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A thin convergent glass lens $(μ_{g} = 1.5)$ has a power of $+ 5.0 D$ . When this lens is immersed in a liquid of refractive index $μ_{l}$ it acts as a divergent lens of focal length $100 c m$ . The value of $μ_{l}$ must be

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$\frac{5}{3}$

$\frac{5}{4}$

$\frac{6}{5}$

$\frac{4}{3}$

answer is B.

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Detailed Solution

P $= + 5.0 D$ .

Therefore $f = \frac{100}{P} = \frac{100}{5.0} = 20 c m$ .

Now
$\frac{1}{20} = (μ_{g} - 1) (\frac{1}{R_{1}} + {\frac{1}{R}}_{2})$
Also

$\frac{1}{- 100} = (\frac{μ_{g} - μ_{l}}{μ_{l}}) (\frac{1}{R_{1}} + {\frac{1}{R}}_{2})$

Dividing we get

$- \frac{1}{20} \times 100 = \frac{(μ_{g} - l)}{(μ_{g} - μ_{l})} \times μ_{l}$
Putting $μ_{g} = 1.5$ we get $μ_{1} = \frac{5}{3}$ .

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