Chapter 19, Problem 39A

To determine

The wavelength of the lasers and rank them in ascending order.

Answer to Problem 39A

The ranking of the wavelengths of the lasers is

  ${\lambda }_A<{\lambda }_D<{\lambda }_C<{\lambda }_B<{\lambda }_E$

Explanation of Solution

Given:

The slit separation is $d=0.035\text{ mm}$

Formula Used:

The wavelength of a light forming interference pattern in a double slit experiment is given by

  $\lambda =\frac{xd}{L}$

  $\lambda$ is the wavelength

  $x$ is the spacing between the bright spots

  $d$ is the slit separation

  $L$ is the distance between the screen and the slits.

Calculation:

  $\lambda =\frac{xd}{L}$

The wavelengths for the lasers are.

  $\begin{array}{l}{\lambda }_A=\frac{0.012\text{ m}\times \left(0.035\times {10}^{-3}\text{ m}\right)}{0.95\text{ m}}=4.42\times {10}^{-7}\text{ m}\\ {\lambda }_B=\frac{0.016\text{ m}\times \left(0.035\times {10}^{-3}\text{ m}\right)}{0.95\text{ m}}=5.89\times {10}^{-7}\text{ m}\\ {\lambda }_C=\frac{0.02\text{ m}\times \left(0.035\times {10}^{-3}\text{ m}\right)}{\text{1}\text{.3 m}}=5.38\times {10}^{-7}\text{ m}\\ {\lambda }_D=\frac{0.04\text{ m}\times \left(0.035\times {10}^{-3}\text{ m}\right)}{\text{2}\text{.8 m}}=5\times {10}^{-7}\text{ m}\\ {\lambda }_E=\frac{0.05\text{ m}\times \left(0.035\times {10}^{-3}\text{ m}\right)}{\text{2}\text{.8 m}}=6.25\times {10}^{-7}\text{ m}\end{array}$

Conclusion:

The ranking of the wavelengths of the lasers is

  ${\lambda }_A<{\lambda }_D<{\lambda }_C<{\lambda }_B<{\lambda }_E$