Chapter 24, Problem 6P

To determine

The difference for second-order fringes.

Answer to Problem 6P

  $0.21mm$

Explanation of Solution

Given info:

First wavelength, ${\lambda }_1=720nm=720\times {10}^{-9}m$

Second wavelength, ${\lambda }_2=660nm=660\times {10}^{-9}m$

Distance between slits, $d=0.58mm=0.58\times {10}^{-3}m$

Distance from slits to the screen, $L=1.0m$

Formula used:

Distance between fringes is equal to product of wavelength and distance of screen from slits over distance between slits. So,

  $y=\frac{\lambda .L}{d}$

  $\lambda$ is wavelength

  $L$ is distance from slits to the screen.

  $d$ is a distance between slits

  $y$ is the distance between the fringes 

Calculation:

Calculation for $y_{720}$ by using the formula, $y_{720}=\frac{\lambda .L}{d}$

  $y_{720}=720\times {10}^{-9}\frac{2}{0.58\times {10}^{-3}}=0.00248m=2.48mm$

Calculation for $y_{660}$ by using the formula $y_{660}=\frac{\lambda .L}{d}$

  $y_{660}=660\times {10}^{-9}\frac{2}{0.58\times {10}^{-3}}=0.00227m=2.27mm$

Now, calculate the difference:

  $y_{720}-y_{660}=2.48mm-2.27mm=0.21mm$

Conclusion: The difference for second-order fringes is $0.21mm$ .