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resources/links/Angular resolution.md
@@ -14,7 +14,7 @@
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It seems that in astronomy, there is a relationship between the image resolution $\theta$ and the wavelength $\lambda$ giving by
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-$$\sqrt{3x-1}+(1+x)^2$$
+$$\sqrt{3x-1} = (1+x)^2$$
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- Where $D$ is the diameter of the telescope.
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