Question

A plane wave of wavelength $6250\,{A^ \circ }$ is incident on a slit of width$2 \times {10^{ - 2\,\,}}\,cm$. The width of the principle maximum of diffraction pattern on a screen at a distance of $50\,\,cm$ will be:
$a)\;312.5 \times {10^{ - 2}}\,\,cm \\\ b)\,312.5 \times {10^{ - 4}}\,\,cm \\\ c)\,312\,\,cm \\\ d)\,312.5 \times {10^{ - 5}}\,\,cm \\\$

Answer 1

Hint Diffraction patterns are observed when light passes through slits. Pattern is in the form of fringes. We are to find the width of the central fringe. The formula for that will be used in question for solving is $\beta = \dfrac{{2\lambda D}}{a}$, where
$\beta$ is width of principal maxima
$\lambda$ is the wavelength
$D$ is the distance between slit and screen
$a$ is the slit width
We are given all the required values of $\lambda ,D$ and $a$ in the question. Substituting them in the above equation simply will give the answer.

Complete step-by-step solution :
When light passes through a small opening of an obstacle, the light spread around the edges of the obstacle. The size of the opening or slit should be comparable in size to the wavelength of light. We call this phenomenon of diffraction of light. When light passes through a slit, we can observe a diffraction pattern on a screen which is placed at a distance which is sufficiently large as compared to the slit width.
We are asked the question to find the principal maxima. Actually the diffraction pattern is in the form of dark and bright fringes positioned alternatively. We have to find the width of central maxima which is given by the formula
$\beta = \dfrac{{2\lambda D}}{a}$ where,
$\beta$ is width of principal maxima
$\lambda$ is the wavelength
$D$ is the distance between slit and screen
$a$ is the slit width
We are given that,
$\lambda = 6250\,{A^ \circ } = 6250 \times {10^{ - 10}}\,m \\\ D = 50\,cm = 0.5\,m \\\ a = 2 \times {10^{ - 2}}\,cm = 2 \times {10^{ - 4}}\,m \\\$
Substituting these values in the formula

$$\beta = \dfrac{{2\lambda D}}{a} = \dfrac{{2(6250 \times {{10}^{ - 10}})(0.5)}}{{(2 \times {{10}^{ - 4}})}} \\\ \beta = 0.03125\,m = 3.125\,cm = 3.125 \times \dfrac{{{{10}^2}}}{{{{10}^2}}}\,cm \\\ \beta = 312.5 \times {10^{ - 2}}\,cm \\\$$

So option A is correct.

Note:- You have noticed the unit conversions done. We, for the ease of our calculations, have converted all the units to the S.I. system of units to bring uniformity. Note that
$1\,{A^ \circ } = 1 \times {10^{ - 10}}\,m \\\ 1\,cm = 1 \times {10^{ - 2}}\,m \\\$
At the last, as per the requirements of the question, we have converted our unit back to centimeters. Be careful while converting units.