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(A) The condition for the $n^{	ext{th}}$ secondary maximum in a single slit diffraction pattern is given by $y_n = (n + \frac{1}{2}) \frac{\lambda D}

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$4300$

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(A) The condition for the $n^{	ext{th}}$ secondary maximum in a single slit diffraction pattern is given by $y_n = (n + \frac{1}{2}) \frac{\lambda D}{a}$.For the $2^{	ext{nd}}$ secondary maximum of red light $(\lambda_1 = 6000 	ext{ Å})$, the position is $y_2 = (2 + \frac{1}{2}) \frac{\lambda_1 D}{a} = 2.5 \frac{\lambda_1 D}{a}$.For the $3^{	ext{rd}}$ secondary maximum of unknown wavelength $\lambda_2$, the position is $y_3 = (3 + \frac{1}{2}) \frac{\lambda_2 D}{a} = 3.5 \frac{\lambda_2 D}{a}$.Since the maxima coincide, $y_2 = y_3$, which implies $2.5 \lambda_1 = 3.5 \lambda_2$.Substituting the values: $2.5 	imes 6000 = 3.5 	imes \lambda_2$.$\lambda_2 = \frac{2.5 	imes 6000}{3.5} = \frac{15000}{3.5} \approx 4285.7 	ext{ Å} \approx 4300 	ext{ Å}$.

$A$ single slit diffraction pattern is formed with white light. For what wavelength of light, the $3^{\text{rd}}$ secondary maximum in the diffraction pattern coincides with the $2^{\text{nd}}$ secondary maximum in the pattern of red light of wavelength $6000 \text{ Å}$ (in $\text{ Å}$)?

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