Question

Coherent light with wavelength 597 mm passes through two very narrow slits, and theinterference pattern is observed on a screen a distance of 3.00m from the slits. The first-order bright fringe is adistance of 4.84 mm from the center of the central bright fringe.

For what wavelength of light will thefirst-order dark fringe be observed at this same point on thescreen?

 

Answer 1

Concepts and reason

The concept required to solve the given problem is position of bright and dark fringes double slit experiment.

Initially, determine the slit separation between the slits by using expression for position of bright fringes in double slit experiment. Then, use the expression for position of dark fringe in double slit experiment to determine the wavelength.

Fundamentals

The position of bright fringes ${y_m}$ on screen in double slit experiment is given by following expression.

${y_m} = \frac{{m\lambda D}}{d}$

Here, m is the order of fringe. d is the slit separation, D is the distance between screen and source, and $\lambda$ is the wavelength.

The position of dark fringes ${y_m}$ on screen in double slit experiment is given by following expression.

${y_m} = \left( {m + \frac{1}{2}} \right)\frac{{\lambda D}}{d}$

Here, m is the order of fringe. d is the slit separation, D is the distance between screen and source, and $\lambda$ is the wavelength.

The position of bright fringes ${y_m}$ on screen in double slit experiment is given by following expression.

${y_m} = \frac{{m\lambda D}}{d}$

Substitute 1 for m in the above equation to solve for d.

$\begin{array}{c}\\{y_1} = \frac{{\left( 1 \right)\lambda D}}{d}\\\\d = \frac{{\lambda D}}{{{y_1}}}\\\end{array}$

Substitute 597 nm for $\lambda ,$ 3.00 m for D, and 4.84 mm for ${y_1}$ in the above equation.

$\begin{array}{c}\\d = \frac{{\left( {597\,{\rm{nm}}} \right)\left( {\frac{{1\,{\rm{m}}}}{{{{10}^9}\,{\rm{nm}}}}} \right)\left( {3.00\,{\rm{m}}} \right)}}{{4.84\,{\rm{mm}}\left( {\frac{{1\;{\rm{m}}}}{{1000\,{\rm{mm}}}}} \right)}}\\\\ = 3.7 \times {10^{ - 4}}\,{\rm{m}}\\\end{array}$

The position of dark fringes ${y_m}$ on screen in double slit experiment is given by following expression.

${y_m} = \left( {m + \frac{1}{2}} \right)\frac{{\lambda D}}{d}$

Substitute 0 for m in the above equation to solve for wavelength.

$\begin{array}{c}\\{y_0} = \left( {0 + \frac{1}{2}} \right)\frac{{\lambda D}}{d}\\\\\lambda = \frac{{2{y_0}d}}{D}\\\end{array}$

Substitute 4.84 mm for ${y_0},$ $3.7 \times {10^{ - 4}}\,{\rm{m}}$ for d, and 3.00 m for D in the above equation.

$\begin{array}{c}\\\lambda = \frac{{2\left( {4.84\,{\rm{mm}}} \right)\left( {\frac{{1\,{\rm{m}}}}{{1000\,{\rm{mm}}}}} \right)\left( {3.7 \times {{10}^{ - 4}}\,{\rm{m}}} \right)}}{{3.00\,{\rm{m}}}}\\\\ = 1.19 \times {10^{ - 6}}\,{\rm{m}}\left( {\frac{{{{10}^6}\,\mu {\rm{m}}}}{{1\,{\rm{m}}}}} \right)\\\\ = 1.19\,\mu {\rm{m}}\\\end{array}$

Ans:

The wavelength required is $1.19\,\mu {\rm{m}}{\rm{.}}$