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Question: In a diffraction pattern due to a single slit of width ‘a’, the first minimum is observed at an angl...

In a diffraction pattern due to a single slit of width ‘a’, the first minimum is observed at an angle 300{30^0} when light of wavelength 5000Ao5000\mathop {\text{A}}\limits^o is incident on the slit. The first secondary maximum is observed at an angle of:
A. sin⁡−1(14) B. sin⁡−1(23) C. sin⁡−1(12) D. sin⁡−1(34)  {\text{A}}{\text{. }}{\sin ^{ - 1}}(\dfrac{1}{4}) \\\ {\text{B}}{\text{. }}{\sin ^{ - 1}}(\dfrac{2}{3}) \\\ {\text{C}}{\text{. }}{\sin ^{ - 1}}(\dfrac{1}{2}) \\\ {\text{D}}{\text{. }}{\sin ^{ - 1}}(\dfrac{3}{4}) \\\

Explanation

Solution

Hint: From the given information of the first minimum, get the value of ‘a’ the slit width. Now, use this value to get the secondary maxima, from the Bragg’s Equation.

Formula used:
Bragg’s Equation:
asin⁡θ=nλa\sin \theta = n\lambda (for minima).
asin⁡θ=(2n+1)λ2a\sin \theta = \dfrac{{(2n + 1)\lambda }}{2} (for maxima).

Complete step-by-step solution -
We see that the conditions for diffraction pattern maxima and minima are opposite to that for interference patterns.
Diffraction is the phenomenon that denotes the wave-like properties of electromagnetic waves. Whenever a wave encounters any objects, diffraction patterns are observed. Basically, light bends around these objects and this we term as diffraction.
Now, the first minima occur at 300{30^0}
So, from the formula:
a\sin {30^0} = (1) \times \lambda \\\ a = 2 \times \lambda \\\
In a diffraction pattern, a central maxima is obtained which is the brightest. Now, as we move away from the center the brightness of the maximas decreases. All other maximas are defined with respect to the central maxima. They are called the secondary maximas. This is why for the first secondary maxima we will have n=1.
Now, putting in the formula:
a\sin \theta = \dfrac{{(2n + 1)\lambda }}{2} \\\ 2\lambda .\sin \theta = \dfrac{{(2*1 + 1)}}{2}\lambda \\\ \sin \theta = \dfrac{3}{4} \\\ \theta = {\sin ^{ - 1}}(\dfrac{3}{4}) \\\
The correct option is (D).

Note: All waves show diffraction patterns on encountering the required conditions. Sound waves also diffract. This is a simple question, for application of Bragg's equation. The equation is also written as 2dsin⁡θ=nλ2d\sin \theta = n\lambda . The ‘a’ should not be confused with ‘d’. Here, ‘a’ represents slit width and so does ‘2d’.