Question

In Young’s double slit experiment for $d = 0.001\,m$ , $D = 1\,m$ interference pattern is obtained for radiations of wavelengths $5000\,{A^ \circ }$ and $5800\,{A^ \circ }$. Which of the following of nearest to third maxima for wavelength $5000\,{A^ \circ }$ :
A. ${2^{nd}}\,\max ima\,for\,5800\,{A^ \circ }$
B. ${3^{rd}}\,\min ima\,for\,5800\,{A^ \circ }$
C. ${3^{rd}}\,\max ima\,for\,5800\,{A^ \circ }$
D. ${3^{rd}}\,\min ima\,for\,5000\,{A^ \circ }$

Answer 1

Here we have to use the Young’s double slit experiment.
The double-slit experiment is a demonstration of modern physics that light and matter will show characteristics of waves and particles both classically defined; however, it demonstrates the essentially probabilistic existence of quantum mechanical phenomena.

Complete step by step answer:
Young's double-slit experiment uses two coherent light sources positioned at a small distance apart, typically just a few orders of magnitude greater than the wavelength of light used. Young's double-slit experiment helped to explain the principle of light waves.
In the simplest version of this experiment, a coherent light source, such as a laser beam, illuminates a plate pierced by two parallel slits, and the light going through the slits is seen on the screen behind the plate.

Let us consider the monochromatic light source being held at a significant distance from two slits ${s_1}$ and ${s_2}$ . $S$ is equidistant to ${s_1}$ and ${s_2}$ . ${s_1}$ and ${s_2}$
serve as two coherent sources, all originating from $S$ .
The light passes through these slits and falls on a screen at a distance of $D$ from the location of the slits ${s_1}$ and ${s_2}$ . Let $d$
be the distinction between the two slits.
If ${s_1}$ is opened and ${s_2}$ is closed, the screen opposite to ${s_1}$ is closed, only the screen opposite to ${s_2}$ is activated. Interference patterns only occur when both slits ${s_1}$ and ${s_2}$ are opened.
The path difference is given by:
$\Delta Z = \dfrac{Y} {D}$
Where $\Delta Z$ is the path difference, $D$ is the distance between the screen and the slit.
$\gamma = \dfrac{{n\lambda D}}{d}$
The ${n^{th}}$ order maxima is given by:

$D$ is the distance between the screen and the slit, $d$ is the distance between two slits, $\lambda$ is the wavelength, $n$ is the number of order of minima and $\gamma$ is the maximum wavelength.
For $n = 2$ and $\lambda = 5800\,{A^ \circ }$
${\gamma _2} = 1.16 \times {10^{ - 9}}$
For $\lambda = 5000\,{A^ \circ }$
${\gamma _3} = 1.5 \times {10^{ - 9}}$

So, the correct answer is “Option A”.

Note:
Here we have to see the units of the given physical quantities. If the units are not in standard form we need to convert them to standard form otherwise the answer would be wrong by a factor of ten or hundred. Then, we have to see which maxima are closer to the given value to get the answer. If the answer is in decimal we need to take the nearest whole number.