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(D) The condition for the $n^{th}$ interference maximum is $d \sin 	heta = n \lambda$. The condition for the first diffraction minimum is $a \sin 	h

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$9$

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(D) The condition for the $n^{th}$ interference maximum is $d \sin 	heta = n \lambda$. The condition for the first diffraction minimum is $a \sin 	heta = \lambda$,which corresponds to the edge of the central diffraction envelope. The angular position of the first diffraction minimum is $\sin 	heta = \frac{\lambda}{a}$. Substituting this into the interference condition: $d \left( \frac{\lambda}{a} ight) = n \lambda$,which gives $n = \frac{d}{a}$. Given $d = 20 \ \mu m$ and $a = 4 \ \mu m$,we get $n = \frac{20}{4} = 5$. This means the $5^{th}$ interference maximum coincides with the first diffraction minimum. The central diffraction envelope contains the central maximum $(n=0)$ and the interference maxima on both sides up to $n=4$. The total number of bright fringes is $2n_{max} + 1 = 2(4) + 1 = 9$.

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