Wave Optics

JEE Main level, numerical answer, medium

125 of 191

Question 125 of 191

In single slit diffraction pattern, the wavelength of light used is 628628 nm and slit width is 0.20.2 mm, the angular width of central maximum is α×10−2\alpha \times 10^{-2} degrees. The value of α\alpha is _______.

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Answer: 36

The angular width of the central maximum in a single slit diffraction pattern is given by Δθ=2λa\Delta \theta = \dfrac{2\lambda}{a}

Substituting the given values:
λ=628×10−9\lambda = 628 \times 10^{-9} m
a=0.2×10−3a = 0.2 \times 10^{-3} m

Δθ=2×628×10−90.2×10−3\Delta \theta = \dfrac{2 \times 628 \times 10^{-9}}{0.2 \times 10^{-3}} rad

Δθ=6280×10−6\Delta \theta = 6280 \times 10^{-6} rad =6.28×10−3= 6.28 \times 10^{-3} rad

To convert radians to degrees, multiply by 180π\dfrac{180}{\pi}:

Δθ=6.28×10−3×1803.14\Delta \theta = 6.28 \times 10^{-3} \times \dfrac{180}{3.14} degrees

Δθ=2×10−3×180\Delta \theta = 2 \times 10^{-3} \times 180 degrees

Δθ=360×10−3\Delta \theta = 360 \times 10^{-3} degrees =36×10−2= 36 \times 10^{-2} degrees

Comparing this with α×10−2\alpha \times 10^{-2} degrees, we get α=36\alpha = 36.

Answer: 3636

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