Question

Aslit $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{mm}$wide is illuminated by light of wavelength $\mathbf{589}\mathbf{}\mathbf{nm}$. We see a diffraction pattern on a screen $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}$away. What is the distance between the first two diffraction minima on the same side of the central diffraction maximum?

Short answer

The distance between first two diffraction minima is$1.77\text{mm}$.

Step-by-step solution

Identification of given data

The wavelength of the light wave is $\lambda =589\text{nm}$

The distance of screen from slit is $D=3\text{m}$.

The width of slit is $d=1\text{mm}$.

The order of first minima is $m_1=1$.

The order of second minima is $m_2=2$.

Concept used

The distance between two successive diffraction minima is called the fringe width. It varies with the wavelength, slit width and distance of screen from slit.

Determination of distance between first two diffraction minima

The distance between first two diffraction minima is given as:

$w=\frac{\left(m_2-m_1\right)\lambda D}{d}$

Substitute all the values in equation.

$$\begin{gathered} w=\frac{\left(2-1\right)\left(589\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\left(3\text{m}\right)}{\left(1\text{mm}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)} \\ =1.77\times {10}^{-3}\text{m} \\ =\left(1.77\times {10}^{-3}\text{m}\right)\left(\frac{1\text{mm}}{{10}^{-3}\text{m}}\right) \\ =1.77\text{mm} \end{gathered}$$

Therefore, the distance between first two diffraction minima is $1.77\text{mm}$.