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$I(\theta)= I_{max} \left[ \dfrac{\sin\left(\frac{\pi a \sin \theta}{\lambda}\right)}{\frac{\pi a \sin \theta}{\lambda}}\right]^2$ where $a$ is the size of aperture. This is the diffraction pattern, now you can easily see that the minimum of diffraction are where $\sin \theta = m \lambda /a$. So as you say the angular width of central maxima is \$2 \lambda ...