JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Wave Optics

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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 ____.

Answer: 36

Solution

The angular width of the central maximum in single-slit diffraction spans the first minimum on each side,

θ = \frac{2 λ}{d}

. Substituting

λ = 628 \times 1 0^{- 9}

m and

d = 0.2 \times 1 0^{- 3}

θ = \frac{2 ( 628 \times 1 0 ^{- 9} )}{0.2 \times 1 0 ^{- 3}} = 6.28 \times 1 0^{- 3}

rad. Converting to degrees,

θ = 6.28 \times 1 0^{- 3} \times \frac{180}{π} = 0.3 6^{\circ} = 36 \times 1 0^{- 2}

degrees. The value of

α

36

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Integer answer

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