Question

The frequency of a light wave in a material is \(4 \times 10^{14} \mathrm{~Hz}\) and wavelength is \(5000 \AA\). The refractive index of material will be take \(\left.\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right)\) (A) \(1.5\) (B) \(1.7\) (C) \(1.33\) (D) None of these

Short answer

The refractive index of the material is 1.5 (option A).

Step-by-step solution

Convert wavelength to meters

First, convert the wavelength from Angstroms to meters. 1 Angstrom (Å) is equal to \(1 \times 10^{-10}\) meters (m). So, the wavelength in meters is: \(λ = 5000 Å × \frac{1 \times 10^{-10} m}{1 Å} = 5 \times 10^{-7} m\)

Calculate the speed of light in the material

Now we can calculate the speed of light in the material (v) using the frequency (f) and wavelength (λ): v = f × λ v = \(4 \times 10^{14} Hz \ × \ 5 \times 10^{-7} m\) v = \(2 \times 10^{8}\) m/s

Find the refractive index

Finally, we can find the refractive index (n) using the speed of light in vacuum (c) and the speed of light in the material (v): n = c / v n = \(\frac{3 \times 10^{8} m/s}{2 \times 10^{8} m/s}\) n = 1.5 Therefore, the refractive index of the material is 1.5 which corresponds to option (A).

Key concepts

Wavelength Conversion

When working with wave properties of light, like wavelength, it's sometimes necessary to convert measurements into units that are more applicable in formulas. In the field of optics, the conversion from Angstroms (Å) to meters is common. One Angstrom is a very small distance, equal to \(1 \times 10^{-10}\) meters. Thus, to convert a wavelength in Angstroms to meters, you multiply the wavelength value by \(1 \times 10^{-10}\) meters per Angstrom.

For example, if we have a wavelength of \(5000\) Å, converting it to meters involves the following calculation:

• Wavelength (λ) in meters: \(5000\) Å \(\times \frac{1 \times 10^{-10} \, \text{m}}{1 \, Å} = 5 \times 10^{-7} \, \text{m}\)

This conversion is crucial because most physics equations involving light use meters as the standard unit of length. By ensuring that your wavelengths are in meters, your calculations remain consistent and free of unit-based errors.

Speed of Light Calculation

The speed of light in a given material can be determined using the relationship between its frequency and wavelength within that material. This speed, often referred to as \(v\), is calculated with the formula \(v = f \times λ\), where \(f\) is the frequency in hertz (Hz) and \(λ\) is the wavelength in meters. Knowing this, you can easily determine the speed of light through the material, which is typically less than its speed in a vacuum (\(c = 3 \times 10^{8} \, \text{m/s}\)).

For this particular exercise, given:

• Frequency (\(f\)) = \(4 \times 10^{14} \, \text{Hz}\)
• Wavelength (\(λ\)) = \(5 \times 10^{-7} \, \text{m}\)

The speed of light in the material is calculated as follows:
\(v = 4 \times 10^{14} \, \text{Hz} \times 5 \times 10^{-7} \, \text{m} = 2 \times 10^{8} \, \text{m/s}\)
Understanding this calculation is crucial for determining how light behaves in different mediums and for further calculations such as finding the refractive index.

Frequency and Wavelength Relationship

The relationship between frequency and wavelength is central to understanding light waves and other types of waves. They are inversely proportional to each other, meaning that as one increases, the other decreases, and vice versa. The formula \(v = f \times λ\) illustrates this relationship, where \(v\) is the speed of the wave (often the speed of light in the context of optics), \(f\) is the frequency, and \(λ\) is the wavelength.

To elaborate, if you have a constant speed of light in a material (as calculated previously), an increase in frequency will result in a corresponding decrease in wavelength, holding the speed constant.

• Higher frequency = Lower wavelength
• Lower frequency = Higher wavelength

Understanding this relationship helps in predicting how changes to one property of wave motion can affect the others. It's fundamental in fields such as telecommunications, radio and TV broadcasting, and optical technologies, all of which rely on the manipulation of wave properties.