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(A) For a single slit diffraction experiment, the condition for the $n^{th}$ minimum is given by $d \sin 	heta = n \lambda$.For the first minimum $(n

Vedclass · Accepted Answer

$4400$

Vedclass · Answer

(A) For a single slit diffraction experiment, the condition for the $n^{th}$ minimum is given by $d \sin 	heta = n \lambda$.For the first minimum $(n=1)$ of red light $(\lambda_R = 660 \, nm)$, the condition is $d \sin 	heta = \lambda_R$.The condition for the $n^{th}$ secondary maximum is given by $d \sin 	heta = (n + \frac{1}{2}) \lambda'$.For the first secondary maximum $(n=1)$, the condition is $d \sin 	heta = (1 + \frac{1}{2}) \lambda' = \frac{3}{2} \lambda'$.Since the first minimum of red light coincides with the first secondary maximum of $\lambda'$, we equate the two expressions:$\lambda_R = \frac{3}{2} \lambda'$$\lambda' = \frac{2}{3} \lambda_R = \frac{2}{3} 	imes 660 \, nm = 440 \, nm$.Converting to $\mathring{A}$: $440 \, nm = 4400 \, \mathring{A}$.

In a single slit diffraction experiment, the first minimum for red light $(\lambda = 660 \, nm)$ coincides with the first secondary maximum of some other wavelength $\lambda'$. The value of $\lambda'$ is ...... $\mathring{A}$.

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