JEE MAIN - Physics (2024 - 29th January Evening Shift - No. 23)

In a single slit diffraction pattern, a light of wavelength 6000$$\mathop A\limits^o$$ is used. The distance between the first and third minima in the diffraction pattern is found to be $$3 \mathrm{~mm}$$ when the screen in placed $$50 \mathrm{~cm}$$ away from slits. The width of the slit is _________ $$\times 10^{-4} \mathrm{~m}$$.
Answer
2

Explanation

For $$\mathrm{n}^{\text {th }}$$ minima

$$\mathrm{b} \sin \theta=\mathrm{n} \lambda$$

($$\lambda$$ is small so $$\sin \theta$$ is small, hence $$\sin \theta \simeq \tan \theta$$)

$$\mathrm{btan} \theta=\mathrm{n} \lambda$$

$$\mathrm{b} \frac{\mathrm{y}}{\mathrm{D}}=\mathrm{n} \lambda$$

$$\Rightarrow \mathrm{y}_{\mathrm{n}}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{b}}\left(\text { Position of } \mathrm{n}^{\mathrm{th}}\right. \text { minima) }$$

JEE Main 2024 (Online) 29th January Evening Shift Physics - Wave Optics Question 24 English Explanation

$$\begin{aligned} & \mathrm{B} \rightarrow 1^{\text {st }} \text { minima, } \mathrm{A} \rightarrow 3^{\mathrm{rd}} \text { minima } \\ & \mathrm{y}_3=\frac{3 \lambda \mathrm{D}}{\mathrm{b}}, \mathrm{y}_1=\frac{\lambda \mathrm{D}}{\mathrm{b}} \\ & \Delta \mathrm{y}=\mathrm{y}_3-\mathrm{y}_1=\frac{2 \lambda \mathrm{D}}{\mathrm{b}} \\ & 3 \times 10^{-3}=\frac{2 \times 6000 \times 10^{-10} \times 0.5}{\mathrm{~b}} \\ & \mathrm{~b}=\frac{2 \times 6000 \times 10^{-10} \times 0.5}{3 \times 10^{-3}} \\ & \mathrm{~b}=2 \times 10^{-4} \mathrm{~m} \\ & x=2 \end{aligned}$$

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