
A boat has green light of wavelength of \[500nm\]on the mast. What wavelength would be measured and what colour would be observed for this light as seen by a diver submerged in water by the side of the boat? Given $\mu_{w}=\dfrac{4}{3}$
\[\begin{align}
& \text{A}\text{. green of wavelength 500nm} \\
& \text{B}\text{. blue of wavelength 376nm} \\
& \text{C}\text{. green of wavelength 376nm} \\
& \text{D}\text{. red of wavelength 665nm} \\
\end{align}\]
Answer
553.5k+ views
Hint: We know from refraction of light that light bends when it travels from one medium to another. From Snell’s law we also know that the angle of the refracted light also depends on the medium i.e. the refractive index of the medium.
Formula:
$\mu_{a}sin (i)=\mu_{w} sin (r)$
Complete answer:
From Snell’s law, we can say that $\mu_{a}sin (i)=\mu_{w} sin (r)$ where $i$ is the angle of incidence of the light ray from the air medium whose refractive index is given as $\mu_{a}$, and $r$ is the angle of refraction of the light ray at the water medium whose refractive index is given as $\mu_{w}$.
Refractive index is the measure of how much a light bends when it passes from one medium to another. The bending of the light depends on the medium if it passes through. Refractive index is a dimensionless number which has no units. And it is usually a constant for a given medium.
Refractive index is also given as $\mu_{w}=\dfrac{c}{v}$, where, $c$ is the speed of light in air and $v$ is the speed of light in medium, here water.
We know that $\nu\lambda=c$ where $\nu$ is the frequency and $\lambda$ is the wavelength.
Since frequency of light doesn’t vary during refraction. Then we can write, $\mu_{w}=\dfrac{\lambda_{i}}{\lambda_{o}}$, where $\lambda_{i}$ is the incident wavelength and $\lambda_{o}$ is the observed wavelength.
Given $\lambda_{i}=500nm$ and $\mu_{w}=\dfrac{4}{3}$ then the observed wavelength is given as $\lambda_{o}=\dfrac{\lambda_{i}}{\mu_{w}}=\dfrac{500}{\dfrac{4}{3}}=376nm$
Thus the answer is \[\text{C}\text{. green of wavelength 376nm}\]
Note:
The refractive index of a material is a property of a material. It depends on the ratio the angle of incident ray to the angle of the refracted ray. It is also taken as the ratio of the speed of the light in medium to the speed of light in vacuum. It is also called the absolute refractive index, if one of the two mediums is air.
Formula:
$\mu_{a}sin (i)=\mu_{w} sin (r)$
Complete answer:
From Snell’s law, we can say that $\mu_{a}sin (i)=\mu_{w} sin (r)$ where $i$ is the angle of incidence of the light ray from the air medium whose refractive index is given as $\mu_{a}$, and $r$ is the angle of refraction of the light ray at the water medium whose refractive index is given as $\mu_{w}$.
Refractive index is the measure of how much a light bends when it passes from one medium to another. The bending of the light depends on the medium if it passes through. Refractive index is a dimensionless number which has no units. And it is usually a constant for a given medium.
Refractive index is also given as $\mu_{w}=\dfrac{c}{v}$, where, $c$ is the speed of light in air and $v$ is the speed of light in medium, here water.
We know that $\nu\lambda=c$ where $\nu$ is the frequency and $\lambda$ is the wavelength.
Since frequency of light doesn’t vary during refraction. Then we can write, $\mu_{w}=\dfrac{\lambda_{i}}{\lambda_{o}}$, where $\lambda_{i}$ is the incident wavelength and $\lambda_{o}$ is the observed wavelength.
Given $\lambda_{i}=500nm$ and $\mu_{w}=\dfrac{4}{3}$ then the observed wavelength is given as $\lambda_{o}=\dfrac{\lambda_{i}}{\mu_{w}}=\dfrac{500}{\dfrac{4}{3}}=376nm$
Thus the answer is \[\text{C}\text{. green of wavelength 376nm}\]
Note:
The refractive index of a material is a property of a material. It depends on the ratio the angle of incident ray to the angle of the refracted ray. It is also taken as the ratio of the speed of the light in medium to the speed of light in vacuum. It is also called the absolute refractive index, if one of the two mediums is air.
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