Question

calculate the Refractive index of diamond w.r.t water. if refractive index of water w. r.t. to air is 4/3 and Diamond w.r.t. air is 2.42.​

Answer 1

• 1.81 is the required refractive index.

Given:-

• Refractive Index of with respect to air ,μwa = 4/3= 1.33
• Refractive index of diamond ,μd = 2.42

To Find :-

• Refractive index of diamond with respect to water

Solution:-

As we have to calculate the refractive index of diamond with respect to water.

It is given that the refractive index of water with respect to air is 1.33 and the refractive index of diamond is 2.42 .

So ,

refractive index is calculated by using the formula.

• Refractive index of Diamond with respect to water = Refractive index of Diamond/Refractive index of water with respect to air.

→ μdg = μd/μwa

where,

• μdg denote refractive index of diamond w.r.t air
• μwa denote refractive index of water w.r.t air.
• μd denote refractive index of diamond

Substitute the value we get

→ μdg = 2.42/1.33

→ μdg = 1.81

• Hence, the refractive index of diamond with respect to air is 1.81 .

Answer 2

$\begin{gathered} \begin{gathered}\begin{gathered}\\\;\underbrace{\underline{\sf{Given \; Question:-}}}\end{gathered}\end{gathered} \end{gathered}$

calculate the Refractive index of diamond with respect to water. if refractive index of water with respect to air is 4/3 and Diamond with respect to air is 2.42.

Let's do it !!

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★ Solution :-

Given ,

• Refractive index of water with respect to air = 4/3 = 1.33

$\begin{gathered}\\\; \Longrightarrow{\sf{{ \blue{ \frac{ \mu \: _w}{ \mu _a} = 1.33} }}}\end{gathered}$

• Refractive index of Diamond with respect to air = 2.42

$\begin{gathered}\\\; \Longrightarrow{\sf{{ \blue{ \frac{ \mu \: _d}{ \mu _a} = 2.42} }}}\end{gathered}$

To find ,

• The Refractive index of diamond with respect to water

Now for finding the Refractive index of diamond with respect to water we need to divide Refractive index of Diamond with respect to air and Refractive index of water with respect to air .

$\begin{gathered}\\\; \Longrightarrow{\sf{{{_w \mu_d} = \blue{ \frac{ \mu_d}{\mu_w} } = \green{ \dfrac{ \dfrac{ \mu_d}{\mu_a} }{ \dfrac{\mu_w}{\mu_a} } } }}}\end{gathered}$

Now it is given to us that , $\begin{gathered}\\ \ {\sf{{ { \frac{ \mu \: _d}{ \mu _a} = 2.42 \: \: \: and \: \: \:\frac{ \mu \: _w}{ \mu _a} = 1.33 } }}}\end{gathered}$

Now after putting the values we get

$\begin{gathered}\\\; \Longrightarrow{\sf{{{_w \mu_d} = \green{ \dfrac{ 2.42}{ 1.33} } }}}\end{gathered}$

$\begin{gathered}\\\; \Longrightarrow{\sf{{{_w \mu_d} = { \dfrac{ 2.42}{ 1.33} = \green{1.819}} }}}\end{gathered}$

$\: \: \boxed{\boxed{\bf{\mapsto \: \: \: hence ,Refractive \: index \: of \: diamond \: with \: respect \: t o \: water = \blue{\underline{1.819}}}}}$