
Why does red light refract the least?
Answer
487.2k+ views
Hint: We will firstly note the wavelengths of the seven colored lights of the spectrum of white light. Then, we will take into consideration the concept of wavelength and velocity of a wave and evaluate an equation which shows the variance of refraction with respect to the wavelength of a wave. Finally, we will answer this question.
Formulae Used:
$\mu {\text{ }} = {\text{ }}\dfrac{c}{v}$
Where, $\mu $ is the refractive index, $c$ is the speed of light in vacuum and $v$ is the speed of light in a medium.
$v{\text{ }} = {\text{ }}\lambda f$
Where, $v$ is the velocity of the wave, $\lambda $ is the wavelength of the wave and $f$ is the frequency of the wave.
Complete step by step solution:
The colours of the spectrum in order from bottom to top is violet, indigo, blue, green, yellow, orange, red and their wavelength are in order of ${\lambda _V} < {\lambda _I} < {\lambda _B} < {\lambda _G} < {\lambda _Y} < {\lambda _O} < {\lambda _R}$.
Now,
We know that the extent of refraction of a wave is determined by the refractive index or in other words the refractive index of the medium through which the wave is travelling depicts the extent of refraction the wave will suffer.
Also,
We know
Refractive Index, $\mu {\text{ }} = {\text{ }}\dfrac{c}{v}$
Now,
Since $c$ is the speed of light in vacuum or air which is a constant.
Thus, we can write,
\[\mu \propto \dfrac{1}{v}{\text{ }} - - - - - - - - - - - - {\text{ }}(i)\]
We also know,
$v = \lambda f$
Now,
Frequency of a wave is a constant which means $f$ is a constant.
Thus, we can write
$v \propto \lambda {\text{ }} - - - - - - - - {\text{ }}(ii)$
Now,
From $(i)$ and $(ii)$, we get
$\mu \propto \dfrac{1}{\lambda }$
Thus,
Higher the wavelength, lower the refraction the wave suffers.
Thus, from the order of wavelength of the spectrum as mentioned prior, we can clearly see the wavelength of red light is the highest out of all the seven colors.
Hence, the refraction of red light is the least.
Note: Students should not memorize this process but concentrate more on the approach as this is not a general relation. Students have to construct a relation on their own depending on the requirement of a given problem. Students often make mistakes while combining the proportionality relations into one. Thus, they have to be very careful while doing the same.
Formulae Used:
$\mu {\text{ }} = {\text{ }}\dfrac{c}{v}$
Where, $\mu $ is the refractive index, $c$ is the speed of light in vacuum and $v$ is the speed of light in a medium.
$v{\text{ }} = {\text{ }}\lambda f$
Where, $v$ is the velocity of the wave, $\lambda $ is the wavelength of the wave and $f$ is the frequency of the wave.
Complete step by step solution:
The colours of the spectrum in order from bottom to top is violet, indigo, blue, green, yellow, orange, red and their wavelength are in order of ${\lambda _V} < {\lambda _I} < {\lambda _B} < {\lambda _G} < {\lambda _Y} < {\lambda _O} < {\lambda _R}$.
Now,
We know that the extent of refraction of a wave is determined by the refractive index or in other words the refractive index of the medium through which the wave is travelling depicts the extent of refraction the wave will suffer.
Also,
We know
Refractive Index, $\mu {\text{ }} = {\text{ }}\dfrac{c}{v}$
Now,
Since $c$ is the speed of light in vacuum or air which is a constant.
Thus, we can write,
\[\mu \propto \dfrac{1}{v}{\text{ }} - - - - - - - - - - - - {\text{ }}(i)\]
We also know,
$v = \lambda f$
Now,
Frequency of a wave is a constant which means $f$ is a constant.
Thus, we can write
$v \propto \lambda {\text{ }} - - - - - - - - {\text{ }}(ii)$
Now,
From $(i)$ and $(ii)$, we get
$\mu \propto \dfrac{1}{\lambda }$
Thus,
Higher the wavelength, lower the refraction the wave suffers.
Thus, from the order of wavelength of the spectrum as mentioned prior, we can clearly see the wavelength of red light is the highest out of all the seven colors.
Hence, the refraction of red light is the least.
Note: Students should not memorize this process but concentrate more on the approach as this is not a general relation. Students have to construct a relation on their own depending on the requirement of a given problem. Students often make mistakes while combining the proportionality relations into one. Thus, they have to be very careful while doing the same.
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