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Physics - Spectrum

Grade 10ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Dispersion of Light: The phenomenon of splitting of a beam of white light into its constituent seven colors (VIBGYOR) when passed through a transparent medium like a glass prism. This happens because different colors of light travel at different speeds in glass, though they travel at the same speed c=3×108 m/sc = 3 \times 10^8 \text{ m/s} in vacuum.

Cause of Dispersion: The refractive index μ\mu of a medium depends on the wavelength λ\lambda of light. According to Cauchy's relation, μA+Bλ2\mu \approx A + \frac{B}{\lambda^2}. Since λred>λviolet\lambda_{red} > \lambda_{violet}, the refractive index for red light is less than that for violet light (μred<μviolet\mu_{red} < \mu_{violet}), causing red to deviate the least and violet to deviate the most.

Electromagnetic Spectrum: The complete range of electromagnetic waves arranged in order of increasing frequency or decreasing wavelength. The sequence is: Radio waves, Microwaves, Infrared (IR), Visible light, Ultraviolet (UV), X-rays, and Gamma rays (γ\gamma-rays).

Visible Spectrum: The narrow band of the EM spectrum visible to the human eye, ranging from approximately 4000 A˚4000 \text{ \AA} (400 nm400 \text{ nm}) for violet to 8000 A˚8000 \text{ \AA} (800 nm800 \text{ nm}) for red light.

Infrared Radiation: These are heat radiations with wavelengths longer than red light (>8000 A˚> 8000 \text{ \AA}). They are detected using a thermopile or a blackened bulb thermometer. They do not affect ordinary photographic film but affect special infrared films.

Ultraviolet Radiation: These have wavelengths shorter than violet light (<4000 A˚< 4000 \text{ \AA}). They can be detected by their chemical effect on silver chloride (AgClAgCl) solution, which turns violet and then black. UV rays are absorbed by the Ozone layer in the atmosphere.

Scattering of Light: The process where light is absorbed by atmospheric particles and re-emitted in all directions. According to Rayleigh's law of scattering, the intensity of scattered light II is inversely proportional to the fourth power of the wavelength: I1λ4I \propto \frac{1}{\lambda^4}. This explains why the sky appears blue (shorter wavelengths scatter more) and the sun appears red at sunrise/sunset (longer wavelengths reach the observer).

📐Formulae

v=fλv = f \lambda

μ=cv\mu = \frac{c}{v}

μ1λ\mu \propto \frac{1}{\lambda}

I1λ4I \propto \frac{1}{\lambda^4}

δ=i1+i2A\delta = i_1 + i_2 - A

1 A˚=1010 m=101 nm1 \text{ \AA} = 10^{-10} \text{ m} = 10^{-1} \text{ nm}

💡Examples

Problem 1:

A beam of monochromatic light has a wavelength of 6000 A˚6000 \text{ \AA} in air. It enters a glass slab of refractive index μ=1.5\mu = 1.5. Calculate the wavelength of light in glass.

Solution:

Given: λair=6000 A˚\lambda_{air} = 6000 \text{ \AA}, μ=1.5\mu = 1.5. We use the formula λmedium=λairμ\lambda_{medium} = \frac{\lambda_{air}}{\mu}. Therefore, λglass=60001.5=4000 A˚\lambda_{glass} = \frac{6000}{1.5} = 4000 \text{ \AA}.

Explanation:

When light enters a denser medium, its speed decreases but its frequency remains constant. This results in a proportional decrease in wavelength.

Problem 2:

Compare the scattering of blue light (4000 A˚4000 \text{ \AA}) and red light (8000 A˚8000 \text{ \AA}) in the atmosphere.

Solution:

According to Rayleigh's Law, I1λ4I \propto \frac{1}{\lambda^4}. The ratio of scattering is IblueIred=(λredλblue)4\frac{I_{blue}}{I_{red}} = \left( \frac{\lambda_{red}}{\lambda_{blue}} \right)^4. Substituting the values: IblueIred=(80004000)4=24=16\frac{I_{blue}}{I_{red}} = \left( \frac{8000}{4000} \right)^4 = 2^4 = 16.

Explanation:

Blue light is scattered 1616 times more than red light because it has a shorter wavelength, which is why the sky appears blue.

Spectrum - Revision Notes & Key Formulas | ICSE Class 10 Science