Chapter 24, Problem 48P

To determine

The refractive index of the liquid.

Answer to Problem 48P

  $1.39$ .

Explanation of Solution

Given:

Initial diameter, $r_1=2.92cm$

Diameter after decrement is $r_2=2.48cm$

Formula used:

It is known that: $\text{n=}\frac{{\text{λ}}_{\text{1}}}{{\text{λ}}_{\text{2}}}\text{=}{\left(\frac{{\text{r}}_{\text{1}}}{{\text{r}}_{\text{2}}}\right)}^{\text{2}}$

Where,

  $r$ is diameter.

  $\text{λ}$ is wavelength of the reflected light.

  $n$ is refractive index.

Calculation:

Consider the figure shown below.

  

The thickness of air space is y at a distance $r$ from the centre from the lens. So, the phase changes and the reflection at the bottom surface is $\varphi =\left(\frac{2y}{\lambda }\right)2\pi +\pi$ .

For the given dark rings:

  $\varphi =\left(\frac{2y}{\lambda }\right)2\pi +\pi =\left(2m+1\right);m=0,1,2,\mathrm{...},ory=\frac{m\lambda }{2};\left(m_1=0,1,2,\mathrm{...},\right)$

Since, $\text{m}=\text{0}$ , then from the triangle diagram,

  $r^2+{\left(R-y\right)}^2=R^{2}or{r}^{2}2yR-y^2\approx 2yR,ify<<R$

Hence,

  $r^2=2\left(\frac{m\lambda }{2}\right)R=m\lambda R;m=0,1,2,\mathrm{...},$

Two conditions are formed for the ratio when apparatus is immersed in the liquid.

  ${\left(\frac{{\text{r}}_{\text{1}}}{{\text{r}}_{\text{2}}}\right)}^{\text{2}}\text{=}\frac{{\text{λ}}_{\text{1}}}{{\text{λ}}_{\text{2}}}\text{=n}$

Therefore, the refractive index is:

  $\text{n=}{\left(\frac{{\text{r}}_{\text{1}}}{{\text{r}}_{\text{2}}}\right)}^{\text{2}}\text{=}{\left(\frac{\text{2}\text{.92}\text{cm}}{\text{2}\text{.48}\text{cm}}\right)}^{\text{2}}\text{=1}\text{.39}$

Conclusion:

The refractive index of the liquid is $1.39$ .