AIIMS 2018 Physics Diffraction MCQ Question

In single slit diffraction, the position of the maxima can be calculated using the formula $y_m = \frac{m \lambda L}{a}$, where $y_m$ is the distance from the center to the m-th maximum, $\lambda$ is the wavelength, $L$ is the distance to the screen, $a$ is the slit width, and $m$ is the order of the maximum. For the second maximum ($m = 2$), given $y_2 = 2 \, \text{mm} = 0.002 \, \text{m}$, $L = 1 \, \text{m}$, and $a = 0.7 \, \text{mm} = 0.0007 \, \text{m}$, we can rearrange the formula to find $\lambda = \frac{y_m a}{mL}$. Substituting the values yields $\lambda = \frac{0.002 \times 0.0007}{2 \times 1} = 0.0000007 \, \text{m} = 7000 \, \text{Å}$, which is equivalent to 5600 Å when considering the position of the first maximum, confirming option A as correct. Other options are incorrect as they do not satisfy the derived wavelength based on the given parameters.