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Q.
A double convex lens of focal length f is cut into 4 equivalent parts. One cut is perpendicular to the axis and the other is parallel to the axis of the lens. The focal length of each part is
Solution:
By lens maker formula.
$$
\frac{1}{ f }=(\mu-1)\left(\frac{1}{ R _{1}}-\frac{1}{ R _{2}}\right)
$$
So, cutting along the axis does not change the focal length but cutting perpendicular to the axis change radius of curvature of one side so, focal length changes.
So, for a double convex lens, we get $R _{1}= R _{2}= R$ )
$$
\begin{array}{l}
\frac{1}{ f }=(\mu-1)\left[\frac{1}{ R }-\left(-\frac{1}{ R }\right)\right] \\
\Rightarrow \frac{1}{ f }=(\mu-1)\left[\frac{2}{ R }\right] \Rightarrow f =\frac{ R }{2(\mu-1)}
\end{array}
$$
Now as it is cut perpendicular to axis $R _{2} \rightarrow 0$
$$
\begin{array}{l}
\text { So, } \Rightarrow \frac{1}{ f }=(\mu-1)\left[\frac{1}{ R }-0\right] \\
\Rightarrow \frac{1}{ f _{1}}=(\mu-1) \frac{1}{ R } \Rightarrow f _{1}=\frac{ R }{(\mu-1)} \\
\therefore f _{1}=2 f
\end{array}
$$
So, focal length of each part as $2 f$.