Question

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

• A. $f/2$
• B. f
• C. 2f
• D. 4f

Answer 1

Answer: (C)

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 $R_2\to 0$ \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 $2f$.