Answer
a) $62.66^\circ$
b) $30.14^\circ$
Work Step by Step
a) In vertical diffraction, we need to use the vertical height of the area which is $1.5\;\mu\rm m$.
And to find the angle between the two first minima, we need to find the angle between the central maxima and the first minima and then double it since $\Delta \theta_{1\; to\;1}=\theta_1$.
$$\sin\theta_1=\dfrac{\lambda}{D_{\rm vertical}}$$
Thus,
$$\theta_1=\sin^{-1}\left[\dfrac{\lambda}{D_{\rm vertical}}\right]=\sin^{-1}\left[\dfrac{780\times 10^{-9}}{1.5\times 10^{-6}}\right]=31.33^\circ$$
Therefore,
$$\Delta \theta_{1\;to\;1}=2\theta_1=2\cdot 31.33=\color{red}{\bf 62.66^\circ}$$
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b) In horizontal diffraction, we need to use the horizontal length of the area which is $3\;\mu\rm m$.
And to find the angle between the two first minima, we need to find the angle between the central maxima and the first minima and then double it since $\Delta \theta_{1\; to\;1}=\theta_1$.
$$\sin\theta_1=\dfrac{\lambda}{D_{\rm horizontal}}$$
Thus,
$$\theta_1=\sin^{-1}\left[\dfrac{\lambda}{D_{\rm horizontal}}\right]=\sin^{-1}\left[\dfrac{780\times 10^{-9}}{3\times 10^{-6}}\right]=15.07^\circ$$
Therefore,
$$\Delta \theta_{1\;to\;1}=2\theta_1=2\cdot 31.33=\color{red}{\bf 30.14^\circ}$$
