Green light of mercury has a wavelength of $5.5 \times {10^{ - 7}}m$ in air.
(a) What is its frequency in $MHz$ and period in $\mu s$?
(b) What is its wavelength in glass, if ${n_{glass}} = 1.5$?
Answer
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Hint: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 speed of light, its corresponding wavelength and frequency and how these quantities vary when passed through another medium other than air. We know that visible white light comprises a band of seven different colors which is given by the term VIBGYOR that specifies each of the colors that it consists of. The light that reaches the Earth has its main source as the closest star to the Earth which is the sun. The light that gets reflected from the sun is often orange or sometimes red and the same way each of the planets in space seem to reflect light of a specific color.
Mercury is said to reflect green colored rays of light which is what is mentioned in the question as well. Since, the light rays are said to be of a single color this type of light is known as monochromatic light which is also said to have a single frequency. This light from mercury is said to have some wavelength which is already given. We are asked to calculate the frequency of the light in the air.
Light travels in a wave like motion and hence these waves that represent light will have a certain wavelength. 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 which is given as $5.5 \times {10^{ - 7}}m$. The speed of light in air is already known as it is a constant and given by the value \[3 \times {10^8}m/s\]. We are asked to find out the frequency of light in the air. Hence:
Given, \[c = 3 \times {10^8}m/s\] and $\lambda = 5.5 \times {10^{ - 7}}m$. 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:
\[f = \dfrac{c}{\lambda }\] ----------($1$)
This formula originates from the speed-distance formula wherein here, the speed of light in air is considered and represented and the distance of light is always measured in terms of its wavelength. The frequency is said to be the number of waves moving through the medium per unit time and hence represents the time quantity.
(a) In order to find the frequency of light in air we substitute the given values in equation ($1$) to get:
\[f = \dfrac{{3 \times {{10}^8}}}{{5.5 \times {{10}^{ - 7}}}}\]
On further simplification we get:
\[f = \dfrac{{3 \times {{10}^8} \times {{10}^7}}}{{5.5}}\]
$ \Rightarrow f = \dfrac{3}{{5.5}} \times {10^{15}}$
By solving this above equation we get:
\[f = 0.545 \times {10^{15}}Hz\]
We now shift the decimal place to the right and this reduces the power of $10$ by $1$. Hence we get:
\[f = 5.45 \times {10^{14}}Hz\]
This is the frequency of light in the air in Hertz. But however we are asked to find out the frequency in terms of the unit MegaHertz. Thus, we need to convert the frequency from Hertz to MegaHertz.
We know that,
$1Hz = {10^{ - 6}}MHz$
Hence, by unitary method we get:
\[f = 5.45 \times {10^{14}} \times {10^{ - 6}}MHz\]
We know that if the base numbers are the same then the powers are added and hence we get:
\[f = 5.45 \times {10^{14 - 6}}MHz\]
On solving further we get:
\[f = 5.45 \times {10^8}MHz\]
Therefore, $5.45 \times {10^8}MHz$ is the frequency of green light from mercury in air in megahertz.
Next, we are asked to find the time period of the light. Hence, we determine the formula for the time period in terms of the frequency. We know that these two quantities are inversely related to each other and hence we have the formula:
$T = \dfrac{1}{f}$ ----------($2$)
We have just determined the value for frequency above. By substituting this value of frequency in equation ($2$) we get:
$T = \dfrac{1}{{5.45 \times {{10}^{14}}}}s$
On solving further we have:
$T = \dfrac{1}{{5.45}} \times {10^{ - 14}}s$
$ \Rightarrow T = 0.1834 \times {10^{ - 14}}s$
However, we are asked to find the time period in terms of the unit microseconds and hence we need to convert the time period from seconds to microseconds.We know that,
$1s = {10^6}\mu s$
Hence, the above time period obtained in seconds can be converted by unitary method.Hence, we get:
$ \Rightarrow 0.1834 \times {10^{ - 14}}s = 0.1834 \times {10^{ - 14}} \times {10^6}\mu s$
On solving further we get:
$ \Rightarrow T = 0.1834 \times {10^{ - 14 + 6}}\mu s$
$ \Rightarrow T = 0.1834 \times {10^{ - 8}}\mu s$
$ \Rightarrow T = 1.834 \times {10^{ - 9}}\mu s$
Therefore, $1.834 \times {10^{ - 9}}\mu s$ is the time period of light from mercury in microseconds.
