What is the position and nature of the image formed by lens combination shown in the figure? (f1, f2 are focal lengths )

                                                 

Answers (1)

Thin lens formula -

\frac{1}{v}-\frac{1}{u}= \frac{1}{f}

- wherein

u \, and \, v are object and image distance from the lens.

For the first lens

u_{1}=-20,\, \, \, f_{1}=5cm

so, \frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f}

\frac{1}{v}-\frac{1}{-\left ( 20 \right )}=\frac{1}{5}

v_{1}=\frac{20}{3}

So  u_{2}  for 2nd  lens is given by

So  u_{2}=\frac{20}{3}-2=\frac{14}{3}

f_{2}=-5cm

So  \frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}

\frac{1}{v_{2}}-\frac{1}{\left ( 1_{\frac{4}{3}} \right )}=\frac{1}{-5}

V_{2}=70cm   and the image is real.

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