Question

In a single slit diffraction the distance between slit & screen is $1\;{\rm{m}}$. The size of the slit is $0.7\;{\rm{mm}}$ & second maximum is formed at a distance of $2\;{\rm{mm}}$ from the center of the screen, then find out the wavelength of light.

Answer 1

Hint: In the solution we will use the general equation of bright fringe for calculating the wavelength of light. They show the relation between the wavelength of light, number of fringes, size of slit and the angle of diffraction. We will also use some basic trigonometry for calculating the angle of diffraction.

Complete step by step solution:
Given:
The distance between slit and screen is, $D = 1\;{\rm{m}}$.
The size of the slit, d is $d = 0.7\;{\rm{mm}} = 0.7\;{\rm{mm}} \times \dfrac{{1\;{\rm{m}}}}{{1000\;{\rm{mm}}}} = 7 \times {10^{ - 4}}\;{\rm{m}}$.
Distance between second maxima and center of screen is, $y = 2\;{\rm{mm}} = \;2\;{\rm{mm}} \times \dfrac{{1\;{\rm{m}}}}{{1000\;{\rm{mm}}}} = 2 \times {10^{ - 3}}\;{\rm{m}}$.

The general equation for calculating the wavelength of n bright fringe is,
$\left( {n + \dfrac{1}{2}} \right)\lambda = d\sin \theta $
Here, $\lambda $ is wavelength of light

For secondary maxima, substitute n=2 in the above expression,
$\begin{array}{l}
\left( {2 + \dfrac{1}{2}} \right)\lambda = d\sin \theta \\
\dfrac{{5\lambda }}{2} = d\sin \theta \\
\lambda = \dfrac{2}{5}d\sin \theta
\end{array}$ … (i)
If $\theta $ is small then $\sin \theta = \tan \theta $.$\tan \theta $ is given by,
$\tan \theta = \dfrac{y}{D}$
If $\theta $ is very small than $\tan \theta = \theta $ and the expression becomes,
 $\theta = \dfrac{y}{D}$
Substitute in equation (i),
$\lambda = \dfrac{{2yd}}{{5D}}$
Substitute the necessary value and calculate wavelength,
$\lambda = \dfrac{{2\left( {7 \times {{10}^{ - 4}}\;{\rm{m}}} \right)\left( {2 \times {{10}^{ - 3}}\;{\rm{m}}} \right)}}{{5\left( {1\;{\rm{m}}} \right)}}$
$\lambda = 56 \times {10^{ - 8\;}}{\rm{m}}$
As, 1 Angstrom is equal to ${10^{ - 10}}\;{\rm{m}}$. Converting the wavelength in angstrom,
$\lambda {\rm{ = 56}} \times {\rm{1}}{{\rm{0}}^{ - 8}}\;{\rm{m}}\; = 56 \times {10^{ - 8}}\;{\rm{m}} \times \dfrac{{1\;\mathop {\rm{A}}\limits^{\rm{0}} }}{{{{10}^{ - 10}}\;{\rm{m}}}} = 5600\;\mathop {\rm{A}}\limits^{\rm{0}} \;$
Therefore, the wavelength of the light used is $5600\;\mathop {\rm{A}}\limits^{\rm{0}} $.

Note: Make sure to use the approximation that the angle of diffraction is very small. Convert the units very carefully. First convert $\sin \theta $ into $\tan \theta $ then use the basic formula of trigonometry. Also, remember that it is the case of maxima, so do not use the formula of minima.