In figure the points $C_{1}$ and $C_{2}$ denote the centres of curvatures then the focal length of the thin lens $(μ = 1.5)$ is

20 c m

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10 c m

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30 c m

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40 c m

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Solution

The correct option is C $40 c m$
Given,
$R_{1} = 10 c m$ $R_{2} = 20 c m$ $μ = 1.5$
We can put these values in the lens makers formula to get the focal length,
$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$\Rightarrow \frac{1}{f} = (1.5 - 1) (\frac{1}{10} - \frac{1}{20})$
$\Rightarrow \frac{1}{f} = \frac{1}{2} \times \frac{1}{20} = \frac{1}{40}$
$∴ f = 40 c m$
Hence the correct answer is option (D).

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Calculate the focal length of the thin lens shown in figure. The points $C_{1}$ and $C_{2}$ denote the centres of curvature. Refractive index of lens is $1.5$ .

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In figure the points C1 and C2 denote the centres of curvatures then the focal length of the thin lens (μ=1.5) is

The correct option is C 40cmGiven,R1=10cm R2=20cm μ=1.5We can put these values in the lens makers formula to get the focal length,1f=(μ−1)(1R1−1R2)⇒1f=(1.5−1)(110−120)⇒1f=12×120=140∴f=40cmHence the correct answer is option (D).

In figure the points $C_{1}$ and $C_{2}$ denote the centres of curvatures then the focal length of the thin lens $(μ = 1.5)$ is