• # question_answer Direction: A point object O is placed at the origin of coordinate system. An equi-convex thin lens$({{\mu }_{g}}=1.5)$ of focal length $f=20\,cm$in air is placed so that its principal axis is along x-axis. Now the lens is cut at the middle (along the principal axis) and upper half is shifted along x-axis and y-axis by 20 cm and 2 mm respectively and right side of lower half is filled with water $({{\mu }_{\omega }}=4/3)$ Coordinates of the image produced by the lens ${{L}_{1}}$ will be A) $\left( \frac{320}{3}cm,\frac{4}{3}mm \right)$    B) $\left( \frac{160}{3}cm,\frac{8}{3}mm \right)$ C) $\left( \frac{320}{3}cm,\frac{8}{3}mm \right)$  D) $\left( \frac{160}{3}cm,\frac{4}{3}mm \right)$

Correct Answer: C

Solution :

 [c] Given, $f=20\text{ }cm$ $\therefore$      $m=\frac{v}{u}$ Now,     $u=-80\,cm$ $\therefore$            $v=\frac{uf}{u+f}=\frac{80}{3}\,cm$ $\left( \text{As},\text{ lens formula is}\,\,\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \right)$
 $\therefore$      x-coordinate $=80+\frac{80}{3}=\frac{320}{3}\,cm$ $\therefore$      y-coordinatemm $=\left( \frac{2}{3}+2 \right)mm=\frac{8}{3}\,mm$ Hence, coordinates of image formed by ${{L}_{1}}=\left( \frac{320}{3}cm,\frac{8}{3}mm \right)$

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