Question

It is observed that the refractive index of diamond with respect to glass is 1.62 and the refractive index of glass with respect to air is 1.50. Find the refractive index of diamond with respect to air.

Answer 1

We can use Snell’s law here to solve this problem. Refractive index can be given as $\mu =\dfrac{\sin i}{\sin r}$
while using Snell’s law the refractive index is given by the ratio of refractive of the refractive index of the second medium to the refractive index of first medium

Complete step by step answer: The refractive index of diamond with respect to glass is 1.62.
$_{d}{{\mu }_{g}}=\dfrac{\sin i}{\sin r}$
$_{g}{{\mu }_{d}}=\dfrac{{{\mu }_{d}}}{{{\mu }_{g}}}=1.62$--(1)
The refractive index of glass with respect to air is 1.50.
$_{a}{{\mu }_{g}}=\dfrac{{{\mu }_{g}}}{{{\mu }_{a}}}=1.50$---(2)
Now, we need to find the refractive index of diamond with respect to air. So, multiplying eq (1) with the eq (2)

& \dfrac{{{\mu }_{d}}}{{{\mu }_{g}}}\times \dfrac{{{\mu }_{g}}}{{{\mu }_{a}}}=1.50\times 1.62 \\\ & \dfrac{{{\mu }_{d}}}{{{\mu }_{a}}}=2.43 \\\ \end{aligned}$$ $${{\Rightarrow }_{a}}{{\mu }_{d}}=2.43$$ **Thus, the correct answer is 2.43** **Additional information-** In refraction of light there always takes place bending of light. The bending of light takes place around normal. If light travels from one medium to another medium, then there is a change in wavelength of light. Reflection and refraction sound the same but they are two different phenomena. Refractive index is an important parameter used while designing lenses **Note:** While using Snell’s law the refractive index is given by the ratio of refractive of the refractive index of the second medium to the refractive index of first medium. Be careful while the angles of reflection, refraction and of incidence are always made with the normal.