8}&=\dfrac{1}{v}+\dfrac{1}{12}\\\\ \dfrac{1}{v}&=\dfrac{1}{8}-\dfrac{1}{12}\\\\ &=\dfrac{3-2}{24}\\\\ \dfrac{1}{v}&=\dfrac{1}{24}\\\\ v&=24\ cm\end{aligned}$$ Magnification (m) $$\begin{aligned}m&=\dfrac{v}{u}\\ &=\dfrac{24}{-12}\\ &=-2\\ m&=\dfrac{h'}{h}\\ -2&=\dfrac{h'}{7}\\ h'&=-14\\h'&=14\\ \end{aligned}$$ The image is formed behind the lens, hence it is real height of the image is -ve, hence it is inverted
(i) 12 cm from lens
(ii) 6cm from lens
Solution:
Height \((h)\) of object = 2cm
Focal length \((f)\) of lens = 8cm
(i) Object Position \((u)\) =-12cm
Using Lens Formula:
$$\begin{aligned}\dfrac{1}{f}&=\dfrac{1}{v}-\dfrac{1}{u}\\\\ \dfrac{1}{8}&=\dfrac{1}{v}-\left( \dfrac{1}{-12}\right) \\\\ &=\dfrac{1}{v}+\dfrac{1}{12}\\\\ \dfrac{1}{8}-\dfrac{1}{12}&=\dfrac{1}{v}\\\\ \Rightarrow \dfrac{1}{v}&=\dfrac{1}{8}-\dfrac{1}{12}\\\\ &=\dfrac{3-2}{24}\\\\ \dfrac{1}{v}&=\dfrac{1}{24}\\\\ v&=24\ cm\end{aligned}$$ Magnification \((m)\) of the lens is given by $$\begin{aligned}m&=\dfrac{v}{u}\\\\ &=\dfrac{-12}{24}\\\\ &=-2\\ m&=\dfrac{h'}{h}\\\\ -2&=\dfrac{h'}{2}\\\\ h'=-4\end{aligned}$$ Because, image position = +24cm (behind Lens)
image size =-4cm (-ve sign indicates that image is inverted)
Therefore, the Image is real and inverted
(ii)
Focal \((f)\) length = 8cm
Object position \((u)\) =-6cm
Using Lens Formula:
$$\begin{aligned}\dfrac{1}{f}&=\dfrac{1}{v}-\dfrac{1}{u}\\\\ \dfrac{1}{8}&=\dfrac{1}{v}-\dfrac{1}{\left( -6\right) }\\\\ \dfrac{1}{8}&=\dfrac{1}{v}+\dfrac{1}{6}\\\\ \dfrac{1}{8}-\dfrac{1}{6}&=\dfrac{1}{v}\\\\ \Rightarrow \dfrac{1}{v}&=\dfrac{1}{8}-\dfrac{1}{6}\\\\ &=\dfrac{3-4}{24}\\\\ \dfrac{1}{v}&=\dfrac{-1}{24}\\\\ v&=-24\end{aligned}$$ Magnification \((m)\) of the Image is given by: $$\begin{aligned}m&=\dfrac{v}{u}\\\\ &=\dfrac{-24}{-6}\\\\ &=4\\ m&=\dfrac{h'}{h}\\\\ 4&=\dfrac{h'}{2}\\\\ h'=8\ cm\end{aligned}$$ Image is formed at the object side, and image size is positive; therefore, the Image is Imaginary and Erect
Find the position, height and nature of Image
b. If the object is moved only 3cm away from the lens, what is the new position height and nature of Image
c. Which of the two cases illustrates the working of a magnifying glass?
