Question

A slit of width $12\times {{10}^{-7}}m$ is illuminated by light of wavelength $6000\text{A}{}^\circ$. The angular width of the central maximum is approximately

& \text{A}\text{. 30}{}^\circ \\\ & \text{B}\text{. 60}{}^\circ \\\ & \text{C}\text{. 90}{}^\circ \\\ & \text{D}\text{. 0}{}^\circ \\\ \end{aligned}$$

Answer 1

Here we have given a slit width and wavelength and we have to find the angular width of the central maxima. As there is one slit that means it is for single slit diffraction for single diffraction the angular width of the central maxima is given as twice of angular width of first minima. Hence by using the formula for the single slit diffraction we can solve the question.

Formula used:
$\lambda =a\sin \theta$

Complete step by step answer:
In a single slit diffraction, a slit is illuminated by a source of light to produce a diffraction pattern on a screen. The diagram for it can be given as
Here a is the width of the slit and D is the distance of slit from the screen. Θ is the angular width of the first minima and y is the linear width. In the single slit diffraction the wavelength can be given as
$\lambda =a\sin \theta$
Hence the angular width for the first minima can be given as
$\theta =\sin {{\left( \dfrac{\lambda }{a} \right)}^{-1}}$
As per question the width of the slit here is $a=12\times {{10}^{-7}}m$and the wavelength is$\lambda =6000\text{A}{}^\circ =6000\times {{10}^{-10}}m$
Substituting these values we get

& \theta =\sin {{\left( \dfrac{6000\times {{10}^{-10}}m}{12\times {{10}^{-7}}m} \right)}^{-1}} \\\ & \theta =\sin {{\left( \dfrac{6\times {{10}^{-7}}}{12\times {{10}^{-7}}} \right)}^{-1}} \\\ & \theta =\sin {{\left( \dfrac{1}{2} \right)}^{-1}} \\\ & \theta =30{}^\circ \\\ \end{aligned}$$ Now the angular width of the central maximum is given by twice of angular width of first minimum $$\begin{aligned} & \text{Angular width of the central maximum}=2\theta \\\ & 2\theta =2\times 30{}^\circ \\\ & 2\theta =60{}^\circ \\\ \end{aligned}$$ **Hence the correct option is B.** **Note:** The units should be converted properly before using to avoid error, like here we have converted the unit of wavelength from angstrom to metres. Similarly, like angular width the linear width of the central maximum is twice of the linear width of the first minimum. The minimum corresponds to the dark spot while the maximum refers to the bright spot.