Chapter 19.2, Problem 16PP

To determine

The width of the central bright band.

Answer to Problem 16PP

  $8.62\times {10}^{-3}\text{m}$

Explanation of Solution

Given:

The wavelength of the monochromatic green light, $\lambda =546\text{}\text{nm}$

The width of the slits, $w=0.095\text{mm}$

The distance of the screen from the slit, $L=75\text{cm}$

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 width of the central bright band will be,

  $\begin{array}{c}2x_1=\frac{2\lambda L}{w}\\ =\frac{2\left(546\times {10}^{-9}\right)\left(0.75\right)}{\left(0.095\times {10}^{-3}\right)}\\ =8.62\times {10}^{-3}\text{m}\end{array}$

Conclusion:

Hence, the width of the central bright band is $8.62\times {10}^{-3}\text{m}$ .