Question

Focal length of plane concave lens is

Answer 1

1fL=(μ−1)[1∞−1−R]1fL=(μ−1)[1∞−1−R]

=μ−1R=μ−1R and fm=(−R)2fm=(−R)2

So, PL=1fL=(μ−1)RPL=1fL=(μ−1)R

and PM=−1fm=2RPM=−1fm=2R

and hence power of system

P=PL+Pm+PLP=PL+Pm+PL

=2PL+PM=2PL+PM

P=2(μ−1)R+2RP=2(μ−1)R+2R

=2μR=2μR

∴F=−1P=−R2μ∴F=−1P=−R2μ

ie the lens will be equivalent to a converging mirror of focal length (R2y)(R2y)

Hence b is the correct answer.