Two lenses are placed in contact with each other and the focal length of combination is `80 cm`. If the focal length of one is `20 cm`, then the power of the other will be

Options

A. `1.66D`
B. `4.00D`
C. `-1.00 D`
D. `-3.75D`

Explanation

To solve the problem step by step, we will use the formulas related to the focal lengths and powers of lenses in contact.

### Step 1: Understand the relationship between focal lengths and power
When two lenses are placed in contact, the equivalent focal length (F_eq) of the combination is given by the formula:
\[ \frac{1}{F_{eq}} = \frac{1}{F_1} + \frac{1}{F_2} \]
where \( F_1 \) and \( F_2 \) are the focal lengths of the individual lenses.

### Step 2: Identify the given values
From the problem, we have:
- \( F_{eq} = 80 \, \text{cm} \)
- \( F_1 = 20 \, \text{cm} \)

We need to find the focal length \( F_2 \) of the second lens.

### Step 3: Substitute the values into the formula
Using the formula from Step 1:
\[ \frac{1}{80} = \frac{1}{20} + \frac{1}{F_2} \]

### Step 4: Solve for \( \frac{1}{F_2} \)
Rearranging the equation gives:
\[ \frac{1}{F_2} = \frac{1}{80} - \frac{1}{20} \]

### Step 5: Calculate \( \frac{1}{80} \) and \( \frac{1}{20} \)
Calculating the fractions:
\[ \frac{1}{80} = 0.0125 \]
\[ \frac{1}{20} = 0.05 \]

### Step 6: Perform the subtraction
Now, subtract the two values:
\[ \frac{1}{F_2} = 0.0125 - 0.05 = -0.0375 \]

### Step 7: Find \( F_2 \)
Taking the reciprocal gives:
\[ F_2 = \frac{1}{-0.0375} \approx -26.67 \, \text{cm} \]

### Step 8: Calculate the power of the second lens
The power \( P \) of a lens is given by:
\[ P = \frac{100}{F} \quad (\text{in cm}) \]
Thus, for the second lens:
\[ P_2 = \frac{100}{F_2} = \frac{100}{-26.67} \approx -3.75 \, \text{D} \]

### Conclusion
The power of the second lens is approximately \(-3.75 \, \text{D}\).

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