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**Hint:**Diffraction is defined as the phenomenon where the light wave bends around a sharp edge or corner, which can be visualised as intrusion of light in the region of the shadow of the object. There are several orders of diffraction wherein the intensity of the light decreases as we proceed away from the first order of diffraction, which is also known as the central maximum. The second minima are the second in order from the central maximum, on either side.

**Complete step by step answer:**

The angular position of nth order diffraction is calculated from the following equation-

$n\lambda = d\sin \theta $

Where,

$\lambda $=wavelength of light ,

d=slit width,

$\theta $= angular position

And n= order of diffraction, for minima $n = \pm 1, \pm 2, \pm 3,.......$ and for maxima $n = 0, \pm \dfrac{3}{2}, \pm \dfrac{5}{2},.............$

Step1:

From the equation of angular position of nth order diffraction we have,

$n\lambda = d\sin \theta $ …………..(1)

Where,

$\lambda $=wavelength of light ,

d=slit width,

$\theta $= angular position

And n= order of diffraction, for minima $n = \pm 1, \pm 2, \pm 3,.......$ and for maxima $n = 0, \pm \dfrac{3}{2}, \pm \dfrac{5}{2},.............$

Since it is given that we have to find angular position of second minima ,

Therefore n=2

Also given that wavelength $\lambda = 550nm = 550 \times {10^{ - 9}}m$

Slit width $d = 22.0 \times {10^{ - 5}}cm = 22 \times {10^{ - 7}}m$

Step2:

Substitute all the values in equation (1) we get,

$2 \times 550 \times {10^{ - 9}} = 22 \times {10^{ - 7}} \times \sin \theta $

$ \Rightarrow \sin \theta = \dfrac{{2 \times 550 \times {{10}^{ - 9}}}}{{22 \times {{10}^{ - 7}}}}$

$ \Rightarrow \sin \theta = \dfrac{1}{2}$

$\therefore \theta = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{6}$

**Hence, the correct answer is option (D).**

**Additional information:**

Diffraction refers to various phenomena that occur when a wave encounters an obstacle or a slit. 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.

Examples: The most striking examples of diffraction are those that involve light; the closely spaced tracks on a CD or DVD act as a diffraction grating to form the familiar rainbow pattern seen when looking at a disc.

**Note:**The students should remember that the maximum intensity of diffraction pattern is at centre (central maxima), on both sides of central maxima there is continuous reduction in the intensity of light. This is the main reason why there is significant reduction in the intensity of the fringes as we move away from the central bright fringe in the Young’s Double Slit experiment.

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