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The wavelength of sodium light in air is \(589 \mathrm{~nm} .\) (a) Find its frequency in air. (b) Find its wavelength in water (refractive index \(=1 \cdot 33\) ). (c) Find its frequency in water. (d) Find its speed in water.

Short Answer

Expert verified
(a) Frequency in air: \(5.09 \times 10^{14} \text{ Hz}\). (b) Wavelength in water: \(443 \text{ nm}\). (c) Frequency in water: \(5.09 \times 10^{14} \text{ Hz}\). (d) Speed in water: \(2.26 \times 10^8 \text{ m/s}\).

Step by step solution

01

Introduction to Concept

The wavelength of light changes when it moves from one medium to another, but its frequency remains constant. The speed of light in a medium is given by the equation: \[ v = \frac{c}{n} \]where \(c\) is the speed of light in a vacuum \( (3 \times 10^8 \text{ m/s}) \) and \(n\) is the refractive index of the medium.
02

Calculate Frequency in Air

The frequency \(f\) can be calculated using the relationship between speed, wavelength, and frequency: \[ v = f \cdot \lambda \] where \(v = c\) in air.Solving for frequency, we have: \[ f = \frac{c}{\lambda} \]Given \( \lambda = 589 \text{ nm} = 589 \times 10^{-9} \text{ m} \):\[ f = \frac{3 \times 10^8}{589 \times 10^{-9}} \approx 5.09 \times 10^{14} \text{ Hz} \]
03

Calculate Wavelength in Water

To find the wavelength in water, use the relationship between wavelength in air \(\lambda_{\text{air}}\) and wavelength in water \(\lambda_{\text{water}}\): \[ \lambda_{\text{water}} = \frac{\lambda_{\text{air}}}{n} \]Given \(n = 1.33\), we calculate:\[ \lambda_{\text{water}} = \frac{589 \times 10^{-9}}{1.33} \approx 443 \times 10^{-9} \text{ m} \text{ or } 443 \text{ nm} \]
04

Calculate Frequency in Water

The frequency of light does not change with the medium. Therefore, the frequency in water is the same as in air: \[ f_{\text{water}} = f_{\text{air}} = 5.09 \times 10^{14} \text{ Hz} \]
05

Calculate Speed in Water

The speed of light in water is given by the equation: \[ v_{\text{water}} = \frac{c}{n} \]Substituting the values, we find:\[ v_{\text{water}} = \frac{3 \times 10^8}{1.33} \approx 2.26 \times 10^8 \text{ m/s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wavelength
Wavelength is the distance between consecutive peaks of a wave and is usually denoted by the Greek letter \( \lambda \). In the context of light, it represents the color or type of light. Wavelengths of light that we can see, also known as visible light, typically range from 380 nm to 750 nm. \(1 \text{ nm} = 1 \times 10^{-9} \text{ m} \). The wavelength of light changes when it enters a different medium, such as moving from air into water, due to its interaction with the medium's atomic structure. Yet, the frequency remains the same across different media. Understanding wavelength helps in identifying the type of wave and how it will behave when passing through different materials.
Frequency
Frequency refers to the number of wave cycles that pass a given point per unit of time and is usually measured in hertz (Hz). In optics, it relates to the color of the light; higher frequencies correspond to blue light, while lower frequencies correspond to red light. The relationship between wavelength \( \lambda \) and frequency \( f \) is given by the formula \( c = f \cdot \lambda \), where \( c \) is the speed of light in a vacuum (\(3 \times 10^8 \text{ m/s}\)). In any medium, the frequency of light does not change when it crosses from one medium to another, which is a fundamental principle in optics. This allows us to predict how light behaves across different media without altering its frequency.
Refractive Index
The refractive index \( n \) of a medium describes how much the light slows down when it enters that medium. It is a dimensionless number indicative of how fast light travels through the material compared to its speed in a vacuum. When light enters a material like water, which has a refractive index of \( 1.33 \), it slows down compared to its speed in air. The equation \( v = \frac{c}{n} \) where \( v \) is the speed of light in the medium, shows how to calculate the new speed of light in the medium. Media with higher refractive indices slow down light more significantly and bend it more sharply.
Speed of Light
The speed of light is a fundamental constant of nature, usually denoted as \( c \). In a vacuum, it is approximately \(3 \times 10^8 \text{ m/s}\). However, when light travels through different media, its speed changes due to interactions with the atoms in the medium. This change in speed is dictated by the refractive index \( n \). Thus, the speed of light in any medium is calculated using \( v = \frac{c}{n} \). For example, in water with a refractive index of \( 1.33 \), the speed of light is slowed to approximately \( 2.26 \times 10^8 \text{ m/s} \). Hence, understanding the speed of light in various media helps in many practical applications including optics and telecommunications.

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Most popular questions from this chapter

A transparent paper (refractive index \(=1 \cdot 45\) ) of thickness \(0.02 \mathrm{~mm}\) is pasted on one of the slits of a Young's double slit experiment which uses monochromatic light of wavelength \(620 \mathrm{~nm}\). How many fringes will cross through the centre if the paper is removed?

A long narrow horizontal slit is placed \(1 \mathrm{~mm}\) above a horizontal plane mirror. The interference between the light coming directly from the slit and that after reflection is seen on a screen \(1 \cdot 0 \mathrm{~m}\) away from the slit. Find the fringe-width if the light used has a wavelength of \(700 \mathrm{~nm}\)

A convex lens of diameter \(8 \cdot 0 \mathrm{~cm}\) is used to focus a parallel beam of light of wavelength \(620 \mathrm{~nm}\). If the light be focused at a distance of \(20 \mathrm{~cm}\) from the lens, what would be the diameter of the central bright spot formed?

In a double slit interference experiment, the separation between the slits is \(1 \cdot 0 \mathrm{~mm}\), the wavelength of light used is \(5 \cdot 0 \times 10^{-7} \mathrm{~m}\) and the distance of the screen from the slits is \(1 \cdot 0 \mathrm{~m} .\) (a) Find the distance of the centre of the first minimum from the centre of the central maximum. (b) How many bright fringes are formed in one centimeter width on the screen?

A mica strip and a polysterene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is \(0-50 \mathrm{~mm}\) and the separation between the slits is \(0 \cdot 12 \mathrm{~cm} .\) The refractive index of mica and polysterene are \(1 \cdot 58\) and \(1 \cdot 55\) respectively for the light of wavelength \(590 \mathrm{~nm}\) which is used in the experiment. The interference is observed on a screen a distance one meter away. (a) What would be the fringe-width? (b) At what distance from the centre will the first maximum be located?

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