Light is electromagnetic radiation belonging to the visible region of the spectrum, with wavelengths ranging from approximately 400 nm to 750 nm . It is the primary medium through which we perceive and interpret the physical world.
Reflection is the phenomenon where light rays strike a boundary separating two different media and return to the medium from which they originated, obeying specific geometric laws.
These laws are universally valid at every point on any reflecting surface, whether plane or curved.
To derive consistent formulas for spherical mirrors and lenses, the Cartesian sign convention is adopted:
For paraxial rays (rays incident close to the pole making small angles with the principal axis), the relation between the focal length ($f$) and the radius of curvature ($R$) of a spherical mirror is given by:
$$f = \frac{R}{2}$$Derivation Trace: Let a ray parallel to the principal axis strike a concave mirror at $M$. The normal $CM$ passes through the centre of curvature $C$. The angle of incidence is $\theta$, and the angle of reflection is also $\theta$. The reflected ray intersects the principal axis at the focus $F$. From geometry, $\angle MCP = \theta$ and $\angle MFP = 2\theta$. For small angles, $\tan \theta \approx \theta = \frac{MD}{CD}$ and $\tan 2\theta \approx 2\theta = \frac{MD}{FD}$. Thus, $2\left(\frac{MD}{CD}\right) = \frac{MD}{FD}$, which simplifies to $FD = \frac{CD}{2}$. Since $D$ is very close to $P$ for paraxial rays, $FD \approx f$ and $CD \approx R$, yielding $f = \frac{R}{2}$.
The relation between the object distance ($u$), image distance ($v$), and focal length ($f$) of a spherical mirror is known as the mirror equation :
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$The linear magnification ($m$) is defined as the ratio of the height of the image ($h'$) to the height of the object ($h$):
$$m = \frac{h'}{h} = -\frac{v}{u}$$Mathematical Problem Statement: An object is placed at (i) 10 cm , (ii) 5 cm in front of a concave mirror of radius of curvature 15 cm . Find the position, nature, and magnification of the image in each case.
Step-by-Step Algorithmic Solution Trace:
- Identify the given parameters: Radius of curvature $R = -15 \text{ cm}$, so focal length $f = \frac{R}{2} = -7.5 \text{ cm}$.
- Case (i): Object distance $u = -10 \text{ cm}$. Use the mirror equation: $\frac{1}{v} + \frac{1}{-10} = \frac{1}{-7.5} \implies \frac{1}{v} = -\frac{1}{7.5} + \frac{1}{10} = \frac{-4 + 3}{30} = -\frac{1}{30} \implies v = -30 \text{ cm}$. Magnification $m = -\frac{v}{u} = -\frac{-30}{-10} = -3$. The image is real, inverted, and magnified.
- Case (ii): Object distance $u = -5 \text{ cm}$. Use the mirror equation: $\frac{1}{v} + \frac{1}{-5} = \frac{1}{-7.5} \implies \frac{1}{v} = -\frac{1}{7.5} + \frac{1}{5} = \frac{-2 + 3}{15} = \frac{1}{15} \implies v = +15 \text{ cm}$. Magnification $m = -\frac{v}{u} = -\frac{15}{-5} = +3$. The image is virtual, erect, and magnified.
Final Evaluation Output: Case (i): $v = -30 \text{ cm}$, $m = -3$ (Real, Inverted); Case (ii): $v = +15 \text{ cm}$, $m = +3$ (Virtual, Erect).
Refraction is the change in direction of propagation of a light ray when it passes obliquely from one transparent medium to another due to a change in its speed.
where $n_{21}$ is the refractive index of medium 2 with respect to medium 1. If $n_{21} > 1$, the refracted ray bends towards the normal (medium 2 is optically denser). If $n_{21} < 1$, the ray bends away from the normal (medium 2 is optically rarer).
