A single slit Fraunhofer diffraction pattern | vedclass

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Question

(C) The condition for the $n^{th}$ secondary maximum in a single slit diffraction pattern is given by $x = \frac{(2n+1) \lambda D}{2a}$,where $n$ is t

Vedclass · Accepted Answer

$4642.8$

Vedclass · Answer

(C) The condition for the $n^{th}$ secondary maximum in a single slit diffraction pattern is given by $x = \frac{(2n+1) \lambda D}{2a}$,where $n$ is the order of the secondary maximum.For the second secondary maximum $(n=2)$ of red light $(\lambda_1 = 6500 \ \mathring{A})$,the position is:$x = \frac{(2 	imes 2 + 1) \lambda_1 D}{2a} = \frac{5 \lambda_1 D}{2a}$.For the third secondary maximum $(n=3)$ of light with wavelength $\lambda_2$,the position is:$x = \frac{(2 	imes 3 + 1) \lambda_2 D}{2a} = \frac{7 \lambda_2 D}{2a}$.Since the positions coincide,we equate them:$\frac{5 \lambda_1 D}{2a} = \frac{7 \lambda_2 D}{2a}$.This simplifies to:$5 \lambda_1 = 7 \lambda_2$.Substituting $\lambda_1 = 6500 \ \mathring{A}$:$\lambda_2 = \frac{5 	imes 6500}{7} = \frac{32500}{7} \approx 4642.8 \ \mathring{A}$.

$A$ single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light does the third secondary maximum in the diffraction pattern coincide with the second secondary maximum in the pattern for red light of wavelength $6500 \ \mathring{A}$?

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