Chapter 06: Spectrum - Vardaan Master Notes

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(A) DEVIATION, DISPERSION AND SPECTRUM

1. Angle of Deviation ($\delta$): Angle between incident ray produced forward and emergent ray produced backward through a prism.

$$\delta = i + e - A \implies A + \delta = i + e$$

Factors Affecting Total Deviation ($\delta$):

Angle of incidence ($i$)
Angle of prism ($A$)
Refractive index of the material of prism ($\mu$)
Wavelength / Colour of the incident light ($\lambda$)

1. Prism Formula for Refractive Index ($\mu$) & Minimum Deviation ($\delta_m$):

$${}_a\mu_g = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$

Where: $A = \text{Angle of Prism}$, $\delta_m = \text{Angle of Minimum Deviation}$, ${}_a\mu_g = \text{Refractive index of glass w.r.t air}$.

2. Small Angle Prism Formula (When Angle $A$ is very small):

$$\delta = (\mu - 1) A$$

3. Fundamental Proportionality Chain:

$$\delta \propto \mu \propto \frac{1}{v} \propto \frac{1}{\lambda}$$

Conditions for Minimum Deviation ($\delta = \delta_m$):

Angle Condition: Angle of incidence equals angle of emergence ($i = e$).
Refraction Angles: Angle of refraction at 1st surface equals angle of incidence at 2nd surface ($r_1 = r_2 = \frac{A}{2}$).
Ray Path: The refracted ray inside an equilateral/isosceles prism travels parallel to the base of the prism.

Variation of Angle of Deviation ($\delta$) with Angle of Incidence ($i$)

As angle of incidence $i$ increases, deviation $\delta$ first decreases, reaches minimum value $\delta_m$ when $i = e$, and then increases.

Fig. 6.1 Deviation produced by a triangular prism

Cause of Dispersion (Wavelength & Speed Link):

In vacuum or air, all colours of light travel with the exact same speed ($c = 3 \times 10^8\text{ m/s}$).
In a transparent medium (like glass): Speed $v$ decreases as wavelength $\lambda$ decreases ($v \propto \lambda$).
Refractive index $\mu = \frac{c}{v} \implies \mu$ increases as $\lambda$ decreases ($\mu \propto 1/\lambda$).
Angle of deviation $\delta \propto \mu \implies$ Violet ($\lambda = 4000\text{ Å}$) deviates MOST and Red ($\lambda = 8000\text{ Å}$) deviates LEAST.

Fig. 6.2 Dispersion by a prism

High-Yield ICSE Exam Definitions:

Dispersion: The phenomenon of splitting of white light by a prism into its constituent seven colours (VIBGYOR).
Spectrum: The coloured band of distinct colours obtained on a screen when white light passes through a glass prism.
Crucial Board Point: Dispersion occurs at the FIRST surface of the prism. The prism does not produce colours, it merely separates existing constituent wavelengths due to their different speeds in glass!

VIBGYOR Spectral Wavelength & Frequency Range

Colour	Wavelength ($\text{Å}$)	Wavelength ($\text{nm}$)	Speed in Glass	Deviation ($\delta$)
Violet	$4000\text{ Å} - 4460\text{ Å}$	$400\text{ nm} - 446\text{ nm}$	Slowest	Maximum
Green	$5000\text{ Å} - 5780\text{ Å}$	$500\text{ nm} - 578\text{ nm}$	Moderate	Mean Deviation
Red	$6200\text{ Å} - 8000\text{ Å}$	$620\text{ nm} - 800\text{ nm}$	Fastest	Minimum

Figure 6.3 Light incident on a prism and on a parallel sided glass slab

$Figure 6.4 Completed ray diagrams showing refraction through prism and parallel sided glass slab$

Figure 6.4 Completed ray diagrams showing refraction through prism and parallel sided glass slab

$Fig. 6.5 & Fig. 6.6 Refraction and Total Internal Reflection of composite beam$

Fig. 6.5 & Fig. 6.6 Refraction and Total Internal Reflection of composite beam (red + blue + yellow) in right-angled isosceles prism

ICSE Board Ray Diagram Explanation ($45^\circ - 90^\circ - 45^\circ$ Prism with $C_{\text{yellow}} = 45^\circ$):

A composite ray of Red, Yellow, and Blue light is incident normally ($i = 0^\circ$) on face $AB$. It enters face $AB$ undeviated ($r = 0^\circ$) and strikes hypotenuse face $AC$ at an angle of incidence $i = 45^\circ$:

