In single slit diffraction pattern, the wavelength of light used is 628 nm and slit width is 0.2 mm

Question

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 $α \times 10^{- 2}$ degrees. The value of $α$ is ______. [2026]

Smart Achievers · Accepted Answer

(36)

$Wavelength (λ) = 628 nm = 628 \times 10^{- 9} m$

$Slit width a = 0.2 mm = 2 \times 10^{- 4} m$

$For Single Slit diffraction,$

$Angular width of central maxima θ = \frac{2 λ}{a}$

$θ = \frac{2 λ}{a} = \frac{2 \times 628 \times 10^{- 9}}{2 \times 10^{- 4}} = 6.28 \times 10^{- 3} radian$

$in degree = \frac{180}{π} ≃ 57.3 °$

$θ = 6.28 \times 10^{- 3} \times 57.3 ≃ 0.36 °$

$θ = 36 \times 10^{- 2} degree$

$θ = 6.28 \times 10^{- 3} \times 57.3 ≃ 0.36 °$

$\Rightarrow θ = 36$

Solution

Q.
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 $α \times 10^{- 2}$ degrees. The value of $α$ is ______. [2026]

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Solution

Q. 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 ______. [2026]

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Q.
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 $α \times 10^{- 2}$ degrees. The value of $α$ is ______. [2026]