Question

The refractive indices of glycerin and diamond with respect to air are 1.4 and 2.4 respectively. Calculate the speed of light in glycerin and diamond. From these results, calculate the refractive index of diamond with respect to glycerin. $\left(c= 3 \times {10}^{8} {m}/{s}\right)$
$A. 2.143\times { 10 }^{ 8 }{ m }/{ s };1.250\times { 10 }^{ 8 }{ m }/{ s};1.714$
$B. 0.2143\times { 10 }^{ 8 }{ m }/{ s };12.50\times { 10 }^{ 5 }{ m }/{ s};0.1714$
$C. 2.143\times { 10 }^{ 6 }{ m }/{ s };1.250\times { 10 }^{ 3 }{ m }/{ s};1.714$
$D. 0.2143\times { 10 }^{ 8 }{ m }/{ s };12.50\times { 10 }^{ 5 }{ m }/{ s};17.14$

Answer 1

To solve this problem, use the formula for refractive index in terms of speed of light in vacuum and speed of light in a medium. Substitute the values in this mentioned equation and find the speed of light in glycerin. Similarly, by substituting the values in this formula find the speed of light in diamond. Now, take the ratio of refractive index of diamond to refractive index of diamond. This will give the refractive index of diamond with respect to glycerin.

Formula used:
$\mu =\dfrac { c }{ v }$

Complete answer:
Given: Refractive index of glycerin, ${ \mu }_{ 1 }=1.4$
Refractive index of diamond, ${ \mu }_{ 2 }=2.4$
$c=3\times { 10 }^{ 8 }{ m }/{ s }$
Refractive index is given by,
$\mu =\dfrac { c }{ v }$ …(1)
Where, v is the speed of light in a medium
c is the speed of light in vacuum
Let ${v}_{1}$ and ${v}_{2}$ be the speed of light in glycerin and diamond respectively.
Using equation. (1), refractive index of glycerin will be given by,
${\mu}_{1}= \dfrac {c}{{v}_{1}}$
Substituting values in the equation. (1) we get,
$1.4=\dfrac {3 \times {10}^{8}}{{v}_{1}}$
$\Rightarrow {v}_{1}=2.143 \times {10}^{8}{m}/{s}$
Similarly, refractive index of diamond will be given by,
${\mu}_{2}= \dfrac {c}{{v}_{2}}$
Substituting values in the equation. (1) we get,
$2.4=\dfrac {3 \times {10}^{8}}{{v}_{2}}$
$\Rightarrow {v}_{2}=1.25 \times {10}^{8}{m}/{s}$
The refractive index of diamond with respect to glycerin will be given by,
$\mu= \dfrac {{\mu}_{1}}{{\mu}_{2}}$
Substituting value in above equation we get,
$\mu= \dfrac {2.4}{1.4}$
$\Rightarrow \mu= 1.714$
Thus, the speed of light in glycerine and diamond are $2.143 \times {10}^{8}{m}/{s}$ and $1.25 \times {10}^{8}{m}/{s}$ respectively. The refractive index of diamond with respect to glycerin is 1.74.

So, the correct answer is option A i.e. $2.143\times { 10 }^{ 8 }{ m }/{ s };1.250\times { 10 }^{ 8 }{ m }/{ s};1.714$

Note:
Students must remember that the refractive index is a unitless quantity as it is a ratio of two quantities having same units. The speed of light in vacuum is faster as compared to in air. Students should know the typical refractive indices as sometimes those are not given in the question. Refractive index of water is 1.33, refractive index of air or vacuum is 1 and the refractive index of glass is 1.52.