A) enlarged and reverse
B) small and erect
C) errect and same size of object
D) reverse and same size of object
Correct Answer: C
Solution :
Equivalent focal length of combination \[\frac{1}{F}=\frac{1}{{{F}_{1}}}=\frac{1}{{{F}_{2}}}=\frac{1}{20}-\frac{1}{20}=0\] \[F=\infty \] Combination will behave as a plane mirror, size of image will be equal to size of object.You need to login to perform this action.
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