A thin convex lens of focal length 50 cm is cut into two pieces 0.5 cm above the principal axis as shown. The parts are now placed on the x-axis. The x co-ordinate of the image of an object placed at (–100, 0) after refraction at both the parts is

$(\frac{355}{6}, - \frac{1}{8})$

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$(\frac{175}{6}, - \frac{1}{8})$

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$(\frac{175}{6}, - \frac{5}{8})$

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$(\frac{355}{6}, - \frac{5}{8})$

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Solution

The correct option is A
$(\frac{355}{6}, - \frac{1}{8})$

Using lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Principal axis for $L_{1}$ is 0.5 cm below x-axis and 0.5 cm above for $L_{2}$
For $L_{1}$
v = 100 cm
For $L_{2}$
v = $\frac{175}{6}$

Therefore image will be at $\frac{175}{6} + 30 = \frac{355}{6}$

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A thin convex lens of focal length 50 cm is cut into two pieces 0.5 cm above the principal axis as shown. The parts are now placed on the x-axis. The x co-ordinate of the image of an object placed at (–100, 0) after refraction at both the parts is

The correct option is A (3556,−18) Using lens formula 1v−1u=1f Principal axis for L1 is 0.5 cm below x-axis and 0.5 cm above for L2 For L1 v = 100 cm For L2 v = 1756 Therefore image will be at 1756+30=3556