Two thin lenses, when in contact, produce a combination of power`+10` diopters. When they are 0.25m apart, the power reduces to `+6`diopters. The focal lengths of the lenses are m and `"________________"` m

Options

A. 0.5 and 0.125
B. 0.5 and 0.75
C. 0.125.and 0.75
D. Botha have the same focal length 1.25

Explanation

To solve the problem, we need to find the focal lengths of two thin lenses given their combined power when in contact and when separated by a distance. Let's break it down step by step.

### Step 1: Understanding the Power of Lenses
When two thin lenses are in contact, the total power \( P \) is given by:
\[ P = P_1 + P_2 \]
where \( P_1 \) and \( P_2 \) are the powers of the individual lenses. The power \( P \) is related to the focal length \( f \) by:
\[ P = \frac{1}{f} \]
Thus, when the lenses are in contact, we have:
\[ P = 10 \text{ diopters} \implies \frac{1}{f_1} + \frac{1}{f_2} = 10 \tag{1} \]

### Step 2: Power When Lenses are Separated
When the lenses are separated by a distance \( d = 0.25 \, \text{m} \), the equivalent power \( P' \) is given by:
\[ P' = P_1 + P_2 - \frac{d}{f_1 f_2} \]
Given that \( P' = 6 \text{ diopters} \), we can write:
\[ \frac{1}{f_1} + \frac{1}{f_2} - \frac{0.25}{f_1 f_2} = 6 \tag{2} \]

### Step 3: Substituting Equation (1) into Equation (2)
From equation (1), we know that:
\[ \frac{1}{f_1} + \frac{1}{f_2} = 10 \]
Substituting this into equation (2):
\[ 10 - \frac{0.25}{f_1 f_2} = 6 \]
Rearranging gives:
\[ \frac{0.25}{f_1 f_2} = 10 - 6 = 4 \]
Thus, we have:
\[ f_1 f_2 = \frac{0.25}{4} = \frac{1}{16} \tag{3} \]

### Step 4: Solving the System of Equations
Now we have two equations:
1. \( \frac{1}{f_1} + \frac{1}{f_2} = 10 \) (Equation 1)
2. \( f_1 f_2 = \frac{1}{16} \) (Equation 3)

From equation (1), we can express \( f_2 \) in terms of \( f_1 \):
\[ \frac{1}{f_2} = 10 - \frac{1}{f_1} \implies f_2 = \frac{f_1}{10f_1 - 1} \]

Substituting this expression for \( f_2 \) into equation (3):
\[ f_1 \left(\frac{f_1}{10f_1 - 1}\right) = \frac{1}{16} \]
This simplifies to:
\[ \frac{f_1^2}{10f_1 - 1} = \frac{1}{16} \]
Cross-multiplying gives:
\[ 16f_1^2 = 10f_1 - 1 \]
Rearranging leads to:
\[ 16f_1^2 - 10f_1 + 1 = 0 \]

### Step 5: Solving the Quadratic Equation
Using the quadratic formula \( f_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ f_1 = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 16 \cdot 1}}{2 \cdot 16} \]
Calculating the discriminant:
\[ = \frac{10 \pm \sqrt{100 - 64}}{32} = \frac{10 \pm \sqrt{36}}{32} = \frac{10 \pm 6}{32} \]
This gives two possible solutions:
1. \( f_1 = \frac{16}{32} = \frac{1}{2} \, \text{m} \)
2. \( f_1 = \frac{4}{32} = \frac{1}{8} \, \text{m} \)

### Step 6: Finding \( f_2 \)
Using \( f_1 f_2 = \frac{1}{16} \):
- If \( f_1 = \frac{1}{2} \, \text{m} \), then \( f_2 = \frac{1}{16} \div \frac{1}{2} = \frac{1}{8} \, \text{m} \).
- If \( f_1 = \frac{1}{8} \, \text{m} \), then \( f_2 = \frac{1}{16} \div \frac{1}{8} = \frac{1}{2} \, \text{m} \).

Thus, the focal lengths of the lenses are:
\[ f_1 = 0.5 \, \text{m} \quad \text{and} \quad f_2 = 0.125 \, \text{m} \]

### Final Answer
The focal lengths of the lenses are \( 0.5 \, \text{m} \) and \( 0.125 \, \text{m} \).

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