$$d \sin \theta = m\lambda$$

For the first minimum, $$m = 1$$ and we know that the angle is $$30^{\circ}$$. Using the given wavelength $$\lambda = 5500 \mathop A\limits^o = 550 \times 10^{-9}$$ meters (since $$1 \mathop A\limits^o = 10^{-10}$$ meters), we can write:

$$d \sin 30^{\circ} = 1 \times 550 \times 10^{-9} \text{ m}$$

$$d \times \frac{1}{2} = 550 \times 10^{-9} \text{ m}$$

$$d = 2 \times 550 \times 10^{-9} \text{ m}$$

$$d = 1100 \times 10^{-9} \text{ m}$$

$$d = 1100 \mathop A\limits^o$$

The secondary maxima occur in between the primary minima. The first secondary maxima (also known as the first 'bright' fringe other than the central maximum) occurs when the path difference is $$3/2$$ times the wavelength (this is the condition for the maximum that lies between the first and second minima, $$m = 1$$ and $$m = 2$$, respectively). This results in the following condition:

$$d \sin \theta = \left(m + \frac{1}{2}\right) \lambda$$

For the first secondary maxima $$m = 1$$:

$$d \sin \theta = \left(1 + \frac{1}{2}\right) 550 \times 10^{-9} \text{ m}$$

$$d \sin \theta = \frac{3}{2} \times 550 \times 10^{-9} \text{ m}$$

$$1100 \sin \theta = 3 \times 550 \times 10^{-9} \text{ m}$$

$$\sin \theta = \frac{3}{2} \times \frac{550 \times 10^{-9}}{1100 \times 10^{-9}}$$

$$\sin \theta = \frac{3}{2} \times \frac{1}{2}$$

$$\sin \theta = \frac{3}{4}$$

Therefore, the correct answer is:

Option B

$$\sin ^{-1}\left(\frac{3}{4}\right)$$

Wave Optics

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