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 $0.2$ m. The lens will behave as a.

Divergent lens of focal length

0.4

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Divergent lens of focal length

0.5

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Convergent lens of focal length

0.4

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Convergent lens of focal length

0.5

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Solution

The correct option is B Divergent lens of focal length $0.5$ m
Power= $\frac{μ_{l e n s} - μ}{μ} (\frac{1}{r} - \frac{1}{r_{1}})$
Power= $\frac{1.6 - 2}{2} (\frac{1}{20} - \frac{- 1}{20}) = - 50 c m$
Divergent lens with focal length 0.5m.

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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 0.2m. The lens will behave as a.

The correct option is B Divergent lens of focal length 0.5mPower=μlens−μμ(1r−1r1)Power=1.6−22(120−−120)=−50cmDivergent lens with focal length 0.5m.

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 $0.2$ m. The lens will behave as a.

The correct option is B Divergent lens of focal length $0.5$ m
Power= $\frac{μ_{l e n s} - μ}{μ} (\frac{1}{r} - \frac{1}{r_{1}})$
Power= $\frac{1.6 - 2}{2} (\frac{1}{20} - \frac{- 1}{20}) = - 50 c m$
Divergent lens with focal length 0.5m.