Question

The wavelength of yellow line of sodium is $589\;nm$
(a) What is its frequency?
(b) What is its wavelength in glass whose refractive index is $1.52$?
(c) From the results from (a) and (b), find its speed in glass

Answer 1

The formula for the frequency in terms of the quantities of speed and wavelength is to be applied in order to find its frequency. All the necessary conversions from one quantity to another are applied to get the required answer in terms of the required unit. The relation between refractive index and the wavelength of light in the differing mediums needs to be found out to find the wavelength of light in glass.

Formula used:
The frequency of light in air is given by the relation:
$f = \dfrac{c}{\lambda }$
Where, $c$ is the speed of light in air and $\lambda$ is the wavelength of light.

Complete step by step answer:
The above problem revolves around the concept of The above problem revolves around the concept of speed of light, its corresponding wavelength and frequency and how these quantities vary when passed through another medium other than air.

Huygens’s wave theory of light states that light travels in a wave-like motion and hence these waves that represent light will have a certain wavelength. This was determined from the various phenomena in nature like the phenomenon where water forms ripples when a stone is thrown into it and it was stated that light also travels in the form of waves.

(a) Let us now extract the data given in the question. The wavelength of the yellow light of sodium is already given to us which is given as $598\;nm$. The frequency is said to be the number of waves moving through the medium per unit time. We are asked to find out this frequency of light in the air.

We will now construct an equation relating these quantities to determine an equation for the frequency of light in terms of its speed and wavelength. Hence, the relation is as follows:
$f = \dfrac{c}{\lambda }$ ----------($1$)
The speed of light in air is already known as it is a constant and given by the value $3 \times {10^8}m/s$.
Given, $c = 3 \times {10^8}m/s$ and $\lambda = 589\;nm$. The wavelength of light is usually measured in terms of the unit meters but the wavelength here is given in nanometers. Thus, we must convert from nanometers to meters. Hence, they are related to each other as follows:
$1nm = {10^{ - 9}}m$
Hence, by unitary method we have:
$\Rightarrow 589\;nm = 589 \times {10^{ - 9}}m$

Thus, we obtain the wavelength of the yellow in meters. We will now substitute these values into the equation ($1$) to obtain the frequency of the yellow light of Sodium in air. Hence we have:
$f = \dfrac{{3 \times {{10}^8}}}{{589 \times {{10}^{ - 9}}}}$
On simplification we get:
$f = \dfrac{3}{{589}} \times {10^{8 + 9}}$
$\Rightarrow f = 0.00509 \times {10^{17}}Hz$
We will now shift the decimal place three times to get:
$\therefore f = 5.09 \times {10^{14}}Hz$
Thus this is the frequency of the light in air.

(b) Wavelength of light is said to be the distance between its two consecutive crests or its two consecutive troughs. We know that light travels as waves and the distance it covers is measured in terms of its wavelength. When the light travels through different mediums it tends to undergo some changes in its path and direction which is governed by a property of light known as refraction.

Refraction is defined as the change in the path of light or more specifically the bending of the light rays due to the difference in the medium it is travelling in which in this case is the glass medium. The amount of refraction of the light rays is dependent on the refractive indices of both the medium through which light is traversing. It measures the ability of the surface to refract light. The medium that the light goes through which is given in our question is glass medium. This refractive index is given to be $1.52$.

Since, the light passes through a different medium, that is, glass it is said to undergo a change in its wavelength as well since there is a change in its speed and we know that the speed and wavelength are related to each other. The frequency is taken to remain constant as we are considering monochromatic light, that is, yellow light which is of single frequency.
We need to determine a relation between the refractive indices of the two mediums and the wavelength of the green light in the two mediums. The refractive index is directly proportional to the speed of light and hence will be inversely proportional to its wavelength.

Therefore, the ratios of the two refractive indices in the two mediums and their corresponding wavelengths will be related inversely to each other. Thus, we can say that:
$\dfrac{{{n_g}}}{{{n_a}}} = \dfrac{{{\lambda _a}}}{{{\lambda _g}}}$ --------($2$)
The subscripts $g$ and $a$ represent glass and air respectively. We are asked to calculate the wavelength of light in the glass medium. We know that the refractive index of air which is said to be $1$ and we are already given the refractive index of light in glass and also the wavelength of light in air.

Given, ${n_{glass}} = 1.52$, ${n_{air}} = 1$ and ${\lambda _a} = 589 \times {10^{ - 9}}m$. We now substitute these values in equation ($2$) to obtain the unknown value which is the wavelength of light in glass. Hence:
$\dfrac{{1.52}}{1} = \dfrac{{589 \times {{10}^{ - 9}}m}}{{{\lambda _g}}}$
Hence, by cross-multiplying the terms we get:
$1.52 \times {\lambda _g} = 589 \times {10^{ - 9}}m \times 1$
By simplifying further to make ${\lambda _g}$ as the subject we get:
${\lambda _g} = \dfrac{{589 \times {{10}^{ - 9}}}}{{1.52}}$

Thus, we get:
${\lambda _g} = 387.5 \times {10^{ - 7}}m$
By shifting the decimal place back by two places to approximate it to three significant figures we get:
${\lambda _g} = 3.87 \times {10^{ - 9}}m$
As per the relation between nanometers and meters we will get:
$\therefore {\lambda _g} = 387\;nm$
Thus, this is the wavelength of yellow light in the glass medium.

(c) The change in speed when the light enters the glass medium is observed and this new speed of light in the glass medium is asked to be found out. The frequency value is constant as the light here is monochromatic which means that the light is of a single colour which is yellow and is of a single frequency. This frequency was obtained from part a). The wavelength of light in glass medium was determined from part b). Hence:

Given, $f = 5.09 \times {10^{14}}Hz$ and ${\lambda _g} = 387\,nm$.We once again apply the formula given in equation ().The speed considered here is not in air but in glass instead and hence the formula becomes:
$f = \dfrac{v}{\lambda }$
Here, $v$ denotes the speed of the sodium light in the glass medium. This is what we are expected to find out so we rearrange the terms to get the equation to find the speed. We get:
$v = f \times \lambda$

Hence, by substituting the given values in the above equation we will get the speed of light in glass medium.
$\Rightarrow v = 5.09 \times {10^{14}} \times 387 \times {10^{ - 9}}$
On further simplification we get:
$\Rightarrow v = 5.09 \times 387 \times \times {10^{14 - 9}}$
$\Rightarrow v = 5.09 \times 387 \times {10^5}$
Hence by solving the equation we get speed to be:
$\therefore v = 1.97 \times {10^5}m/s$
This is the speed of the yellow light from sodium in glass.

Note: The common error which can be made in this problem can be when determining the relation between the refractive index and wavelength wherein the mistake is make when the inverse relation between the two terms is not taken into consideration which will lead to the wrong value as the answer.