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 a thin glass lens in a liquid medium with a refractive index of 1.6, we will use the lens maker's formula. The steps are as follows:

### Step 1: Understand the lens maker's formula
The lens maker's formula in air is given by:
\[ \frac{1}{f} = (n_g - n_a) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
Where:
- \( f \) is the focal length of the lens,
- \( n_g \) is the refractive index of the lens material (glass),
- \( n_a \) is the refractive index of air (approximately 1).

### Step 2: Calculate the power in air
The optical power \( P \) of the lens is given by:
\[ P = \frac{1}{f} \]
Given that the optical power in air is \( -8D \), we can write:
\[ P = n_g - n_a \]
Substituting the values:
\[ -8 = (1.5 - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
This implies:
\[ -8 = 0.5 \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
From this, we can deduce that:
\[ \frac{1}{R_1} - \frac{1}{R_2} = -16 \]

### Step 3: Write the lens maker's formula for the liquid medium
Now, we need to find the power \( P' \) of the lens in the liquid medium with a refractive index \( n_l = 1.6 \):
\[ P' = \frac{1}{f'} = (n_g - n_l) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
Substituting the values:
\[ P' = (1.5 - 1.6) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
This gives:
\[ P' = (-0.1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

### Step 4: Substitute the value of \(\frac{1}{R_1} - \frac{1}{R_2}\)
From Step 2, we know:
\[ \frac{1}{R_1} - \frac{1}{R_2} = -16 \]
Substituting this into the equation for \( P' \):
\[ P' = (-0.1)(-16) = 1.6 \]

### Step 5: Conclusion
Thus, the optical power of the lens in the liquid medium with a refractive index of 1.6 is:
\[ P' = 1.6D \]

### Final Answer
The optical power in the liquid medium is approximately **1.6 Diopters**.

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