Question

The figure below shows the light intensity on a screen 2.7m behind a double slit. The wavelength of the lightis 519 nm .

What is the spacing between the slits?

Answer 1

Concepts and reason

The concept required for this problem are condition for interference.

First write the condition for interference and obtain an expression for the spacing between the slits. From the expression for the condition of the interference, the distance between the plate is depends on the wavelength and order of spectrum.

Fundamentals

Write the condition for interference.

$d\sin \theta = m\lambda$

Here, $d$ is the spacing between the slits , $\theta$ is the angular spacing between the fringes, $m$ is the order of spectrum and $\lambda$ is the wavelength of the light.

Draw the figure which shows the interference pattern.

Here, $L$ is the distance between the screen and the slits, $y$ is the spacing between the fringes and $\theta$ is the angular spacing between two fringes.

Write the condition for interference.

$d\sin \theta = m\lambda$ …... (1)

Since, $y \ll L$

Therefore, $\sin \theta = \tan \theta = \frac{y}{L}$ .

Here, $L$ is the distance between the screen and the slits and $y$ is the spacing between the fringes.

Substitute $\sin \theta = \frac{y}{L}$ in equation (1)

$d\left( {\frac{y}{L}} \right) = m\lambda$

Rearrange for $d$ .

$d = m\lambda \left( {\frac{L}{y}} \right)$

Substitute $1$ for $m$ (for first order of spectrum), $519\;{\rm{nm}}$ for $\lambda$ , $0.6\;{\rm{cm}}$ for $y$ and $2.7\;{\rm{m}}$ for $L$ .

$\begin{array}{c}\\d = m\lambda \left( {\frac{L}{y}} \right)\\\\ = \left( 1 \right)\left( {519\;{\rm{nm}}} \right)\left( {\frac{{2.7\;{\rm{m}}}}{{0.6\;{\rm{cm}}}}} \right)\\\\ = \left( {519\;{\rm{nm}}\left( {\frac{{{{10}^{ - 9}}\;{\rm{m}}}}{{1\;{\rm{nm}}}}} \right)} \right)\left( {\frac{{2.7\;{\rm{m}}}}{{0.6\;{\rm{cm}}\left( {\frac{{{{10}^{ - 2}}\;{\rm{m}}}}{{1\;{\rm{cm}}}}} \right)}}} \right)\\\\ = \left( {519 \times {{10}^{ - 9}}\;{\rm{m}}} \right)\left( {\frac{{2.7\;{\rm{m}}}}{{0.6 \times {{10}^{ - 2}}\;{\rm{m}}}}} \right)\\\end{array}$

Further solving,

$\begin{array}{c}\\\;\;\; = \left( {519 \times {{10}^{ - 9}}\;{\rm{m}}} \right)\left( {4.5 \times {{10}^2}} \right)\\\\ = 0.2335 \times {10^{ - 3}}\;{\rm{m}}\\\end{array}$

Ans:

The spacing between the slits is $0.2335 \times {10^{ - 3}}\;{\rm{m}}$ .