Question

Two slits spaced 0.500 mm apart are placed 80.0 cm from a screen.

What is the distance between the second and third dark lines of the interference pattern on the screen when the slits are illuminated with coherent light with a wavelength of 510 nm?

delta_y=__________________mm

Answer 1

Concept and reason

The concept needed to solve this problem is the equation for fringe space in double slit interference experiment.

Write the equation for fringe width in double slit interference experiment. Substitute all the known values and solve for fringe width.

Fundamentals

The equation for fringe space ( $\Delta y$ ) in double slit interference experiment is,

$\Delta y = \frac{{\lambda L}}{d}$

Here, $\lambda$ is the wavelength of the light used, $L$ is the distance of the screen from the center of the slits, and d is the spacing between the slits.

The fringe width is the space which is between any two successive bright or dark interference lines.

The equation for fringe space ( $\Delta y$ ) in double slit interference experiment is,

$\Delta y = \frac{{\lambda L}}{d}$

Substitute 510 nm for $\lambda$ , 80.0 cm for L, and 0.500 mm for d.

$\begin{array}{c}\\\Delta y = \frac{{\left( {510{\rm{ nm}}} \right)\left( {80.0{\rm{ cm}}} \right)}}{{\left( {0.500{\rm{ mm}}} \right)}}\\\\ = \frac{{\left( {510{\rm{ nm}}\frac{{1{\rm{ mm}}}}{{{{10}^6}{\rm{ nm}}}}} \right)\left( {80.0{\rm{ cm}}\frac{{10{\rm{ mm}}}}{{1{\rm{ cm}}}}} \right)}}{{\left( {0.500{\rm{ mm}}} \right)}}\\\\ = {\rm{0}}{\rm{.816 mm}}\\\end{array}$

Ans:

The distance between second and third dark interference lines is 0.816 mm.