Question

A thin lens of focal length +12cm is immersed in water (u=1.33). what is it's new focal length.​

Answer 1

the new focal length will be 9 cm

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Answer 2

Answer:

f' = 48 cm.

Explanation:

we have,

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

When the lens is placed in air, f = 12 cm.

Thus, $\frac{1}{12cm} = (1.5 - 1)(\frac{1}{ R_{1} } - \frac{1}{ R_{2} } )$

Or,

$\frac{1}{ R_{1} } - \frac{1}{ R_{2} } = \frac{1}{6cm}$

If the focal length Becomes f' when placed in water,

$\frac{1}{f'} = ( \frac{1.5}{1.33} - 1)(\frac{1}{ R_{1} } - \frac{1}{ R_{2} })$

$= \frac{1}{8} \times \frac{1}{6cm} = \frac{1}{48cm}$

Or f' = 48 cm.