A thin glass (refractive index `1.5`) lens has optical power of `-8D` in air, its optical power in a liquid medium with refractive index `1.6` will be

Options

A. `1 D`
B. `-1 D`
C. `25 D`
D. `-25 D`

Explanation

To find the optical power of the lens in a liquid medium with a refractive index of 1.6, we can use the formula for the power of a lens in different media. The power of a lens in air (P_air) is given by:

\[ P_{\text{air}} = \frac{(n_l - n)}{f} \]

Where:
- \( n_l \) is the refractive index of the lens material (1.5 for glass),
- \( n \) is the refractive index of the medium (1.0 for air),
- \( f \) is the focal length of the lens.

Since we know the power of the lens in air is -8D, we can relate this to the focal length:

\[ P_{\text{air}} = \frac{1}{f} \]
\[ -8 = \frac{1}{f} \]
\[ f = -\frac{1}{8} \text{ m} = -0.125 \text{ m} \]

Now, we need to find the power of the lens in the liquid medium with a refractive index of 1.6. The new power (P_liquid) can be calculated using the same formula:

\[ P_{\text{liquid}} = \frac{(n_l - n_{\text{liquid}})}{f} \]

Where:
- \( n_{\text{liquid}} = 1.6 \)

Substituting the values into the formula:

\[ P_{\text{liquid}} = \frac{(1.5 - 1.6)}{-0.125} \]
\[ P_{\text{liquid}} = \frac{-0.1}{-0.125} \]
\[ P_{\text{liquid}} = 0.8 \text{ D} \]

Thus, the optical power of the lens in the liquid medium with a refractive index of 1.6 is \( 0.8 \text{ D} \).