JEE MAIN - Physics (2024 - 29th January Evening Shift - No. 23)
Explanation
For $$\mathrm{n}^{\text {th }}$$ minima
$$\mathrm{b} \sin \theta=\mathrm{n} \lambda$$
($$\lambda$$ is small so $$\sin \theta$$ is small, hence $$\sin \theta \simeq \tan \theta$$)
$$\mathrm{btan} \theta=\mathrm{n} \lambda$$
$$\mathrm{b} \frac{\mathrm{y}}{\mathrm{D}}=\mathrm{n} \lambda$$
$$\Rightarrow \mathrm{y}_{\mathrm{n}}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{b}}\left(\text { Position of } \mathrm{n}^{\mathrm{th}}\right. \text { minima) }$$
$$\begin{aligned} & \mathrm{B} \rightarrow 1^{\text {st }} \text { minima, } \mathrm{A} \rightarrow 3^{\mathrm{rd}} \text { minima } \\ & \mathrm{y}_3=\frac{3 \lambda \mathrm{D}}{\mathrm{b}}, \mathrm{y}_1=\frac{\lambda \mathrm{D}}{\mathrm{b}} \\ & \Delta \mathrm{y}=\mathrm{y}_3-\mathrm{y}_1=\frac{2 \lambda \mathrm{D}}{\mathrm{b}} \\ & 3 \times 10^{-3}=\frac{2 \times 6000 \times 10^{-10} \times 0.5}{\mathrm{~b}} \\ & \mathrm{~b}=\frac{2 \times 6000 \times 10^{-10} \times 0.5}{3 \times 10^{-3}} \\ & \mathrm{~b}=2 \times 10^{-4} \mathrm{~m} \\ & x=2 \end{aligned}$$
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