Yellow Light ($i = 45^\circ = C_{\text{yellow}}$): Since angle of incidence equals its critical angle ($C_{\text{yellow}} = 45^\circ$), yellow light refracts grazing along the surface AC at an angle of refraction $r = 90^\circ$.
Red Light ($\lambda_{\text{red}} > \lambda_{\text{yellow}} \implies \mu_{\text{red}} < \mu_{\text{yellow}}$): Its critical angle is greater than $45^\circ$ ($C_{\text{red}} > 45^\circ$). Thus $i = 45^\circ < C_{\text{red}}$, so red light refracts OUT of face $AC$ into air, bending away from the normal.
Blue Light ($\lambda_{\text{blue}} < \lambda_{\text{yellow}} \implies \mu_{\text{blue}} > \mu_{\text{yellow}}$): Its critical angle is less than $45^\circ$ ($C_{\text{blue}} < 45^\circ$). Thus $i = 45^\circ > C_{\text{blue}}$, so blue light suffers Total Internal Reflection (TIR) at face $AC$, reflects down by $90^\circ$, and emerges normally out of base $BC$ undeviated!

(B) ELECTROMAGNETIC SPECTRUM AND ITS BROAD CLASSIFICATION

Wave Equation:

$$c = f \lambda \implies f = \frac{c}{\lambda}$$

Where $c = 3 \times 10^8\text{ m/s}$ (speed of ALL EM waves in vacuum).

Complete Electromagnetic Spectrum (In Increasing Order of Wavelength $\lambda$ / Decreasing Order of Frequency $f$):

Gamma ($\gamma$) rays (Shortest $\lambda < 0.1\text{ Å}$, Highest Frequency & Energy)
X-rays ($\lambda = 0.1\text{ Å} - 100\text{ Å}$)
Ultraviolet (UV) rays ($\lambda = 100\text{ Å} - 4000\text{ Å}$)
Visible light ($\lambda = 4000\text{ Å} - 8000\text{ Å}$)
Infrared (IR) radiations ($\lambda = 8000\text{ Å} - 10^7\text{ Å}$)
Microwaves ($\lambda = 10^7\text{ Å} - 10^{11}\text{ Å}$)
Radio waves (Longest $\lambda > 10^{11}\text{ Å}$, Lowest Frequency & Energy)

⚡ Golden Board Rule: These EM waves have different wavelengths, but each wave travels with the exact SAME speed equal to $3 \times 10^8\text{ m s}^{-1}$ in vacuum (or air)!

Fig. 6.10 Electromagnetic spectrum

ICSE Board EM Spectrum Classification Table

Radiation	Wavelength Range ($\lambda$)	Detection Method	High-Yield Board Uses
Gamma ($\gamma$) Rays	$< 0.1\text{ Å}$	Penetrating power / Geiger Counter	Medical radiotherapy (killing cancer cells), inspecting welding joints.
X-Rays	$0.1\text{ Å} - 100\text{ Å}$	Fluorescence on ZnS / Photographic plate	Radiography for detecting bone fractures, CAT scans, studying crystal structures.
Ultraviolet (UV)	$100\text{ Å} - 4000\text{ Å}$	Silver chloride solution (turns dark brown/black)	Sterilising surgical instruments, detecting fake currency/gems, Vitamin D synthesis. Requires Quartz Prism!
Visible Light	$4000\text{ Å} - 8000\text{ Å}$	Human retina / Photographic film	Enables human vision, photography, photosynthesis in green plants.
Infrared (IR)	$8000\text{ Å} - 10^7\text{ Å}$	Thermopile / Blackened bulb thermometer	Photography in dark/fog, TV remote controls, signals in war, heat therapy. Requires Rock-Salt Prism!
Microwaves	$10^7\text{ Å} - 10^{11}\text{ Å}$	Oscillatory tuned electrical circuit	Radar communication, satellite TV, microwave cooking ovens.
Radio Waves	$> 10^{11}\text{ Å}$	Receiving antenna / Aerials	Radio and television broadcasting transmissions.

ICSE Board Reasons to Remember:

Why Quartz Prism for UV? Ordinary glass absorbs ultraviolet light. Quartz does not absorb UV radiations, so a quartz prism is used to obtain the UV spectrum.
Why Rock-Salt Prism for IR? Ordinary glass absorbs infrared radiations. Rock-salt does not absorb IR radiations, so a rock-salt prism is required.
Why IR for Fog Photography? Infrared radiations have longer wavelengths ($\lambda$), so they are scattered very little by atmospheric fog particles ($I \propto 1/\lambda^4$) and can penetrate deep through mist and fog.

(C) SCATTERING OF LIGHT AND ITS APPLICATIONS

Scattering of Light: The phenomenon by which fine dust particles and air molecules in atmosphere absorb light energy and re-emit it in all random directions.

