Answer
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Hint: Use Snell’s laws of refraction. The ratio of the sin of the incident angle and the refraction angle gives the refractive index of the medium. We can also express it in terms of velocity of light in the media. Obtain that mathematical expression. Put the velocity of light in the two media in the mathematical formula to find your answer.
Complete step by step answer:
Snell’s law describes that the ratio of sine of incident angle to the sign of refracted angle is a constant for the given media.
$\dfrac{\sin i}{\sin r}={{n}_{21}}$
Where, i is the angle of incidence and r is the angle of refraction.
${{n}_{21}}$ is a constant and is called the refractive index of the second medium with respect to the first medium.
Refractive index can be defined as the ratio of velocity of light in the two media through the interface of which light is passed.
${{n}_{12}}=\dfrac{1}{{{n}_{21}}}$
Where, ${{n}_{12}}$ is the refractive index of the first medium with respect to the second medium.
Relative refractive index can be defined as the ratio of velocity of light in the given media with respect to the velocity of light in vacuum. We can write
$n=\dfrac{c}{v}$
Now, let the velocity of light in the rarer medium is v.
The velocity of light in denser medium will decrease by $30\%$ of the velocity at the rarer medium.
Let this velocity will be,
$\begin{align}
& {v}'=v-30\%\text{ of }v \\
& {v}'=v-\dfrac{30}{100}\times v \\
& {v}'=v-\dfrac{3v}{10} \\
& {v}'=\dfrac{7v}{10} \\
\end{align}$
So, the refractive index of the denser medium with respect to the rarer medium will be,
$\begin{align}
& n=\dfrac{v}{{{v}'}} \\
& n=\dfrac{v}{\dfrac{7v}{10}} \\
& n=\dfrac{10}{7}=1.428 \\
\end{align}$
The correct option is (D)
Note: We can find the refractive index with respect to vacuum by directly dividing the velocity of light in vacuum by the velocity of light in the medium. This is also called the relative refractive index. Mathematically, $n=\dfrac{c}{v}$
Complete step by step answer:
Snell’s law describes that the ratio of sine of incident angle to the sign of refracted angle is a constant for the given media.
$\dfrac{\sin i}{\sin r}={{n}_{21}}$
Where, i is the angle of incidence and r is the angle of refraction.
${{n}_{21}}$ is a constant and is called the refractive index of the second medium with respect to the first medium.
Refractive index can be defined as the ratio of velocity of light in the two media through the interface of which light is passed.
${{n}_{12}}=\dfrac{1}{{{n}_{21}}}$
Where, ${{n}_{12}}$ is the refractive index of the first medium with respect to the second medium.
Relative refractive index can be defined as the ratio of velocity of light in the given media with respect to the velocity of light in vacuum. We can write
$n=\dfrac{c}{v}$
Now, let the velocity of light in the rarer medium is v.
The velocity of light in denser medium will decrease by $30\%$ of the velocity at the rarer medium.
Let this velocity will be,
$\begin{align}
& {v}'=v-30\%\text{ of }v \\
& {v}'=v-\dfrac{30}{100}\times v \\
& {v}'=v-\dfrac{3v}{10} \\
& {v}'=\dfrac{7v}{10} \\
\end{align}$
So, the refractive index of the denser medium with respect to the rarer medium will be,
$\begin{align}
& n=\dfrac{v}{{{v}'}} \\
& n=\dfrac{v}{\dfrac{7v}{10}} \\
& n=\dfrac{10}{7}=1.428 \\
\end{align}$
The correct option is (D)
Note: We can find the refractive index with respect to vacuum by directly dividing the velocity of light in vacuum by the velocity of light in the medium. This is also called the relative refractive index. Mathematically, $n=\dfrac{c}{v}$
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