
When a ray of light passes from air to glass, for what angle of incidence, the ray will not be deviated?
Answer
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Hint: Refraction is the process of bending of light. When light travels from one medium to another medium, it bends either towards the normal or away from the normal, depending upon the mediums. The angle of refraction or the amount of bending of light depends on the refractive index $\mu $ of the two media.
Formulae used:
Snell’s law $\dfrac{{\sin i}}{{\sin r}} = \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{{\mu _2}}}{{{\mu _1}}}$ , where ${v_1}$ and ${v_2}$ are the speeds of light in medium $1$ and medium $2$ respectively, ${\mu _1}$ and ${\mu _2}$ are the refractive indices of medium $1$ and $2$ respectively and $i$ is the angle of incidence and $r$ is the angle of refraction.
Complete step by step solution:
When light passes through an optically rarer medium (low refractive index), like air to an optically denser medium (high refractive index), like water, its speed decreases and it bends towards the normal, making the angle of refraction less than the angle of incidence.
For no deviation, we need $i = r$
From Snell’s law ,
$\dfrac{{\sin i}}{{\sin r}} = \dfrac{{{\mu _2}}}{{{\mu _1}}}$
$ \Rightarrow \dfrac{{\sin i}}{{\sin r}} = {\mu _1}^2$ here, ${\mu _1}^2$ is not square of the refractive index, but a way of writing the refractive index of medium 2 with respect to medium 1
$ \Rightarrow \sin i = {\mu _1}^2\sin r$
Since the refractive index cannot be zero here, then $i$ must be zero for no deviation. ( if angle of incidence becomes zero, then ray will pass through normal. Therefore, it can no longer bend towards the normal as it is already travelling through the normal)
Therefore, for angle of incidence $i = 0^\circ $ , the ray of light suffers no deviation.
Note: Refracted ray will not be deviated at angle of incidence \[0^\circ \] and \[90^\circ \] i.e. when the incident ray is passing through normal or is parallel to the reflecting surface. We ignore the value of $90^\circ $ in the above question. This is because the question has asked for the ray of light to pass through the air-glass intersection.
Formulae used:
Snell’s law $\dfrac{{\sin i}}{{\sin r}} = \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{{\mu _2}}}{{{\mu _1}}}$ , where ${v_1}$ and ${v_2}$ are the speeds of light in medium $1$ and medium $2$ respectively, ${\mu _1}$ and ${\mu _2}$ are the refractive indices of medium $1$ and $2$ respectively and $i$ is the angle of incidence and $r$ is the angle of refraction.
Complete step by step solution:
When light passes through an optically rarer medium (low refractive index), like air to an optically denser medium (high refractive index), like water, its speed decreases and it bends towards the normal, making the angle of refraction less than the angle of incidence.
For no deviation, we need $i = r$
From Snell’s law ,
$\dfrac{{\sin i}}{{\sin r}} = \dfrac{{{\mu _2}}}{{{\mu _1}}}$
$ \Rightarrow \dfrac{{\sin i}}{{\sin r}} = {\mu _1}^2$ here, ${\mu _1}^2$ is not square of the refractive index, but a way of writing the refractive index of medium 2 with respect to medium 1
$ \Rightarrow \sin i = {\mu _1}^2\sin r$
Since the refractive index cannot be zero here, then $i$ must be zero for no deviation. ( if angle of incidence becomes zero, then ray will pass through normal. Therefore, it can no longer bend towards the normal as it is already travelling through the normal)
Therefore, for angle of incidence $i = 0^\circ $ , the ray of light suffers no deviation.
Note: Refracted ray will not be deviated at angle of incidence \[0^\circ \] and \[90^\circ \] i.e. when the incident ray is passing through normal or is parallel to the reflecting surface. We ignore the value of $90^\circ $ in the above question. This is because the question has asked for the ray of light to pass through the air-glass intersection.
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