Question

A beam of light of wavelength 600 nm from a distant source falls on a single slit 1.0 mm wide and the resulting diffraction pattern is observed on a screen 2m away. The distance between the first dark fringes on either side of the central bright fringe is?

Answer 1

Hint Diffraction refers to various phenomena that occur when a wave encounters an obstacle or opening. It is defined as the bending of waves around the corners of an obstacle or through an aperture into the region of geometrical shadow of the obstacle/aperture.
For the diffraction pattern the condition for ${n^{th}}$ minimum, i.e., the ${n^{th}}$ dark fringe is given by,
$b\sin \theta = n\lambda $
Where,
$\Rightarrow$ $b$ is width of slit
$\Rightarrow$ $\lambda $ is wavelength of light
$\Rightarrow$ $n$ is order of diffraction

Complete Step by step solution
Given: $\lambda = 600nm$
              $b = 1mm$
             $D = 2m$
Let us say the distance of 1st dark fringe from the centre is ${d_{}}$
Now we know that the condition for ${n^{th}}$ dark fringe is given by,
$b\sin \theta = n\lambda $
For first order diffraction or say for first dark fringe
$n = 1$
Therefore, we have
$
  b\sin \theta = \lambda \\
  \sin \theta = \dfrac{\lambda }{b} \\
 $
On putting the values, we get
$\sin \theta = \dfrac{{600 \times {{10}^{ - 9}}}}{{{{10}^{ - 3}}}}$
On solving above equation, we get
$\sin \theta = 6 \times {10^{ - 4}}......(1)$
Now we know that,
$\Rightarrow$ $\sin \theta = \dfrac{d}{D}$
Where, $d$ is the distance of 1st dark fringe from the centre
and $D$ is the distance between slit and screen
 $\therefore d = D\sin \theta ......(2)$
Now using equation (1) and (2), we get
$
$\Rightarrow$ d = 2 \times 6 \times {10^{ - 4}} \\
$\Rightarrow$ d = 12 \times {10^{ - 4}} \\
$\Rightarrow$ d = 1.2 \times {10^{ - 3}} \\
  d = 1.2mm \\
 $
 As the dark fringes are on the both side of central maxima and are equidistant from the centre.
Hence, we have
The width of dark fringe $
   = 1.2mm + 1.2mm \\
   = 2.4mm \\
 $

Hence, the required distance between the two dark fringes is $2.4mm$

Note The intensity of the diffraction pattern decreases for higher orders. The intensity at the first maximum after the central one is only \[\dfrac{1}{{22}}\] of the intensity of the central maximum. This divergence in $\theta $ is inversely proportional to width $b$ of the slit. If the slit width $b$ is increased, the divergence is increased and the light is diffracted in the wider cone. On the other hand, if the slit-width is large compared to the wavelength, $\dfrac{\lambda }{b} = 0$ and the light continues, undiffracted in the direction $\theta = 0$. This clearly indicates that the diffraction effects are observable only when the obstacle or the opening has the dimensions comparable to the wavelength of the wave.