in \theta = n \lambda $$

where $a$ is the width of the slit, $\theta$ is the angle of diffraction, $\lambda$ is the wavelength of light, and $n = 1, 2, 3, \ldots$ for the first, second, third minima, respectively.

For the first minimum ($n=1$), the given angle is $30^{\circ}$. So,

$$ a \sin 30^{\circ} = 1 \cdot \lambda $$

We are given $\sin 30^{\circ} = 0.5$.

$$ a (0.5) = \lambda $$

$$ a = \frac{\lambda}{0.5} $$

$$ a = 2\lambda \quad \ldots(1) $$

2. Condition for the first secondary maximum:

The condition for the $n$-th secondary maximum (other than the central maximum) is given by:

$$ a \sin \theta' = \left(n + \frac{1}{2}\right) \lambda $$

or equivalently,

$$ a \sin \theta' = \frac{(2n+1)}{2} \lambda $$

where $n = 1, 2, 3, \ldots$ for the first, second, third secondary maxima, respectively.

For the first secondary maximum ($n=1$), let the angle be $\theta'$.

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

$$ a \sin \theta' = \frac{3}{2} \lambda \quad \ldots(2) $$

3. Calculate the angle for the first secondary maximum:

Now, substitute the value of $a$ from equation (1) into equation (2):

$$ (2\lambda) \sin \theta' = \frac{3}{2} \lambda $$

Since $\lambda$ is non-zero, we can cancel it from both sides:

$$ 2 \sin \theta' = \frac{3}{2} $$

Solve for $\sin \theta'$:

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

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

Therefore, the angle at which the first secondary maximum is observed is:

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

The correct option is D.

Wave Optics

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