In a single slit diffraction experiment, the | vedclass

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(A) For a single slit diffraction experiment, the condition for the $n^{th}$ minimum is given by $a \sin 	heta = n \lambda_1$, where $n = 1, 2, 3, ..

Vedclass · Accepted Answer

$4400$

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(A) For a single slit diffraction experiment, the condition for the $n^{th}$ minimum is given by $a \sin 	heta = n \lambda_1$, where $n = 1, 2, 3, ...$.For the first minimum $(n=1)$, the position is $a \sin 	heta = 1 	imes \lambda_1 = 660 \ nm$.The condition for the $m^{th}$ secondary maximum is given by $a \sin 	heta = (2m + 1) \frac{\lambda_2}{2}$, where $m = 1, 2, 3, ...$.For the first secondary maximum $(m=1)$, the position is $a \sin 	heta = (2(1) + 1) \frac{\lambda_2}{2} = \frac{3}{2} \lambda_2$.Since the positions coincide, we equate the two expressions:$660 \ nm = \frac{3}{2} \lambda_2$.Solving for $\lambda_2$:$\lambda_2 = 660 	imes \frac{2}{3} = 440 \ nm$.Converting to $\mathring{A}$s $(\mathring{A})$:$440 \ nm = 440 	imes 10 \mathring{A} = 4400 \mathring{A}$.

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

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