Question

The refractive index of diamond is 2.0, the velocity of light in diamond in centimeter per second is approximate?
A. $2.0 \times {10^{10}}$
B. $6 \times {10^{10}}$
C. $1.5 \times {10^{10}}$
D. $3 \times {10^{10}}$

Answer 1

Hint: First, write the known or given values in the question. Then write the known formula for the refractive index that is $\eta = \dfrac{c}{v}$ and then substitute the given values of $c = 3 \times {10^8}$ and $\eta = 2$ in the equation. After that solve for the value and convert m/s to cm/s by multiplying the result by${10^2}$.

Complete step by step answer:
 We know that the velocity of light in a vacuum is equal to $c = 3 \times {10^8}$ m/s.
Given that the value of the refractive index of Diamond is equal to $\eta = 2$
We know that the refractive index $\eta $ of any medium is defined as the ratio of the speed of light in vacuum$c$, and the phase velocity $v$ of light in the medium that is
$\eta = \dfrac{c}{v}$ ------------------------------------ (1)
Now taking $v$ to the Left hand side and taking $\eta $ to the right hand side we can write equation (1) as
\[v = \dfrac{c}{\eta }\] ------------------------------------ (2)
Now substituting the values of $c$ and $\eta $ in the equation (2), we will get
$v = \dfrac{{3 \times {{10}^8}}}{2}$--------------------------------- (3)
Now after solving equation (3) we will get
$v = 1.5 \times {10^8}$ m/s.
But in the question, it is mentioned that we need to write the answer in cm/s. So to convert meter to centimeter we need to multiply the value with ${10^2}$, we get
$v = 1.5 \times {10^8}$m/s $ = 1.5 \times {10^8} \times {10^2}$cm/s $ = 1.5 \times {10^{10}}$ cm/s
So the velocity of light in diamond in centimeter per second is $ = 1.5 \times {10^{10}}$ cm/s and hence option C is correct.

Note: For these types of questions, we first need to recall the formula that is to be used in the question. Then we will write the value of the variable given in the formula, after that we will substitute the values and solve for the result. Then we do the conversion of the unit if needed for the question.