
Orange light has a wavelength of $6\times {{10}^{-7}}\text{ m}$. What is its frequency? The speed of light is $3\times {{10}^{8}}\text{ m}{{\text{s}}^{-1}}$
a.) $2\times {{10}^{15}}Hz$
b.) $2\times {{10}^{-15}}Hz$
c.) $5\times {{10}^{14}}Hz$
d.) $5\times {{10}^{-14}}Hz$
e.) $2\times {{10}^{14}}Hz$
Answer
533.4k+ views
Hint: Light is an electromagnetic wave, what we see is a form of Visible light and that is just a small part of the Electromagnetic spectrum. Visible light is also known as VIBGYOR where each of the alphabet denotes the color of the light such as “R” stands for Red and “V” stands for violet and “G” stands for the green color.
Complete step-by-step solution:
The relationship between the wavelength and the frequency of the light can be given by
$\lambda =\dfrac{c}{v}$ ………………… (1)
Where $\lambda $ is the wavelength of the light.
“c” denotes the speed of the light in vacuum.
And $v$ denotes the frequency of light
Now, In the question, we have been given that
$\lambda $ = $6\times {{10}^{-7}}\text{ m}$
C = $3\times {{10}^{8}}\text{ m}{{\text{s}}^{-1}}$
Putting the values in equation (1), we get
$\lambda =\dfrac{c}{v}$
$6\times {{10}^{-7}}=\dfrac{3\times {{10}^{8}}}{v}$
$v=5\times {{10}^{14}}Hz$
Therefore, we can conclude that the frequency of light is $5\times {{10}^{14}}Hz$
Hence, we can say that the option (c) $v=5\times {{10}^{14}}Hz$ is the correct answer for this question.
Note: Electromagnetic waves are distinguished by the type of their wavelength and frequency. Wavelength and frequency both are inversely proportional to each other and hence, a light of higher wavelength will have a low frequency and a wave of low wavelength will have higher frequency. And hence, we can say that the red color will have higher wavelength but low frequency and the violet color will have high frequency but low wavelength. We can use the VIBGYOR pattern to understand and remember about which color will come at what position, but this only works for visible spectrum only. After Reed color we will have infra-red and before violet color we will have ultra violet.
Complete step-by-step solution:
The relationship between the wavelength and the frequency of the light can be given by
$\lambda =\dfrac{c}{v}$ ………………… (1)
Where $\lambda $ is the wavelength of the light.
“c” denotes the speed of the light in vacuum.
And $v$ denotes the frequency of light
Now, In the question, we have been given that
$\lambda $ = $6\times {{10}^{-7}}\text{ m}$
C = $3\times {{10}^{8}}\text{ m}{{\text{s}}^{-1}}$
Putting the values in equation (1), we get
$\lambda =\dfrac{c}{v}$
$6\times {{10}^{-7}}=\dfrac{3\times {{10}^{8}}}{v}$
$v=5\times {{10}^{14}}Hz$
Therefore, we can conclude that the frequency of light is $5\times {{10}^{14}}Hz$
Hence, we can say that the option (c) $v=5\times {{10}^{14}}Hz$ is the correct answer for this question.
Note: Electromagnetic waves are distinguished by the type of their wavelength and frequency. Wavelength and frequency both are inversely proportional to each other and hence, a light of higher wavelength will have a low frequency and a wave of low wavelength will have higher frequency. And hence, we can say that the red color will have higher wavelength but low frequency and the violet color will have high frequency but low wavelength. We can use the VIBGYOR pattern to understand and remember about which color will come at what position, but this only works for visible spectrum only. After Reed color we will have infra-red and before violet color we will have ultra violet.
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