Q:Spy planes fly at extremely high altitudes (25.3 km) to avoid interception. Their cameras are reportedly able to discern features as small as 5.20 cm. What must be the minimum aperture of the camera lens to afford this resolution? (Use lambda = 550 nm.) I first found theta.. .052m/2 = .026m (since .026 above and below the center = .052m) tan theta = .026m/25300m, and theta = 4.578x10^-5 Then plugging into the rayleigh criterion for resolution limit of theta = 1.22lamda/D, where D is the diameter of the objective lens (same as aperture of camera??)... I get theta = 1.22lamda/D 4.578x10^-5 = 1.22(550x10^-9)/D D= .0147m, which is incorrect.. anyone know where I went wrong? Thanks!