Chapter 24, Problem 52P

To determine

The refractive index of the gas.

Answer to Problem 52P

  $1.00554$ .

Explanation of Solution

Given:

Distance, $d=1.30\times {10}^{-2}m$

Number of fringes shift, $N=236$

Wavelength, ${\text{λ=610×10}}^{\text{-9}}\text{m}$

Formula used:

It is known that: ${\text{n}}_{\text{gas}}\text{=}\frac{\text{N×λ}}{\text{2d}}\text{+1}$

Where,

  $\text{d}$ is distance.

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

  $\text{N}$ is the number of fringes shift.

  ${\text{n}}_{\text{gas}}$ is the refractive index of the gas.

Calculation:

The cavity has depth of $\text{d}$ . So, the number of wavelengths in vacuum is $\frac{\text{d}}{\text{λ}}$ .

Hence, the number of wavelengths when gas present is $\frac{\text{d}}{{\text{λ}}_{\text{gas}}}\text{=}\frac{{\text{n}}_{\text{gas}}\text{d}}{\text{λ}}$

  $\therefore \text{N=2}\left[\left(\frac{{\text{n}}_{\text{gas}}\text{d}}{\text{λ}}\right)\text{-}\left(\frac{\text{d}}{\text{λ}}\right)\right]\text{=2}\left(\frac{\text{d}}{\text{λ}}\right)\left({\text{n}}_{\text{gas}}\text{-1}\right)$

Therefore, the refractive index of the gas will be: ${\text{n}}_{\text{gas}}\text{=}\frac{\text{N×λ}}{\text{2d}}\text{+1}$

  ${\text{n}}_{\text{gas}}=\left(\frac{\text{N}\times \lambda }{2d}+1\right)=\left(\frac{236\times \left(610\times {10}^{-9}\text{m}\right)}{2\left(1.30\times {10}^{-2}\text{m}\right)}+1\right)=1.00554$

Conclusion:

The refractive index of the gas will be $1.00554$ .