(b) Next, we are asked to find out the wavelength of light in a medium which is given to be glass, also given the refractive index of light in glass to be $1.5$. 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. Refractive index is the ratio between the speeds of light in a vacuum to that of a medium and hence given as the relation:
$n = \dfrac{\text{Speed of light in vacuum}}{\text{Speed of light in medium}}$
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 considered monochromatic light of single frequency as mentioned earlier.
Thus 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:
$\dfrac{{{n_g}}}{{{n_a}}} = \dfrac{{{\lambda _a}}}{{{\lambda _g}}}$ --------($3$)
The subscripts g and a represent glass and air respectively. The ratios of refractive indices of light of the medium of air and glass is related to the respective wavelengths in this way. 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. Hence:
Given, ${n_{glass}} = 1.5$ and ${n_{air}} = 1$. We now substitute these values in equation ($3$) to obtain the unknown value which is the wavelength of light in glass. Hence:
$\dfrac{{1.5}}{1} = \dfrac{{5.5 \times {{10}^{ - 7}}}}{{{\lambda _g}}}$
Hence, by cross-multiplying the terms we get:
$1.5 \times {\lambda _g} = 5.5 \times {10^{ - 7}} \times 1$
By simplifying further to make ${\lambda _g}$ as the subject we get:
${\lambda _g} = \dfrac{{5.5 \times {{10}^{ - 7}}}}{{1.5}}$
Thus, we get: ${\lambda _g} = 3.67 \times {10^{ - 7}}m$
Thus, $3.67 \times {10^{ - 7}}m$ is the wavelength of light in the glass medium.
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. The wavelengths differ for different colors that are in the visible light spectrum as per the concept of EM waves as visible light is also said to be an example of EM waves. The red color was the highest wavelength which means that it is visible through a long distance while violet has the shortest wavelength.
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 speed of light, its corresponding wavelength and frequency and how these quantities vary when passed through another medium other than air. We know that visible white light comprises a band of seven different colors which is given by the term VIBGYOR that specifies each of the colors that it consists of. The light that reaches the Earth has its main source as the closest star to the Earth which is the sun. The light that gets reflected from the sun is often orange or sometimes red and the same way each of the planets in space seem to reflect light of a specific color.
Mercury is said to reflect green colored rays of light which is what is mentioned in the question as well. Since, the light rays are said to be of a single color this type of light is known as monochromatic light which is also said to have a single frequency. This light from mercury is said to have some wavelength which is already given. We are asked to calculate the frequency of the light in the air.
Light travels in a wave like motion and hence these waves that represent light will have a certain wavelength. 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 which is given as $5.5 \times {10^{ - 7}}m$. The speed of light in air is already known as it is a constant and given by the value \[3 \times {10^8}m/s\]. We are asked to find out the frequency of light in the air. Hence:
Given, \[c = 3 \times {10^8}m/s\] and $\lambda = 5.5 \times {10^{ - 7}}m$. 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:
\[f = \dfrac{c}{\lambda }\] ----------($1$)
This formula originates from the speed-distance formula wherein here, the speed of light in air is considered and represented and the distance of light is always measured in terms of its wavelength. The frequency is said to be the number of waves moving through the medium per unit time and hence represents the time quantity.
(a) In order to find the frequency of light in air we substitute the given values in equation ($1$) to get:
\[f = \dfrac{{3 \times {{10}^8}}}{{5.5 \times {{10}^{ - 7}}}}\]
On further simplification we get:
\[f = \dfrac{{3 \times {{10}^8} \times {{10}^7}}}{{5.5}}\]
$ \Rightarrow f = \dfrac{3}{{5.5}} \times {10^{15}}$
By solving this above equation we get:
\[f = 0.545 \times {10^{15}}Hz\]
We now shift the decimal place to the right and this reduces the power of $10$ by $1$. Hence we get:
\[f = 5.45 \times {10^{14}}Hz\]
This is the frequency of light in the air in Hertz. But however we are asked to find out the frequency in terms of the unit MegaHertz. Thus, we need to convert the frequency from Hertz to MegaHertz.
