Thank you for reporting, we will resolve it shortly
Q.
An optician prescribes a lens of power $+2.5\, D.$ The focal length of the lens in water is (Refractive indices of the lens and water are respectively 1.5 and 1.3)
Given, power of lens $=+2.5 D , \mu_{\text {water }}=1.3$ 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'}=\left(\frac{\mu_{\text {lens }}}{\mu_{\text {water }}}-1\right) \frac{2}{R} $
$\Rightarrow \frac{1}{f'}=\left(\frac{1.5}{1.3}-1\right) 2 \times 25$
$\Rightarrow f'=\frac{1}{0.15 \times 2 \times 25}=1.333\, m =133.3\, cm$