Light with a wavelength of 410 nm shines through a lens with an aperture diameter of 0.60 cm.Use Rayleigh's criterion to determine the limiting angle of resolution.(1 nm = 10−⁹ m) A)8.3 × 10−⁵ rad B)2.8 × 10−⁹ rad C)1.8 × 10−⁴ rad D)7.9 × 10−⁷ rad E)0.6 × 10−⁷ rad

Question

Quizplus · Accepted Answer

Rayleigh's criterion for the limit of resolution $\theta$ is given by $\theta = 1.22 \frac{\lambda}{D}$, where $\lambda$ is the wavelength of light and $D$ is the diameter of the aperture. Given $\lambda = 410 \times 10^{-9}$ m (or $410 \, \text{nm}$) and $D = 0.60 \times 10^{-2}$ m (or $0.60 \, \text{cm}$), substituting these values gives $\theta = 1.22 \times \frac{410 \times 10^{-9}}{0.60 \times 10^{-2}} = 8.3 \times 10^{-5}$ rad.

Light with a Wavelength of 410 Nm Shines Through a Lens