In figure, an air lens of radius of curvature of each surface equal to $10 c m$ is cut into a cylinder of glass of refractive index $1.5.$ The focal length and the nature of lens are

15 c m

diverging

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15 c m

converging

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10 c m

diverging

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10 c m

converging

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Solution

The correct option is A $15 c m$ diverging
$\frac{1}{F} = (\frac{μ_{a i r}}{μ_{g}} - 1) (\frac{1}{R_{1}} + \frac{1}{R_{2}})$
$= \frac{0.5}{1.5} \times \frac{2}{10}$
$F = 15 c m$

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In figure, an air lens of radius of curvature of each surface equal to 10cm is cut into a cylinder of glass of refractive index 1.5. The focal length and the nature of lens are

The correct option is A 15cm diverging1F=(μairμg−1)(1R1+1R2)=0.51.5×210F=15cm

In figure, an air lens of radius of curvature of each surface equal to $10 c m$ is cut into a cylinder of glass of refractive index $1.5.$ The focal length and the nature of lens are

The correct option is A $15 c m$ diverging
$\frac{1}{F} = (\frac{μ_{a i r}}{μ_{g}} - 1) (\frac{1}{R_{1}} + \frac{1}{R_{2}})$
$= \frac{0.5}{1.5} \times \frac{2}{10}$
$F = 15 c m$