Light is fundamental in science laboratory technology for analyzing materials and conducting experiments involving optical instruments. Understanding how light propagates and its properties enables laboratory technologists to accurately measure and interpret optical phenomena, essential in fields such as spectroscopy, microscopy, and photometry. This chapter focuses on the principles governing light’s behavior and characteristics to equip technologists with the skills to measure optical properties reliably.

5.1 Propagation of light

Propagation of light explains how light travels through different media, which is crucial when using optical instruments in laboratory settings such as spectrophotometers and refractometers. Knowledge of light propagation assists in understanding phenomena like reflection, refraction, and diffraction, which affect measurement accuracy in science laboratories.

5.1.1 Nature of Light Propagation

Light propagates as an electromagnetic wave, traveling in straight lines in a homogeneous medium unless it encounters a boundary or obstacle. The speed of light in vacuum is constant, approximately \(3 \times 10^8\) meters per second, but it slows down when passing through different media due to interaction with atoms.

$$ v = \frac{c}{n} $$

Where:
- \(v\) = speed of light in the medium (m/s)
- \(c\) = speed of light in vacuum (\(3 \times 10^8\) m/s)
- \(n\) = refractive index of the medium (dimensionless)

Worked Examples

Example 1: A light beam travels through water with refractive index \(n = 1.33\). Calculate the speed of light in water.

Given:
\(c = 3 \times 10^8\) m/s,
\(n = 1.33\)

$$ v = \frac{c}{n} $$

$$ v = \frac{3 \times 10^8}{1.33} $$

$$ v = 2.2556 \times 10^8 \text{ m/s} $$

Answer: \(2.26 \times 10^8\) m/s

Example 2: Calculate the speed of light in a glass medium with refractive index 1.5.

Given:
\(c = 3 \times 10^8\) m/s,
\(n = 1.5\)

$$ v = \frac{3 \times 10^8}{1.5} $$

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

Answer: \(2.0 \times 10^8\) m/s

Example 3: A light ray travels through a medium with refractive index 2. Calculate the speed of light in this medium.

Given:
\(c = 3 \times 10^8\) m/s,
\(n = 2\)

$$ v = \frac{3 \times 10^8}{2} $$

$$ v = 1.5 \times 10^8 \text{ m/s} $$

Answer: \(1.5 \times 10^8\) m/s

5.1.2 Reflection of Light

Reflection occurs when light strikes a surface and bounces back into the original medium. The angle of incidence equals the angle of reflection, a principle used in optical instruments such as mirrors and lasers.

$$ \theta_i = \theta_r $$

Where:
- \(\theta_i\) = angle of incidence
- \(\theta_r\) = angle of reflection

Worked Examples

Example 1: A light ray strikes a mirror at an angle of 30°. Calculate the angle of reflection.

Given:
\(\theta_i = 30^\circ\)

$$ \theta_r = \theta_i $$

$$ \theta_r = 30^\circ $$

Answer: 30°

Example 2: If a light beam hits a plane mirror at 45°, what is the angle between the incident and reflected rays?

Given:
\(\theta_i = 45^\circ\)

The angle between incident and reflected rays = \(2 \times \theta_i\)

$$ 2 \times 45^\circ = 90^\circ $$

Answer: 90°

Example 3: A light ray hits a mirror at an angle of 60°. Determine the angle between the reflected ray and the surface of the mirror.

Given:
\(\theta_i = 60^\circ\)

Angle between reflected ray and surface = \(90^\circ - \theta_r\)

Since \(\theta_r = \theta_i = 60^\circ\),

$$ 90^\circ - 60^\circ = 30^\circ $$

Answer: 30°

5.1.3 Refraction of Light

Refraction is the bending of light when it passes from one medium to another with a different refractive index. It is governed by Snell’s Law and is critical in lens design and optical measurements.

$$ n_1 \sin \theta_1 = n_2 \sin \theta_2 $$

Where:
- \(n_1\), \(n_2\) = refractive indices of medium 1 and medium 2
- \(\theta_1\), \(\theta_2\) = angles of incidence and refraction

Worked Examples

Example 1: A light ray passes from air (\(n_1 = 1.0\)) into water (\(n_2 = 1.33\)) at an angle of incidence 30°. Find the angle of refraction.

