In Young's double slit experiment, carried out with light of wavelength 5000 Å, the distance between the slits is 0.3 mm and the screen is at 200 cm from the slits. The central maximum is at x = 0 cm. The value of x for third maxima is __________ mm. Numerical answer type. Work it out, then open the answer.

Answer: 10. x₃ = (3λ D)/(d) x₃ = 3 × 5×10⁻⁷ × 20.3×10⁻³ = 1×10⁻² m So x = 10 mm.

In Young's double slit experiment, carried out with light of… | JEE Main 2024 (5 Apr, Shift 1)

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YDSE: Fringe Width & Fringe Positions

In Young's double slit experiment, carried out with light of wavelength $5000\ \text{Å}$ , the distance between the slits is $0.3\text{ mm}$ and the screen is at $200\text{ cm}$ from the slits. The central maximum is at $x = 0\text{ cm}$ . The value of $x$ for third maxima is __________ $\text{mm}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 10

$x_3 = \frac{3\lambda D}{d}$
$x_3 = \frac{3 \times 5\times10^{-7} \times 2}{0.3\times10^{-3}} = 1\times10^{-2}\text{ m}$
So $x = 10$ mm.

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