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A diverging meniscus lens of 1.5 refractive index has surface of radii 4 and 5cm. The position of the image, if an object is placed 10cm in front of the lens is

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$7 c m$

$- 8 c m$

$9 c m$

$10 c m$

answer is B.

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Detailed Solution

$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$(1.5 - 1) (\frac{1}{5} - \frac{1}{4})$

$\frac{1}{f} = \frac{- 0.5}{20} = \frac{5}{200} = \frac{- 1}{40}$

$f = - 40 c m$

$\frac{1}{f} = \frac{1}{v} - \frac{1}{- 4} \Rightarrow \frac{- 1}{40} = \frac{1}{v} + \frac{1}{10}$

$= \frac{1}{v} = \frac{- 1}{40} - \frac{1}{10} = \frac{- 5}{40}$

$v = - 8 c m$

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