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Physics - Spectrum

Grade 10ICSE

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

🔑Concepts

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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.

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Cause of Dispersion: The refractive index μ\mu of a medium depends on the wavelength λ\lambda of light. According to Cauchy's relation, μ≈A+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.

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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).

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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.

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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.

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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.

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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: I∝1λ4I \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}

μ∝1λ\mu \propto \frac{1}{\lambda}

I∝1λ4I \propto \frac{1}{\lambda^4}

δ=i1+i2−A\delta = i_1 + i_2 - A

1 A˚=10−10 m=10−1 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, I∝1λ4I \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 Class 10 Notes & Examples | ICSE Science