Rayleigh's Law of Scattering ($d \ll \lambda$): The intensity of scattered light ($I$) is inversely proportional to the 4th power of wavelength ($\lambda$).

$$I \propto \frac{1}{\lambda^4}$$

Short wavelength blue/violet light scatters nearly 16 times more than long wavelength red light!

Fig. 6.11 Red colour of sun at sunrise and sunset

Rayleigh Scattering Curve ($I \propto 1/\lambda^4$)

6 Must-Know ICSE Board Full Exam-Ready Explanations

1. Why the Sun appears Red at Sunrise and Sunset:

At sunrise and sunset, the Sun is near the horizon. Therefore, sunlight has to travel through a maximum distance (thickness) of the Earth's atmosphere to reach an observer. As light passes through the atmosphere, shorter wavelengths (violet, indigo, blue) are scattered away in all directions according to Rayleigh's Law ($I \propto 1/\lambda^4$). Light of longest wavelength (Red, $\lambda \approx 8000\text{ Å}$) is scattered the least and reaches the observer's eyes directly. Thus, the Sun appears red.

2. Why the Sky appears Blue during a Clear Day:

The Earth's atmosphere contains fine air molecules ($\text{N}_2, \text{O}_2$) whose size ($d$) is much smaller than the wavelength of visible light ($d \ll \lambda$). According to Rayleigh's Law ($I \propto 1/\lambda^4$), short wavelength blue light is scattered nearly 16 times more strongly than long wavelength red light in all directions. When we look at the sky, this scattered blue light enters our eyes, making the sky appear blue.

3. Why Clouds appear White:

Clouds consist of water droplets, ice crystals, and dust aggregates whose size ($d$) is much larger than the wavelength of visible light ($d \gg \lambda$). When particle size is large, Rayleigh's Law fails, and light of all wavelengths is scattered equally. Since all 7 constituent colours of white light are scattered in equal proportions, clouds appear white.

4. Why the Sky appears Black to an Astronaut in Space or on the Moon:

Outer space and the surface of the Moon do not possess an atmosphere. In the absence of air molecules or dust particles, no scattering of sunlight takes place. Since no scattered light enters the astronaut's eyes from the surrounding space, the sky appears completely dark or jet black.

5. Why Danger Signals and Stop Lights are Always Red:

Among all colours of visible light, red light has the longest wavelength ($\lambda \approx 8000\text{ Å}$). By Rayleigh's Law ($I \propto 1/\lambda^4$), red light is scattered the least by atmospheric fog, smoke, or dust particles. It can penetrate through fog over maximum distances without losing intensity, remaining visible from far away.

6. Why the Sun appears White at Noon:

At noon, the Sun is directly overhead. Sunlight travels the shortest distance through the atmosphere to reach an observer. Due to this short path length, only a negligible amount of light of any colour gets scattered. Sunlight reaching our eyes directly contains all 7 constituent colours in almost equal proportions, so the Sun appears white.

High-Yield Board Numericals

Numerical 1

NUMERICAL Yellow light wavelength $\lambda = 5800\text{ Å}$. Calculate frequency $f$ ($c = 3 \times 10^8\text{ m/s}$).

$$\lambda = 5800 \times 10^{-10}\text{ m} = 5.8 \times 10^{-7}\text{ m}$$ $$f = \frac{c}{\lambda} = \frac{3 \times 10^8}{5.8 \times 10^{-7}} = \mathbf{5.17 \times 10^{14}\text{ Hz.}}$$

Numerical 2

NUMERICAL Radio station frequency $f = 100\text{ MHz}$. Find wavelength $\lambda$.

$$f = 100 \times 10^6\text{ Hz} = 10^8\text{ Hz}$$ $$\lambda = \frac{c}{f} = \frac{3 \times 10^8}{10^8} = \mathbf{3\text{ m.}}$$

ICSE Board Past Year Questions (PYQs)

ICSE 2015

BOARD (a) Name waves used to study crystal structure. (b) State one more use.

Ans: (a) X-Rays. (b) Bone fracture detection in radiography.

ICSE 2014

BOARD Name radiation used for: (i) photography in dark/fog, (ii) sterilizing surgical tools.

Ans: (i) Infrared radiations. (ii) Ultraviolet radiations.

ICSE 2013

BOARD Why quartz prism for UV light and rock-salt prism for IR light?

Ans: Glass absorbs both UV and IR. Quartz is transparent to UV, rock-salt is transparent to IR.

ICSE 2010

BOARD Name radiation detected by: (i) thermopile, (ii) silver chloride solution.

Ans: (i) Infrared radiation. (ii) Ultraviolet radiation.