Chapter 24, Problem 12P

To determine

The distance between the fringes.

Answer to Problem 12P

  $2.41mm$

Explanation of Solution

Given info:

The wavelength of the light, $\lambda =480nm$

The distance between the slits, $d=6\times {10}^{-2}mm=6\times {10}^{-5}m$

The distance between the slits to screen, $L=40cm=0.400m$

Refractive index of water, $n=1.33$

Formula used:

  $\Delta y=\frac{L\lambda \Delta m}{d}$

  $m$ is an integer

  $\lambda$ is wavelength

  $L$ is distance from slits to the screen.

  $d$ is a distance between slits

  $y$ is the distance between the fringes 

Calculation:

According to the question,

  $\begin{array}{c}{\lambda }_{water}=\frac{\lambda }{n_{water}}\\ =\frac{480nm}{1.33}\\ =361nm=361\times {10}^{-9}m\end{array}$

  $d\sin \theta =m{\lambda }_{water},m=0,1,2,3,\mathrm{...}$

  $y=L\tan \theta$

  $sin\theta \approx \tan \theta$

Now, $y=L\left(\frac{m{\lambda }_{water}}{d}\right)=\frac{mL{\lambda }_{water}}{d}$

  $\Delta y=\frac{L{\lambda }_{water}\Delta m}{d}$

Substitute the values in formula, $\Delta y=\frac{L{\lambda }_{water}\Delta m}{d}$ ,

  $\Delta y=\frac{\left(0.400m\right)\left(361\times {10}^{-9}m\right)\left(1\right)}{\left(6.00\times {10}^{-5}\right)}$

  $\Delta y=2.41\times {10}^{-3}m$

  $\Delta y=2.41mm$

Conclusion: The required value is $2.41mm$ .