Light of wavelength 589.3 , nm is incident no | vedclass

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(A) The angular width of the central diffraction maximum for a single slit is given by the formula: $\beta = \frac{2\lambda}{a}$,where $\lambda$ is th

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$0.68$

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(A) The angular width of the central diffraction maximum for a single slit is given by the formula: $\beta = \frac{2\lambda}{a}$,where $\lambda$ is the wavelength and $a$ is the slit width.Given: $\lambda = 589.3 \, nm = 589.3 	imes 10^{-9} \, m$ and $a = 0.1 \, mm = 0.1 	imes 10^{-3} \, m$.Substituting the values: $\beta = \frac{2 	imes 589.3 	imes 10^{-9}}{0.1 	imes 10^{-3}} \, rad$.$\beta = 2 	imes 589.3 	imes 10^{-5} \, rad = 1178.6 	imes 10^{-5} \, rad = 0.011786 \, rad$.To convert radians to degrees,multiply by $\frac{180}{\pi}$:$\beta_{deg} = 0.011786 	imes \frac{180}{3.14159} \approx 0.675^\circ \approx 0.68^\circ$.

Light of wavelength $589.3 \, nm$ is incident normally on a slit of width $0.1 \, mm$. What will be the angular width of the central diffraction maximum in degrees?

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