Chapter 19, Problem 2STP

To determine

To Choose: The correct option.

Answer to Problem 2STP

D. $0.063\text{m}$

Explanation of Solution

Given:

The wavelength of the light, $\lambda =410\text{nm}$

The width of the slits, $w=3.8\times {10}^{-6}\text{m}$

From the figure, the distance of the screen from the slit, $L=0.29\text{m}$

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,

  $\begin{array}{c}2x_1=\frac{2\lambda L}{w}\\ =\frac{2\left(410\times {10}^{-9}\right)\left(0.29\right)}{\left(3.8\times {10}^{-6}\right)}\\ =6.26\times {10}^{-2}\text{m}\\ =\text{0}\text{.063 m}\end{array}$

Conclusion:

Hence, the width of the central bright band is $0.063\text{m}$ .