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A concave lens made of material of refractive index $1.6$ is immersed in a medium of refractive index $2.0$. The two surfaces of the concave lens have the same radius of curvature $0.2 \mathrm{~m}$. The lens will behave as a
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Convergent lens of focal length $0.5 \mathrm{~m}$

$\frac{1}{F}=\left(\frac{1.6}{2}-1\right)\left(\frac{1}{-0.2}-\frac{1}{0.2}\right)$
$=\frac{0.4}{2} \times \frac{1}{0.1}$
$F=0.5$ converging lens
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