Answer
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Hint Refraction will take place at the boundary between two media irrespective of wavelength of the light incident on it. The Snell’s law of refraction determines the condition at which refraction takes place.
In this solution we will be using the following formula;
$\Rightarrow {n_1}\sin {\theta _1} = {n_2}\sin {\theta _2}$, where ${n_1}$ is the index of refraction of the incident medium, ${\theta _1}$ is the angle of incidence of the light ray, ${n_2}$ is the index of refraction of the incident medium, and ${\theta _2}$ is the angle of refraction of the refractive medium.
Complete step by step answer
Refraction, as popularly defined, is the bending of light rays as it crosses the boundary between two media. This bending of light rays is always accompanied by an increase or decrease of velocity depending on whether it is moving to a denser medium or a rarer medium. Naturally when the light rays are moving from a rarer medium to a denser medium, the light velocity decreases, when it moves from a denser medium to a rarer medium, light velocity increases.
In general, for refraction to occur, both mediums have to be permissive of light, that is to say they must allow light to pass through them without absorbing or reflecting it away.
To answer the question, let’s investigate Snell’s law which state mathematically that${n_1}\sin {\theta _1} = {n_2}\sin {\theta _2}$where ${n_1}$ is the index of refraction of the incident medium, ${\theta _1}$ is the angle of incidence of the light ray, ${n_2}$ is the index of refraction of the incident medium, and ${\theta _2}$ is the angle of refraction of the refractive medium.
Hence, we see from the equation that if the refractive indices of the two media are equal i.e. ${n_1} = {n_2}$, then $\sin {\theta _1} = \sin {\theta _2}$, hence ${\theta _1} = {\theta _2}$ which is interprets no refraction since there is no bending of the rays. This eliminates option B
Also, ${\theta _1} = 0$ then
$\Rightarrow {n_2}\sin {\theta _2} = 0$
$\Rightarrow {\theta _2} = 0$
Hence, ${\theta _1} = {\theta _2} = 0$ which still implies no refraction, eliminating option C
However, nothing in Snell’s law or any law, confirms that a refracted light must be of one wavelength.
Hence, we can conclude that “the light must be of one wavelength” is thus a false statement.
Thus, option D is the correct answer.
Note
For clarity, note that when we speak of dense and rare medium, we speak of optical density and not the mass density of the substance. Higher optical density, higher refractive index (dense medium) and lower optical density, lower refractive index hence a rare medium.
In this solution we will be using the following formula;
$\Rightarrow {n_1}\sin {\theta _1} = {n_2}\sin {\theta _2}$, where ${n_1}$ is the index of refraction of the incident medium, ${\theta _1}$ is the angle of incidence of the light ray, ${n_2}$ is the index of refraction of the incident medium, and ${\theta _2}$ is the angle of refraction of the refractive medium.
Complete step by step answer
Refraction, as popularly defined, is the bending of light rays as it crosses the boundary between two media. This bending of light rays is always accompanied by an increase or decrease of velocity depending on whether it is moving to a denser medium or a rarer medium. Naturally when the light rays are moving from a rarer medium to a denser medium, the light velocity decreases, when it moves from a denser medium to a rarer medium, light velocity increases.
In general, for refraction to occur, both mediums have to be permissive of light, that is to say they must allow light to pass through them without absorbing or reflecting it away.
To answer the question, let’s investigate Snell’s law which state mathematically that${n_1}\sin {\theta _1} = {n_2}\sin {\theta _2}$where ${n_1}$ is the index of refraction of the incident medium, ${\theta _1}$ is the angle of incidence of the light ray, ${n_2}$ is the index of refraction of the incident medium, and ${\theta _2}$ is the angle of refraction of the refractive medium.
Hence, we see from the equation that if the refractive indices of the two media are equal i.e. ${n_1} = {n_2}$, then $\sin {\theta _1} = \sin {\theta _2}$, hence ${\theta _1} = {\theta _2}$ which is interprets no refraction since there is no bending of the rays. This eliminates option B
Also, ${\theta _1} = 0$ then
$\Rightarrow {n_2}\sin {\theta _2} = 0$
$\Rightarrow {\theta _2} = 0$
Hence, ${\theta _1} = {\theta _2} = 0$ which still implies no refraction, eliminating option C
However, nothing in Snell’s law or any law, confirms that a refracted light must be of one wavelength.
Hence, we can conclude that “the light must be of one wavelength” is thus a false statement.
Thus, option D is the correct answer.
Note
For clarity, note that when we speak of dense and rare medium, we speak of optical density and not the mass density of the substance. Higher optical density, higher refractive index (dense medium) and lower optical density, lower refractive index hence a rare medium.
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