A thin glass (refractive index $$1.5$$) lens has optical power of $$-5D$$ in air. Its optical power in a liquid medium with refractive index $$1.6$$ will be

Optical power is the reciprocal of focal length in metres: $$P=\frac{1}{f}$$.

For a thin lens surrounded by a medium of refractive index $$n_m$$, the Lens-Maker’s formula is
$$\frac{1}{f}=\left(\frac{n_\ell}{n_m}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$ where $$n_\ell$$ is the refractive index of the lens material.

Step 1 : Determine the curvature factor $$K$$ of the lens
In air ($$n_m=1$$) the given power is $$P_a=-5\text{ D}$$, so $$f_a=-0.2\text{ m}$$.
Using the formula with $$n_m=1$$:
$$\frac{1}{f_a}=(n_\ell-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right).$$
Hence
$$\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{1/f_a}{n_\ell-1}.$$

With $$n_\ell=1.5$$ and $$f_a=-0.2\text{ m}$$:
$$\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{-5}{0.5}=-10.$$
Denote this constant curvature factor by $$K=-10\text{ m}^{-1}$$.

Step 2 : Find the focal length in the liquid medium
The surrounding liquid has $$n_m=1.6$$. Putting this into the Lens-Maker’s formula:
$$\frac{1}{f_\ell}=\left(\frac{n_\ell}{n_m}-1\right)K.$$

Compute the bracketed term:
$$\frac{n_\ell}{n_m}-1=\frac{1.5}{1.6}-1=0.9375-1=-0.0625.$$

Therefore
$$\frac{1}{f_\ell}=(-0.0625)(-10)=0.625\text{ m}^{-1}.$$

Step 3 : Optical power in the liquid
$$P_\ell=\frac{1}{f_\ell}=0.625\text{ dioptres}.$$

This value is neither $$1\text{ D}$$, $$-1\text{ D}$$, nor $$25\text{ D}$$. Hence none of the listed options matches the correct power.

Option D which is: None of these