Question

If 720-nm and 660-nm light passes through two slits 0.68 mm apart, how far apart are the second-order fringes for these two wavelengths on a screen 1.0 m away?

Answer 1

Concepts and reason

The concept of interference is used here.

Initially, use the expression of the distance of bright fringe from central maxima for both the wavelength i.e. 720 nm and 660 nm. Find the position value for both the wavelength and subtract them to obtain the separation between the two fringes.

Fundamentals

Interference is the phenomena in which two waves superimpose to form a resultant wave of greater, smaller or same amplitude. The two sources should be coherent in nature.

The distance of the ${{\rm{n}}^{{\rm{th}}}}$bright fringe from the central maxima is,

${x_{\rm{n}}} = \frac{{n\lambda D}}{d}$

Here, n is the order of fringes, $\lambda$is the wavelength of the incident light, d is the separation between the slits and D is the distance between slit and screen.

The distance of the ${{\rm{n}}^{{\rm{th}}}}$bright fringe from the central maxima is,

${x_{\rm{n}}} = \frac{{n\lambda D}}{d}$

The distance of second order fringe for wavelength ${\lambda _1}$is,

${x_2} = \frac{{n{\lambda _1}D}}{d}$

Substitute 2 for n, 720 nm for ${\lambda _1}$, 1 m for D and 0.68 mm for d in equation ${x_2} = \frac{{n{\lambda _1}D}}{d}$.

$\begin{array}{c}\\{x_2} = \frac{{2\left( {720{\rm{ nm}}\left( {\frac{{{{10}^{ - 9}}{\rm{ m}}}}{{1.0{\rm{ nm}}}}} \right)} \right)\left( {1.0{\rm{ m}}} \right)}}{{0.68{\rm{ mm}}\left( {\frac{{{{10}^{ - 3}}{\rm{ m}}}}{{1.0{\rm{ mm}}}}} \right)}}\\\\ = 2.12 \times {10^{ - 3}}{\rm{ m}}\\\end{array}$

The distance of second order fringe for wavelength ${\lambda _2}$is,

${x'_2} = \frac{{n{\lambda _2}D}}{d}$

Substitute 2 for n, 660 nm for ${\lambda _1}$, 1 m for D and 0.68 mm for d in equation ${x'_2} = \frac{{n{\lambda _2}D}}{d}$.

$\begin{array}{c}\\{{x'}_2} = \frac{{2\left( {{\rm{660 nm}}\left( {\frac{{{{10}^{ - 9}}{\rm{ m}}}}{{1.0{\rm{ nm}}}}} \right)} \right)\left( {1.0{\rm{ m}}} \right)}}{{0.68{\rm{ mm}}\left( {\frac{{{{10}^{ - 3}}{\rm{ m}}}}{{1.0{\rm{ mm}}}}} \right)}}\\\\ = 1.94 \times {10^{ - 3}}{\rm{ m}}\\\end{array}$

The separation between the two fringes is calculated as:

$\Delta x = {x_2} - {x'_2}$

Substitute $1.94 \times {10^{ - 3}}{\rm{ m}}$for ${x'_2}$and $2.12 \times {10^{ - 3}}{\rm{ m}}$for ${x_2}$in equation $\Delta x = {x_2} - {x'_2}$.

$\begin{array}{c}\\\Delta x = 2.12 \times {10^{ - 3}}{\rm{ m}} - 1.94 \times {10^{ - 3}}{\rm{ m}}\\\\ = {\rm{0}}{\rm{.18}} \times {\rm{1}}{{\rm{0}}^{ - 3}}{\rm{ m}}\left( {\frac{{1.0{\rm{ mm}}}}{{{{10}^{ - 3}}{\rm{ m}}}}} \right)\\\\ = 0.18{\rm{ mm}}\\\end{array}$

Ans:

The separation between the two fringes is 0.18 mm.