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Physics Question 46 – JEE-MAIN 2026

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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×10−2 degrees. The value of α is _______.

The central maximum in a single-slit diffraction pattern is bounded by the first minima on either side.

Step 1: Identify the formula for the angular position of the first minimum✦ Active

For single-slit diffraction, the first minimum occurs at an angle θ such that asin⁡θ=λ. For small angles, sin⁡θ≈θ (in radians), so the angular position of the first minimum is θ=λa.

Step 2: Calculate the angular width of the central maximum in radians○ Expand

The angular width of the central maximum is 2θ. Substitute the given values: wavelength λ=628 nm=628×10−9 m and slit width a=0.2 mm=0.2×10−3 m.

2θ=2×λa=2×628×10−9 m0.2×10−3 m=2×3140×10−6=6280×10−6=6.28×10−3 radians
💡 Teacher's Secret Hint

Ensure all units are consistent (e.g., SI units) before calculation.

Step 3: Convert the angular width to degrees and find α○ Expand

Convert the angular width from radians to degrees using the conversion factor 1 radian=180π degrees. Using π≈3.14:

2θdegrees=(6.28×10−3)×180π≈(6.28×10−3)×1803.14=(2×3.14×10−3)×1803.14=2×10−3×180=360×10−3=0.36 degrees

The angular width is given as α×10−2 degrees. Comparing 0.36 degrees with α×10−2 degrees:

0.36=α×10−2⟹α=0.3610−2=0.36×100=36
💡 Teacher's Secret Hint

Pay attention to the required format of the final answer, which includes a power of 10.

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