We know that,
$1Hz = {10^{ - 6}}MHz$
Hence, by unitary method we get:
\[f = 5.45 \times {10^{14}} \times {10^{ - 6}}MHz\]
We know that if the base numbers are the same then the powers are added and hence we get:
\[f = 5.45 \times {10^{14 - 6}}MHz\]
On solving further we get:
\[f = 5.45 \times {10^8}MHz\]
Therefore, $5.45 \times {10^8}MHz$ is the frequency of green light from mercury in air in megahertz.
Next, we are asked to find the time period of the light. Hence, we determine the formula for the time period in terms of the frequency. We know that these two quantities are inversely related to each other and hence we have the formula:
$T = \dfrac{1}{f}$ ----------($2$)
We have just determined the value for frequency above. By substituting this value of frequency in equation ($2$) we get:
$T = \dfrac{1}{{5.45 \times {{10}^{14}}}}s$
On solving further we have:
$T = \dfrac{1}{{5.45}} \times {10^{ - 14}}s$
$ \Rightarrow T = 0.1834 \times {10^{ - 14}}s$
However, we are asked to find the time period in terms of the unit microseconds and hence we need to convert the time period from seconds to microseconds.We know that,
$1s = {10^6}\mu s$
Hence, the above time period obtained in seconds can be converted by unitary method.Hence, we get:
$ \Rightarrow 0.1834 \times {10^{ - 14}}s = 0.1834 \times {10^{ - 14}} \times {10^6}\mu s$
On solving further we get:
$ \Rightarrow T = 0.1834 \times {10^{ - 14 + 6}}\mu s$
$ \Rightarrow T = 0.1834 \times {10^{ - 8}}\mu s$
$ \Rightarrow T = 1.834 \times {10^{ - 9}}\mu s$
Therefore, $1.834 \times {10^{ - 9}}\mu s$ is the time period of light from mercury in microseconds.
(b) Next, we are asked to find out the wavelength of light in a medium which is given to be glass, also given the refractive index of light in glass to be $1.5$. 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. Refractive index is the ratio between the speeds of light in a vacuum to that of a medium and hence given as the relation:
$n = \dfrac{\text{Speed of light in vacuum}}{\text{Speed of light in medium}}$
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 considered monochromatic light of single frequency as mentioned earlier.
Thus 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:
$\dfrac{{{n_g}}}{{{n_a}}} = \dfrac{{{\lambda _a}}}{{{\lambda _g}}}$ --------($3$)
The subscripts g and a represent glass and air respectively. The ratios of refractive indices of light of the medium of air and glass is related to the respective wavelengths in this way. 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. Hence:
Given, ${n_{glass}} = 1.5$ and ${n_{air}} = 1$. We now substitute these values in equation ($3$) to obtain the unknown value which is the wavelength of light in glass. Hence:
$\dfrac{{1.5}}{1} = \dfrac{{5.5 \times {{10}^{ - 7}}}}{{{\lambda _g}}}$
Hence, by cross-multiplying the terms we get:
$1.5 \times {\lambda _g} = 5.5 \times {10^{ - 7}} \times 1$
By simplifying further to make ${\lambda _g}$ as the subject we get:
${\lambda _g} = \dfrac{{5.5 \times {{10}^{ - 7}}}}{{1.5}}$
Thus, we get: ${\lambda _g} = 3.67 \times {10^{ - 7}}m$
Thus, $3.67 \times {10^{ - 7}}m$ is the wavelength of light in the glass medium.
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. The wavelengths differ for different colors that are in the visible light spectrum as per the concept of EM waves as visible light is also said to be an example of EM waves. The red color was the highest wavelength which means that it is visible through a long distance while violet has the shortest wavelength.
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