
The velocity of light in ice is \[2.3\times {{10}^{8}}m{{s}^{-1}}\]. Its refractive index is –
A) 1.512
B) 1.414
C) 1.305
D) 1.23
Answer
457.2k+ views
Hint: We need to understand the relation between the refractive index of a medium and the velocity with which the light travels through it. The ratio between a standard refractive index, i.e., in air will help us find the required refractive index of ice.
Complete Step-by-Step Solution:
We are given that the light passes through ice at a speed of \[2.3\times {{10}^{8}}m{{s}^{-1}}\]. We can find the refractive index of ice using this data on the speed of light through the ice.
The refractive index is the relative measure of the velocities with which light can travel through different media. The absolute refractive index is the ratio of the velocity of the light in vacuum or air to the velocity of the light through the given media.
We often use a relative refractive index which is given by the ratio between the velocity of light in the first medium to the velocity of light in the second medium. It is equal to the ratio of the absolute refractive index of the second medium to the absolute refractive index of the first medium.
If n is the refractive index and v is the velocity of the light in a medium, then for the medium ice and air, we can find the absolute refractive index of the ice as –
\[\begin{align}
& \dfrac{{{n}_{ice}}}{{{n}_{air}}}=\dfrac{{{v}_{air}}}{{{v}_{ice}}} \\
& \text{but,} \\
& {{v}_{air}}=c=3\times {{10}^{8}}m{{s}^{-1}}, \\
& {{v}_{ice}}=2.3\times {{10}^{8}}m{{s}^{-1}}, \\
& {{n}_{air}}=1 \\
& \Rightarrow {{n}_{ice}}=\dfrac{c}{{{v}_{ice}}}\times {{n}_{air}} \\
& \Rightarrow {{n}_{ice}}=\dfrac{3\times {{10}^{8}}m{{s}^{-1}}}{2.3\times {{10}^{8}}m{{s}^{-1}}}\times 1 \\
& \therefore {{n}_{ice}}=1.305 \\
\end{align}\]
The refractive index of ice is 1.305.
Henec, the correct answer is option C.
Note:
The refractive index is a relative measure of how fast light travels through a medium or how much deviation is affected by the light ray when moving from a particular medium to another when incident at an angle which is given explicitly by the Snell’s law.
Complete Step-by-Step Solution:
We are given that the light passes through ice at a speed of \[2.3\times {{10}^{8}}m{{s}^{-1}}\]. We can find the refractive index of ice using this data on the speed of light through the ice.
The refractive index is the relative measure of the velocities with which light can travel through different media. The absolute refractive index is the ratio of the velocity of the light in vacuum or air to the velocity of the light through the given media.
We often use a relative refractive index which is given by the ratio between the velocity of light in the first medium to the velocity of light in the second medium. It is equal to the ratio of the absolute refractive index of the second medium to the absolute refractive index of the first medium.
If n is the refractive index and v is the velocity of the light in a medium, then for the medium ice and air, we can find the absolute refractive index of the ice as –
\[\begin{align}
& \dfrac{{{n}_{ice}}}{{{n}_{air}}}=\dfrac{{{v}_{air}}}{{{v}_{ice}}} \\
& \text{but,} \\
& {{v}_{air}}=c=3\times {{10}^{8}}m{{s}^{-1}}, \\
& {{v}_{ice}}=2.3\times {{10}^{8}}m{{s}^{-1}}, \\
& {{n}_{air}}=1 \\
& \Rightarrow {{n}_{ice}}=\dfrac{c}{{{v}_{ice}}}\times {{n}_{air}} \\
& \Rightarrow {{n}_{ice}}=\dfrac{3\times {{10}^{8}}m{{s}^{-1}}}{2.3\times {{10}^{8}}m{{s}^{-1}}}\times 1 \\
& \therefore {{n}_{ice}}=1.305 \\
\end{align}\]
The refractive index of ice is 1.305.
Henec, the correct answer is option C.
Note:
The refractive index is a relative measure of how fast light travels through a medium or how much deviation is affected by the light ray when moving from a particular medium to another when incident at an angle which is given explicitly by the Snell’s law.
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