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Let the radius of curvature of the concave part of the lens be R. Then when using the lens makers Equation, $$ \cfrac{1}{f} = \left(\cfrac{\mu_2}{\mu_1} - 1\right)\left(\cfrac{1}{R_1} - \cfrac{1}{R_2}\right)$$ $ R_2$ is clearly $\infty$ but is $R_1 = -R$ (as convention would assume, since the focal point is ont the left) or is $R_1 = R$.

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  • $\begingroup$ You can assume two cases when the planar face is on the left, and the one where it is on the right. Both will give the same answer, given you follow the sign conventions. $\endgroup$ – Eagle Feb 27 at 15:13
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R1 is the Radius of curvature of first surface and R2 of the second. Keeping Plane side as first surface $\frac{1}{R1}$=0 while R2=-R. Keeping Concave side as first surface , $\frac{1}{R2}$=0 while R1=+R. Where R is radius of curvature of concave part.

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That depend how you place, the lens and your sign convention let say we have this lens enter image description here, Let you have surface ET draw a circle containing ET part taking CT as a line passes through origin, you can see it is on left hand side of origin, hence you can take $R_1=(-)$, as $R_2$ is infinite.

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Yes you are correct, this formula is correct for piano convex and you can proceed with the general equation for refraction for both surfaces separately, then combining them you will surely reach there.

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    $\begingroup$ This isn't a yes or no question- they are asking if $R_1=R$ or $-R$. $\endgroup$ – Chris Oct 10 at 15:08

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