In a single slit diffraction pattern, a light of wavelength 6000 is used. The distance between the first and third minima in the diffraction pattern is found to be 3 mm when the screen in placed 50 cm away from slits. The width of the slit is _________ × 10⁻⁴ m. Numerical answer type. Work it out, then open the answer.

Answer: 2. Minima on the screen: yn = (nλ D)/(a) y₃ - y₁ = (2λ D)/(a) = 3 × 10⁻³ m a = 2 × 6 × 10⁻⁷ × 0.53 × 10⁻³ = 2 × 10⁻⁴ m So the answer is 2 (i.e. 2 × 10⁻⁴ m).

In a single slit diffraction pattern, a light of wavelength 6000 is… | JEE Main 2024 (29 Jan, Shift 2)

Q33JEE Main 2024 · 29 JAN · SHIFT 2Medium

Diffraction at a Single Slit

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

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

Show answer and solution

Answer: 2

Minima on the screen: $y_n = \dfrac{n\lambda D}{a}$
$y_3 - y_1 = \dfrac{2\lambda D}{a} = 3 \times 10^{-3}$ m
$a = \dfrac{2 \times 6 \times 10^{-7} \times 0.5}{3 \times 10^{-3}} = 2 \times 10^{-4}$ m
So the answer is $2$ (i.e. $2 \times 10^{-4}$ m).

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