Chapter 19.2, Problem 17PP

To determine

The distance of the slits from the screen.

Answer to Problem 17PP

  $2.43\text{m}$

Explanation of Solution

Given:

The width of the central bright band, $2x_1=2.60\times {10}^{-2}\text{m}$

The wavelength of yellow slits, $\lambda =589\text{nm}$

The width of the slits, $w=0.110\text{mm}$

Formula used:

The width of bright band in Single-Slit Diffraction is given by,

  $2x_1=\frac{2\lambda L}{w}$

Where,

L = distance between screen and slit

  $\begin{array}{l}\lambda =\text{wavelength}\\ w=\text{ width of the slits}\end{array}$

Calculation:

Using the above mentioned formula and the given values, the distance of the slits from the screen will be,

  $\begin{array}{c}2x_1=\frac{2\lambda L}{w}\\ 2.60\times {10}^{-2}=\frac{2\left(589\times {10}^{-9}\right)L}{\left(0.11\times {10}^{-3}\right)}\\ L=2.43\text{m}\end{array}$

Conclusion:

Hence, the distance of the slits from the screen is $2.43\text{m}$ .