The half angular width of the central maxima | vedclass

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Question

(A) In Fraunhofer diffraction,the condition for the first minimum (which defines the half angular width of the central maximum) is given by $d \sin

Vedclass · Accepted Answer

$600$

Vedclass · Answer

(A) In Fraunhofer diffraction,the condition for the first minimum (which defines the half angular width of the central maximum) is given by $d \sin 	heta = n \lambda$.For the first minimum,$n = 1$,so $d \sin 	heta = \lambda$.Given slit width $d = \frac{1200}{\sqrt{2}} \ \mu m$ and half angular width $	heta = 45^o$.Substituting the values into the formula:$\lambda = \left( \frac{1200}{\sqrt{2}} ight) \sin(45^o)$Since $\sin(45^o) = \frac{1}{\sqrt{2}}$,we have:$\lambda = \left( \frac{1200}{\sqrt{2}} ight) 	imes \left( \frac{1}{\sqrt{2}} ight)$$\lambda = \frac{1200}{2} = 600 \ \mu m$.

The half angular width of the central maxima in the Fraunhofer diffraction due to a slit of width $\frac{1200}{\sqrt{2}} \ \mu m$ is $45^o$. Then the wavelength of the light is ...... $\mu m$.

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