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QJEEM22Ph4295: JEE Main 2022 Physics - Class 12 JEEM22Ph4295 Solution: Sodium light of wavelengths 650 nm and 655 nm is used to study…

Sodium light of wavelengths 650 nm and 655 nm is used to study diffraction at a single slit of aperture 0.5 mm. The distance between the slit and the screen is 2.0 m. The separation between the positions of the first maxima of diffraction pattern obtained in the two cases is ___________ $\times$ 10^{$-$ 5} m.

Step-by-Step Verified Solution

JEE Main 2022 (Online) 24th June Morning Shift Physics - Wave Optics Question 58 English Explanation

Condition for diffraction maximum is

a \sin \theta=(2 n+1) \frac{\pi}{2}

For first Maxima,

n=1, a \sin \theta=\frac{3 \lambda}{2}

Since

\theta

is very less, so

\begin{aligned} \sin θ \approx \tan θ \\ ∴ a \tan θ = \frac{3 λ}{2} \end{aligned}

\begin{array}{r} a (\frac{y}{D}) = \frac{3 λ}{2} \\ \Rightarrow y = \frac{3 λ D}{2 a} \end{array}

For the two cases given

\begin{aligned} y_{2} - y_{1} & = \frac{3 D}{2 a} (λ_{2} - λ_{1}) \\ = \frac{3}{2} \times \frac{2}{0 \cdot 5 \times 10^{- 3}} (655 - 650) \times 10^{- 9} \\ = 6 \times 10^{3} \times 5 \times 10^{- 9} \\ = 3 \times 10^{- 5} m \end{aligned}

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