Optical density is a measure of the speed of light in a medium (higher optical density means lower speed of light), whereas mass density is mass per unit volume. They are not directly proportional. For example, turpentine has a lower mass density than water but a higher optical density, meaning light travels slower in turpentine than in water.
When a ray of light travels from an optically denser medium to an optically rarer medium, and the angle of incidence exceeds a specific threshold, the light is completely reflected back into the denser medium. This phenomenon is called total internal reflection .
The angle of incidence in the denser medium for which the angle of refraction in the rarer medium is $90^\circ$ is called the critical angle ($i_c$). From Snell's law:
$$\sin i_c = n_{21} = \frac{1}{n_{12}}$$| Substance Medium | Refractive Index ($n$) | Critical Angle ($i_c$) |
|---|---|---|
| Water | 1.33 | $48.75^\circ$ |
| Crown Glass | 1.52 | $41.14^\circ$ |
| Dense Flint Glass | 1.62 | $37.31^\circ$ |
| Diamond | 2.42 | $24.41^\circ$ |
Refraction at curved boundaries forms the basis of image formation by lenses.
For a spherical interface of radius of curvature $R$ separating two media of refractive indices $n_1$ and $n_2$, the relation between object distance $u$ and image distance $v$ is:
$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$By applying the single spherical surface refraction formula successively to the two surfaces of a thin lens, we obtain the Lens Maker's Formula :
$$\frac{1}{f} = \left(\frac{n_2}{n_1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$where $R_1$ and $R_2$ are the radii of curvature of the two surfaces, and $f$ is the focal length. The Thin Lens Formula relates $u$, $v$, and $f$ as:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$The linear magnification ($m$) produced by a lens is:
$$m = \frac{h'}{h} = \frac{v}{u}$$The power of a lens ($P$) is a measure of its ability to converge or diverge light, defined as the reciprocal of its focal length (in metres):
$$P = \frac{1}{f}$$The SI unit of power is the dioptre ($D$), where $1 \text{ D} = 1 \text{ m}^{-1}$. Power is positive for converging (convex) lenses and negative for diverging (concave) lenses.
When several thin lenses of focal lengths $f_1, f_2, f_3, \dots$ are placed in contact, the effective focal length ($f$) of the combination is given by:
$$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} + \dots$$In terms of power, the net power ($P$) is the algebraic sum of individual powers:
$$P = P_1 + P_2 + P_3 + \dots$$The total magnification ($m$) is the product of individual magnifications:
$$m = m_1 \times m_2 \times m_3 \times \dots$$A prism is a wedge-shaped transparent medium bounded by two refracting surfaces inclined at an angle called the angle of the prism ($A$).
When a ray of light passes through a triangular prism, it suffers deviation. The total angle of deviation ($\delta$) is given by:
$$\delta = i + e - A$$where $i$ is the angle of incidence, $e$ is the angle of emergence, and $A$ is the angle of the prism. At the position of minimum deviation ($\delta = D_m$), we have $i = e$ and $r_1 = r_2 = A/2$. The refractive index ($n_{21}$) of the prism material is given by:
$$n_{21} = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$For a thin prism (small angle $A$), the minimum deviation is simplified to:
$$D_m = (n_{21} - 1)A$$Optical instruments utilize the principles of reflection and refraction to enhance human vision or capture images.
A simple microscope is a converging lens of short focal length. The magnifying power ($m$) is:
To achieve much higher magnification, a compound microscope uses two converging lenses: the objective (near the object, short focal length $f_o$) and the eyepiece (near the eye, short focal length $f_e$). The total magnification ($m$) when the final image is at infinity is:
$$m = m_o \times m_e \approx \left(\frac{L}{f_o}\right)\left(\frac{D}{f_e}\right)$$where $L$ is the tube length (distance between the second focal point of the objective and the first focal point of the eyepiece).
Telescopes are used to provide angular magnification of distant objects.
The length of the telescope tube is $f_o + f_e$.