Solution:
a. Height \((h)\) of object = 3cm
Focal length \((f)\) of lens = 8cm
Object distance \((u)\) =-24cm
Using Lens Formula: $$\begin{aligned}\dfrac{1}{f}&=\dfrac{1}{v}-\dfrac{1}{u}\\\\ \dfrac{1}{8}&=\dfrac{1}{v}-\dfrac{1}{\left( -24\right) }\\\\ \dfrac{1}{8}&=\dfrac{1}{v}+\dfrac{1}{24}\\\\ \dfrac{1}{8}-\dfrac{1}{24}&=\dfrac{1}{v}\\\\ \Rightarrow \dfrac{1}{v}&=\dfrac{1}{8}-\dfrac{1}{24}\\\\ &=\dfrac{3-1}{24}\\\\ \dfrac{1}{v}&=\dfrac{2}{24}\\\\ \dfrac{1}{v}&=\dfrac{1}{12}\\\\ v=12\ cm\end{aligned}$$ Magnification \((m)\) of the image is given by $$\begin{aligned}m&=\dfrac{v}{u}\\\\ &=\dfrac{+12}{-24}\\\\ &=-\dfrac{1}{2}\\\\ \dfrac{h'}{h}&=\dfrac{-1}{2}\\\\ h'&=\dfrac{-1\times 3}{2}\\ &=-1.5\ cm\end{aligned}$$ b. object new position \((u)\) = -3cm
Using Lens Formula:
$$\begin{aligned}\dfrac{1}{f}&=\dfrac{1}{v}-\dfrac{1}{u}\\\\ \dfrac{1}{8}&=\dfrac{1}{v}-\dfrac{1}{\left( -3\right) }\\\\ \dfrac{1}{8}&=\dfrac{1}{v}+\dfrac{1}{3}\\\\ \dfrac{1}{8}-\dfrac{1}{3}&=\dfrac{1}{v}\\\\ \Rightarrow \dfrac{1}{v}&=\dfrac{1}{8}-\dfrac{1}{3}\\\\ &=\dfrac{3-8}{24}\\\\ \dfrac{1}{v}&=\dfrac{-5}{24}\\\\ v&=\dfrac{-24}{5}\\\\ &=-4.8\ cm\end{aligned}$$ Magnification \((m)\) of the Image can be given by : $$\begin{aligned}m&=\dfrac{v}{u}\\\\ &=\dfrac{-4.8}{-3}\\\\ &=1.6\\ \dfrac{h'}{h}&=1.6\\ h'&=3\times 1.6\\ &=4.8\ cm\end{aligned}$$ Image is virtual and Erect
(i) 0.50m (ii) 0.25m
Solution:
(i) Focal length \((f)\) of lens = 0.20m = 20cm
Object position \((u)\) = 0.50m = 50cm
Using Lens Formula: $$\begin{aligned}\dfrac{1}{f}&=\dfrac{1}{v}-\dfrac{1}{u}\\\\ \dfrac{1}{20}&=\dfrac{1}{v}-\dfrac{1}{\left( -50\right) }\\\\ \dfrac{1}{20}&=\dfrac{1}{v}+\dfrac{1}{50}\\\\ \dfrac{1}{20}-\dfrac{1}{50}&=\dfrac{1}{v}\\\\ \Rightarrow \dfrac{1}{v}&=\dfrac{1}{20}-\dfrac{1}{50}\\\\ &=\dfrac{5-2}{100}\\\\ &=\dfrac{3}{100}\\\\ v&=\dfrac{100}{3}\\\\ &=33.33\ cms\end{aligned}$$ Magnification \((m)\):is given by" $$\begin{aligned}m&=\dfrac{v}{u}\\\\ &=\dfrac{100}{3\times \left( -50\right) }\\\\ &=-\dfrac{2}{3}\end{aligned}$$ The image is real and inverted (ii) object position \((u)\) =-0.25m =-25cm
$$\begin{aligned}\dfrac{1}{f}&=\dfrac{1}{v}-\dfrac{1}{u}\\\\ \dfrac{1}{20}&=\dfrac{1}{v}-\dfrac{1}{\left( -25\right) }\\\\ \dfrac{1}{20}&=\dfrac{1}{v}+\dfrac{1}{25}\\\\ \dfrac{1}{20}-\dfrac{1}{25}&=\dfrac{1}{v}\\\\ \Rightarrow \dfrac{1}{v}&=\dfrac{1}{20}-\dfrac{1}{25}\\\\ &=\dfrac{5-4}{100}\\\\ \dfrac{1}{v}&=\dfrac{1}{100}\\\\ v&=100cm\end{aligned}$$ Magnification \((m)\) is given by: $$\begin{aligned}m&=\dfrac{v}{u}\\\\ &=\dfrac{100}{-25}\\\\ &=-4\end{aligned}$$ The image will be real and inverted.
Important Points
- Light travels in straight lines.
- Mirrors and lenses form images. These images can be real or virtual depending on the object's position.
- All reflecting surfaces obey the laws of reflection. Refracting surfaces obey the laws of refraction.
- New Cartesian Sign Conventions are followed for spherical mirrors and lenses.
- The focal length of a spherical mirror is equal to half its radius of curvature.
- The magnification produced by a spherical mirror is the ratio of the image height to the object height.
- A light ray travelling obliquely from a denser medium to a rarer medium bends away from the normal. When it travels from a rarer to a denser medium, it bends towards the normal.
- Light travels in vacuum at a speed of \(3 \times 10^8~\mathrm{ms}^{-1}\). Its speed varies in different media.
- The refractive index of a transparent medium is the ratio of the speed of light in a vacuum to that in the medium.
- In a rectangular glass slab, refraction takes place at both air–glass and glass–air interfaces. The emergent ray is parallel to the incident ray.
- The power of a lens is the reciprocal of its focal length. The SI unit of power is the dioptre (D).
