Question

In a single slit diffraction pattern, the distance between the first maximum on the left and the first maximum on the right is $\;5mm$. The screen on which the diffraction pattern is displaced is at a distance of $\;80cm$from the slit. The wavelength is $6000angstrom$. The slit width is mm is about
A) $0.576$
B) $0.348$
C) $0.192$
D) $0.096$

Answer 1

Diffraction: It is the phenomenon of bending of light around the corners of small obstacles or apertures and the consequent spreading into the regions of geometrical shadows.

Formula used:
${\text{d = }}\dfrac{{{\text{m}}\lambda {\text{D}}}}{{\text{y}}}$, here, d=slit width, $\lambda =$wavelength of the light wave, $D$=distance of the screen from the slit, y=distance between the first maxima from the center, $m$= order of the fringes.

Complete step-by-step solution:
Given details: The order of the fringes, $m = 1$,
Distance between the first maximum (or minimum) and central fringe is, $y = \dfrac{5}{2} = 2.5cm$,
Distance between the screen and the slit, $D = 80cm = 800mm,$
Wavelength if the light wave,$\lambda = 6000angstorm$
Single slit diffraction pattern:

Single slit diffraction can be performed through a single slit whose width is on the order of the wavelength of light, then the light wave is diffracted on the edge of the aperture forming a single slit diffraction pattern on the screen that is placed at a certain distance from the apparatus. Therefore, we can find the value of slit width using the above formula,
${\text{d = }}\dfrac{{{\text{m}}\lambda {\text{D}}}}{{\text{y}}}$,
Substitution of the values in the given equations give,
$\Rightarrow {\text{d = }}\dfrac{{(1)(800)(6000)}}{{2.5}}$
Simplifying the given equations we get,
$\Rightarrow (1.92{\text{x1}}{{\text{0}}^5}){10^{ - 7}}$
=$\therefore 0.192mm$

Hence, The correct option is (C).

Note: There are two types of a single slit diffraction pattern,
Fresnel’s diffraction pattern: In this diffraction pattern, the screen and the slit are placed close to the aperture or the obstacles and the light after diffraction appears converging towards the screen.
Fraunhofer’s diffraction pattern: In this diffraction pattern, the source and the screen are placed at a large distance from the aperture or the obstacles.