In the figure- an air lens of radii of curvature 10 cmR1-R2-10 cm is cut in a cylinder ofglass -1-5- The focal length and the nature of the lens is

The correct option is C 15 cm, concave 1f=(_gμ_a 1) (1R_1 1R_2)=(23 1 ) (210) ⇒ f= 15 cm , so behaves as concave lens. ...

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Byju's Answer

Standard XII

Physics

Moving Object, Stationary Lens

In the figure...

Question

In the figure, an air lens of radii of curvature 10 $c m (R_{1} = R_{2} = 10 c m)$ is cut in a cylinder of
glass $(μ = 1.5)$ . The focal length and the nature of the lens is

15 c m,

concave

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15 c m,

convex

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\infty,

neither concave nor convex

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0,

concave

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Solution

The correct option is C $15 c m,$ concave
$\frac{1}{f} = (_{g} μ_{a} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = (\frac{2}{3} - 1) (\frac{2}{10})$
$\Rightarrow f = - 15 c m$ , so behaves as concave lens.

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In the figure, an air lens of radii of curvature 10 cm(R1=R2=10cm) is cut in a cylinder ofglass (μ=1.5). The focal length and the nature of the lens is

The correct option is C 15cm, concave1f=(gμa−1)(1R1−1R2)=(23−1)(210)⇒f=−15 cm, so behaves as concave lens.

In the figure, an air lens of radii of curvature 10 $c m (R_{1} = R_{2} = 10 c m)$ is cut in a cylinder of
glass $(μ = 1.5)$ . The focal length and the nature of the lens is

The correct option is C $15 c m,$ concave
$\frac{1}{f} = (_{g} μ_{a} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = (\frac{2}{3} - 1) (\frac{2}{10})$
$\Rightarrow f = - 15 c m$ , so behaves as concave lens.