Given:
\(n_1 = 1.0\),
\(n_2 = 1.33\),
\(\theta_1 = 30^\circ\)

$$ 1.0 \times \sin 30^\circ = 1.33 \times \sin \theta_2 $$

$$ \sin \theta_2 = \frac{\sin 30^\circ}{1.33} = \frac{0.5}{1.33} = 0.3759 $$

$$ \theta_2 = \sin^{-1} 0.3759 = 22.09^\circ $$

Answer: 22.1°

Example 2: Light passes from glass (\(n_1 = 1.5\)) to air (\(n_2 = 1.0\)) with an angle of incidence 40°. Calculate the angle of refraction.

Given:
\(n_1 = 1.5\),
\(n_2 = 1.0\),
\(\theta_1 = 40^\circ\)

$$ 1.5 \times \sin 40^\circ = 1.0 \times \sin \theta_2 $$

$$ \sin \theta_2 = 1.5 \times \sin 40^\circ = 1.5 \times 0.6428 = 0.9642 $$

Since \(\sin \theta_2\) cannot be greater than 1, total internal reflection occurs; no refraction.

Answer: Total internal reflection occurs

Example 3: Light moves from air into diamond (\(n=2.42\)) at 25°. Calculate the refracted angle.

Given:
\(n_1 = 1.0\),
\(n_2 = 2.42\),
\(\theta_1 = 25^\circ\)

$$ 1.0 \times \sin 25^\circ = 2.42 \times \sin \theta_2 $$

$$ \sin \theta_2 = \frac{\sin 25^\circ}{2.42} = \frac{0.4226}{2.42} = 0.1747 $$

$$ \theta_2 = \sin^{-1} 0.1747 = 10.06^\circ $$

Answer: 10.1°

5.1.4 Diffraction and Interference of Light

Diffraction is the bending of light waves around obstacles or through small openings, while interference is the phenomenon of superposition of two or more light waves leading to patterns of constructive and destructive interference. These effects are essential in optical experiments and measurement techniques like interferometry.

The diffraction angle \(\theta\) for a slit of width \(a\) and wavelength \(\lambda\) satisfies:

$$ a \sin \theta = m \lambda $$

Where \(m = 0, \pm 1, \pm 2, ...\) is the order of the diffraction maximum.

Worked Examples

Example 1: Light of wavelength 600 nm passes through a slit 0.3 mm wide. Calculate the angle for the first order diffraction maximum.

Given:
\(\lambda = 600 \times 10^{-9}\) m,
\(a = 0.3 \times 10^{-3}\) m,
\(m=1\)

$$ a \sin \theta = m \lambda $$

$$ \sin \theta = \frac{1 \times 600 \times 10^{-9}}{0.3 \times 10^{-3}} = 0.002 $$

$$ \theta = \sin^{-1} 0.002 = 0.1146^\circ $$

Answer: 0.11°

Example 2: For the same slit, find the angle of the second order maximum.

Given:
\(m=2\)

$$ \sin \theta = \frac{2 \times 600 \times 10^{-9}}{0.3 \times 10^{-3}} = 0.004 $$

$$ \theta = \sin^{-1} 0.004 = 0.229^\circ $$

Answer: 0.23°

Example 3: A laser with wavelength 500 nm produces interference fringes on a screen 2 m away. If the fringe spacing is 2 mm, calculate the slit separation \(d\).

Using the interference fringe formula:

$$ \Delta y = \frac{\lambda L}{d} $$

Where:
\(\Delta y = 2 \times 10^{-3}\) m,
\(L = 2\) m,
\(\lambda = 500 \times 10^{-9}\) m

Rearranged:

$$ d = \frac{\lambda L}{\Delta y} $$

$$ d = \frac{500 \times 10^{-9} \times 2}{2 \times 10^{-3}} $$

$$ d = 5 \times 10^{-4} \text{ m} = 0.5 \text{ mm} $$

Answer: 0.5 mm

Chapter Summary

This chapter explored the propagation of light and its fundamental properties, setting the foundation for understanding optical behavior. It detailed the laws of reflection, explaining how light interacts with surfaces, and introduced calculations involving focal lengths, object distances, image distances, and magnification using the mirror formula. The process of image formation by lenses was examined alongside the laws of refraction, which govern the bending of light as it passes between different media. Further, the chapter covered calculations related to lenses using the lens formula and explained key concepts such as refractive index, critical angle, and total internal reflection. Key optical quantities including luminous flux and luminous intensity were defined and their significance discussed. Finally, guidance on report writing for practical measurements of optical properties was provided to ensure accurate documentation and analysis.