Diffraction by a Circular Aperture — Question 3

Question 3

Numerical aperture combines immersion index and collection angle

Use the object-space Rayleigh estimate $s_R=0.61\lambda_0/(n\sin\alpha)$, with vacuum wavelength $\lambda_0=600\,\mathrm{nm}$. A microscope collects to half-angle $\alpha=30^\circ$. Compare air ($n=1$) with an immersion medium ($n=1.50$) at the same angle.

1. Calculate the numerical aperture and resolution in both cases.

2. Find the air collection half-angle needed to match the immersion resolution, if geometrically possible.

Question 3 - Solution

Numerical aperture combines immersion index and collection angle

(a) Method and calculation

$$\mathrm{NA}=n\sin\alpha\Longrightarrow\boxed{\mathrm{NA}_{\mathrm{air}}=0.50,\quad\mathrm{NA}_{\mathrm{imm}}=0.75}.$$$$\boxed{s_{R,\mathrm{air}}=732\,\mathrm{nm},\quad s_{R,\mathrm{imm}}=488\,\mathrm{nm}}.$$

(b) Interpretation and further calculation

$$\mathrm{NA}_{\mathrm{air,new}}=0.75\Longrightarrow\sin\alpha_{\mathrm{new}}=0.75.$$$$\boxed{\alpha_{\mathrm{new}}=48.5904^\circ}.$$

Physical check

The wavelength in the medium is shorter, but using numerical aperture already includes the index factor; it must not be counted twice.