Question

An optician prescribes a lens of power $+2.5D.$ The focal length of the lens in water is (Refractive indices of the lens and water are respectively 1.5 and 1.3)

• A. 40 cm
• B. $\frac{2660}{17}$ cm
• C. $\frac{17}{2660}$ cm
• D. $\frac{3000}{17}$ cm
• E. $\frac{17}{3000}$ cm

Answer 1

Answer: (B)

Given, power of lens $=+2.5D,{\mu }_{\text{water }}=1.3$ and
${\mu }_{\text{lens }}=1.5$
Lense maker's formula for equiconvex lense
$\frac{1}{f}=\left({\mu }_{rel}-1\right)\frac{2}{R}$
Since, power of lens, $P=\frac{1}{f}$
So, for air, lense medium
$25=\left(\frac{1.5}{1}-1\right)\frac{2}{R}\left(\because {\mu }_{\text{air}}=1\right)$
$\Rightarrow 25=\frac{0.5\times 2}{R}$
$\Rightarrow R=\frac{1}{25}m$
(where, $R=$ radius of curvature $)$
New focal length, when lens in water
$\frac{1}{f^{\prime }}=\left(\frac{{\mu }_{\text{lens }}}{{\mu }_{\text{water }}}-1\right)\frac{2}{R}$
$\Rightarrow \frac{1}{f^{\prime }}=\left(\frac{1.5}{1.3}-1\right)2\times 25$
$\Rightarrow f^{\prime }=\frac{1}{0.15\times 2\times 25}=1.333m=133.3cm$