A) \[1:1\]
B) \[1:2\]
C) \[2:1\]
D) \[2:\frac{1}{2}\]
Correct Answer: A
Solution :
Focal length of lens \[\frac{1}{f}=(\mu -1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] After cutting lens along a plane perpendicular to principal axis, we get two plano-convex lens. For them\[{{R}_{1}}=R,\,\,{{R}_{2}}=\infty \]. \[\therefore \]New focal length\[\frac{1}{{{f}_{1}}}=(\mu -1)\left( \frac{1}{R}-\frac{1}{\infty } \right)\] \[\frac{1}{{{f}_{1}}}=\frac{(\mu -1)}{R}\] ? (i) For second part \[\frac{1}{{{f}_{2}}}=(\mu -1)\left( \frac{1}{\infty }+\frac{1}{R} \right)\] \[\frac{1}{{{f}_{2}}}=\frac{(\mu -1)}{R}\] ... (ii) From Eqs. (i) and (ii),\[{{f}_{1}}:{{f}_{2}}=1:1\]
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