Answer
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Hint: To solve this question, recall what is the basic cause and meaning of refraction and refractive index respectively. Refraction happens due to the fact that the speed of light changes from medium to medium. Refraction index is nothing but the ratio of the speed of light in vacuum to the speed of light in the particular medium.
Complete step by step answer:
As we told in the hint section of the solution to this question, refraction happens due to the variance of speed in light when it changes the medium since the speed of light varies from medium to medium. In fact, refractive index is defined as the ratio of speed of light in vacuum to the speed of light in the particular medium. Mathematically, we can easily write that:
$\mu = \dfrac{c}{v}$
Where, $\mu $ is the refractive index of the medium,
$c$ is the speed of light in vacuum and,
$v$ is the speed of light in the medium.
As given in the question, for medium A, refractive index is ${\mu _A} = 1.52$
Using the above-mentioned equation, we can write:
${\mu _A} = \dfrac{c}{{{v_A}}}$
After, transposing and substituting ${\mu _A} = 1.52$ , we get:
$\Rightarrow {v_A} = \dfrac{c}{{1.52}}$
For medium B, refractive index is given as ${\mu _B} = 1.33$
Using the equation, we can write:
${\mu _B} = \dfrac{c}{{{v_B}}}$
After substituting and transposing, we get:
$\Rightarrow {v_B} = \dfrac{c}{{1.33}}$
For medium C, it is given as ${\mu _C} = 1.71$
Using the equation, we can write:
${\mu _C} = \dfrac{c}{{{v_C}}}$
After transposing and substituting, we get:
$\Rightarrow {v_C} = \dfrac{c}{{1.71}}$
Now, for medium D, it is given that ${\mu _D} = 2.42$
Using the equation, we can write:
${\mu _D} = \dfrac{c}{{{v_D}}}$
After transposing and substituting, we get:
$\Rightarrow {v_D} = \dfrac{c}{{2.42}}$
Now, we have found out the values of the velocities of all the mediums: A, B, C and D.
Let’s compare them now:
We can clearly see that the numerator part of all the velocities is exactly same and equal to the speed of light in vacuum $\left( c \right)$
As for the denominator part, we can see that the denominators of the velocities can be arranged as:
${D_D} > {D_C} > {D_A} > {D_B}$
Where, ${D_i}$ means the denominator of velocity ${v_i}$
Since, this is the order of the denominators, we can confidently say that:
${v_D} < {v_C} < {v_A} < {v_B}$
Hence, the speed of light in the medium B is the maximum, hence, option (B) is the correct answer to the question.
Note: A major error that students do is that instead of considering refractive index as the ratio of speed of light in vacuum to the speed of light in medium, they consider it to be the other way around and reach at an exact opposite answer with speed in medium B as the least and in D as the maximum. Always remember which part is the numerator part and which one is the denominator part.
Complete step by step answer:
As we told in the hint section of the solution to this question, refraction happens due to the variance of speed in light when it changes the medium since the speed of light varies from medium to medium. In fact, refractive index is defined as the ratio of speed of light in vacuum to the speed of light in the particular medium. Mathematically, we can easily write that:
$\mu = \dfrac{c}{v}$
Where, $\mu $ is the refractive index of the medium,
$c$ is the speed of light in vacuum and,
$v$ is the speed of light in the medium.
As given in the question, for medium A, refractive index is ${\mu _A} = 1.52$
Using the above-mentioned equation, we can write:
${\mu _A} = \dfrac{c}{{{v_A}}}$
After, transposing and substituting ${\mu _A} = 1.52$ , we get:
$\Rightarrow {v_A} = \dfrac{c}{{1.52}}$
For medium B, refractive index is given as ${\mu _B} = 1.33$
Using the equation, we can write:
${\mu _B} = \dfrac{c}{{{v_B}}}$
After substituting and transposing, we get:
$\Rightarrow {v_B} = \dfrac{c}{{1.33}}$
For medium C, it is given as ${\mu _C} = 1.71$
Using the equation, we can write:
${\mu _C} = \dfrac{c}{{{v_C}}}$
After transposing and substituting, we get:
$\Rightarrow {v_C} = \dfrac{c}{{1.71}}$
Now, for medium D, it is given that ${\mu _D} = 2.42$
Using the equation, we can write:
${\mu _D} = \dfrac{c}{{{v_D}}}$
After transposing and substituting, we get:
$\Rightarrow {v_D} = \dfrac{c}{{2.42}}$
Now, we have found out the values of the velocities of all the mediums: A, B, C and D.
Let’s compare them now:
We can clearly see that the numerator part of all the velocities is exactly same and equal to the speed of light in vacuum $\left( c \right)$
As for the denominator part, we can see that the denominators of the velocities can be arranged as:
${D_D} > {D_C} > {D_A} > {D_B}$
Where, ${D_i}$ means the denominator of velocity ${v_i}$
Since, this is the order of the denominators, we can confidently say that:
${v_D} < {v_C} < {v_A} < {v_B}$
Hence, the speed of light in the medium B is the maximum, hence, option (B) is the correct answer to the question.
Note: A major error that students do is that instead of considering refractive index as the ratio of speed of light in vacuum to the speed of light in medium, they consider it to be the other way around and reach at an exact opposite answer with speed in medium B as the least and in D as the maximum. Always remember which part is the numerator part and which one is the denominator part.
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