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(B) The condition for the $n^{	ext{th}}$ secondary maximum in a single slit diffraction pattern is given by: $a \sin 	heta = (n + \frac{1}{2}) \lamb

Vedclass · Accepted Answer

$\frac{7 \lambda}{9}$

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(B) The condition for the $n^{	ext{th}}$ secondary maximum in a single slit diffraction pattern is given by: $a \sin 	heta = (n + \frac{1}{2}) \lambda'$,where $n$ is the order of the secondary maximum.For the $4^{	ext{th}}$ secondary maximum of wavelength $\lambda_1$,the condition is: $a \sin 	heta_1 = (4 + \frac{1}{2}) \lambda_1 = \frac{9}{2} \lambda_1$.For the $3^{	ext{rd}}$ secondary maximum of wavelength $\lambda$,the condition is: $a \sin 	heta_2 = (3 + \frac{1}{2}) \lambda = \frac{7}{2} \lambda$.Since the maxima coincide,$	heta_1 = 	heta_2$,so $a \sin 	heta_1 = a \sin 	heta_2$.Equating the two expressions: $\frac{9}{2} \lambda_1 = \frac{7}{2} \lambda$.Solving for $\lambda_1$: $\lambda_1 = \frac{7}{9} \lambda$.

$A$ single slit diffraction pattern is formed with white light. For what wavelength of light the $4^{\text{th}}$ secondary maximum in the diffraction pattern coincides with the $3^{\text{rd}}$ secondary maximum in the pattern of light of wavelength $\lambda$?

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