Question

A monochromatic beam of light is travelling from medium $A$ of refractive index $n_1$ to a medium $B$ of refractive index $n_2$. In the medium $A$, there are $x$ number of waves in certain distance. In the medium $B$, there are $y$ number of waves in the same distance. Then, refractive index of medium $A$ with respect to medium $B$ is

• A. $\frac {y}{x}$
• B. $\sqrt{\frac{x}{y}}$
• C. $ \frac{x}{y-x}$
• D. $ \frac{x}{y}$

Answer 1

The Correct Option is D

If x number of waves of wavelength $\lambda_{1}$ and y number of waves of wavelength $\lambda_{2}$ are present in same distance s, then
$s=x \lambda_{1}=y \lambda_{2}$
or $\frac{x}{y}=\frac{\lambda_{2}}{\lambda_{1}}=\frac{n_{1}}{n_{2}} $
$\frac{n_{1}}{n_{2}}=\frac{x}{y}$

Answer 2

Step 1: Expressing the wavelength of the beam of light in the mediums A and B.

It is already given that medium A has x number of waves and medium B has y number of waves in the same distance which is that of in A.

So let the d be that distance in medium A and medium B. So, then the wavelength of light in the medium A will be \(\lambda _{A}=\frac{d}{x}\)------(1)

Alike, we have the wavelength of light in the medium B as \(\lambda _{B}=\frac{d}{y}\)------(2)

So, now it is given that n1 and n2 are the refractive indices of the mediums A and B respectively.

 Step 2: Expressing the refractive indices of the two mediums A and B.

The refractive index of medium A can be expressed as: \(n _{A}=\frac{\lambda _{0}}{\lambda _{A}}\); λ₀ is the wavelength of light in the vacuum and it is a constant.

⇒ \(n _{A}\propto\frac{1}{\lambda _{A}}\)-------(3)

So, now the refractive index of medium B can be expressed as \(n _{B}=\frac{\lambda _{0}}{\lambda _{B}}\).

⇒ \(n _{B}\propto\frac{1}{\lambda _{B}}\)-------(4)

 Step 3: Using equations (3) and (4) we can obtain the relative refractive index of medium A with that of medium B.

So the refractive index of medium A relative to that of B can be expressed as: \(n_{AB}=\frac{n_{A}}{n_{B}}\)-------(5)

So, now using the equation (3), (4) and equation (5) becomes, \(n _{AB}=\frac{\lambda _{B}}{\lambda _{A}}\)-------(6)

After substituting equations (1) and (2) in equation (6) we get, 

\(n_{AB}=\frac{(\frac{d}{y})}{(\frac{d}{x})}\)

⇒ \(n_{AB}=\frac{x}{y}\)

Therefore, the refractive index of medium A with respect to that of medium B is \(n_{AB}=\frac{x}{y}\)

Therefore, the correct option is ‘D’.