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Feb 20 at 1:05 vote accept uhoh
Feb 14 at 22:09 comment added pela Observed angles of deflection are typically several arcmin. If they get much larger, the image would be too faint. So you could play around with different sets of $D_\mathrm{L}$ and $D_\mathrm{S}$ in @PM2Ring's equation to see what would give you such values of $\theta$. However, those distances are angular diameter distances, so you would need to convert to physical (comoving) distances to get your ratio.
Feb 14 at 18:05 answer added eshaya timeline score: 2
Feb 14 at 1:07 comment added PM 2Ring Relevant: en.wikipedia.org/wiki/Einstein_ring which has the equation $$\theta_1 = \sqrt{\frac{4GM}{c^2}\;\frac{D_{LS}}{D_S D_L}}$$
Feb 14 at 0:50 history asked uhoh CC BY-SA 4.0