JEE Main 2026 — Apr 2, Shift 1 — Physics solution

JEE Main2026Apr 2, Shift 1Physics

In single slit diffraction pattern, the wavelength of light used is $628$ nm and slit width is $0.2$ mm, the angular width of central maximum is $α \times 1 0^{- 2}$ degrees. The value of $α$ is _____.

Your answer

Solution

The angular width of the central maximum in single-slit diffraction is $θ_{c m} = \frac{2 λ}{d}$ .

$θ_{c m} = \frac{2 \times 628 \times 1 0 ^{- 9}}{0.2 \times 1 0 ^{- 3}} = 6.28 \times 1 0^{- 3}$ rad

Converting to degrees by multiplying by $180/ π$ :

$θ_{c m} = 6.28 \times 1 0^{- 3} \times \frac{180}{π} \approx 0.3 6^{\circ} = 36 \times 1 0^{- 2}$ degrees

$∴ α = 36$

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