For example, ns can twin the distance of focal size on the principle axis to uncover the facility of curvature that concave mirror.

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Yes, the radius the curvature is double the focal size from the pole (A in top diagram, ns in reduced diagram) : $R=2f$.You can achieve an approximate worth of $f$ by detect the point at i m sorry a remote object is focussed, as said by the 2nd diagram. Or friend can acquire $R$ together the street at which an object is focussed in ~ the same distance (inverted).You can acquire a much more accurate value of $f$ by plotting a graph of image vs object distances, in the form $1/v$ against $1/u$. The intercept is $1/f$.

A concave mirror used for focusing light is parabolic, not spherical. There thus isn"t a "center that curvature".

If you understand the thing distance and image distance, you deserve to solve because that the focal size f using the mirror equation. Climate you will additionally know the center of curvature due to the fact that it is 2 times f.

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What will certainly be the result on the image developed by a mirascope if spherical mirrors room used rather of parabolic mirrors?

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