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JEE Main 2026 · 2 Apr · Shift 1, Physics question 46

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\alpha\times10^{-2} degrees. The value of α\alpha is ________.

Answer and worked solution

Official answer: 36

Angular width of the central maximum =2λa=2(628×10−9)2×10−4=6.28×10−3=\dfrac{2\lambda}a=\dfrac{2(628\times10^{-9})}{2\times10^{-4}}=6.28\times10^{-3} rad.

In degrees. 6.28×10−3×180π=0.36∘=36×10−26.28\times10^{-3}\times\dfrac{180}\pi=0.36^\circ=36\times10^{-2} degrees.

Answer: 36

Topics:Diffraction
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