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Question
An equiconvex lens of focal length 'f' is cut into two identical plane convex lenses. How will the power of each part be related to the focal length of the original lens ?
Solution
\[\text{The focal length of original equiconvex lens is f } . \]
\[\text { Let the focal length of each part after cutting be F } . \]
\[\text { Here }, \]
\[\frac{1}{f} = \frac{1}{F} + \frac{1}{F}\]
\[ \Rightarrow \frac{1}{f} = \frac{2}{F}\]
\[ \Rightarrow f = \frac{F}{2}\]
\[ \Rightarrow F = 2f\]
\[\text { Power of each part will be given by }\]
\[P = \frac{1}{F}\]
\[ \Rightarrow P = \frac{1}{2f}\]
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