Two thin lenses have a combined power of +9D.When they are separated by a distance of 20 cm, then their equivalent power becomes `+27/5` D. Their individual powers (in dioptre) are

Options

A. 1,8
B. 2,7
C. 3,6
D. 4,5

Explanation

To solve the problem, we need to find the individual powers of two thin lenses given their combined power and the equivalent power when they are separated by a distance. Let's break it down step by step.

### Step 1: Define the variables
Let the power of the first lens be \( P_1 \) and the power of the second lens be \( P_2 \).

### Step 2: Write the equations based on the problem statement
1. The combined power of the two lenses when they are in contact is given by:
\[ P_1 + P_2 = 9 \quad \text{(Equation 1)} \]

2. When the lenses are separated by a distance of 20 cm (which is 0.2 m), the equivalent power is given by:
\[ P_{eq} = P_1 + P_2 - \frac{D \cdot P_1 \cdot P_2}{100} \quad \text{where } D = 20 \text{ cm} \]
Given that \( P_{eq} = \frac{27}{5} \), we can write:
\[ P_1 + P_2 - \frac{20 \cdot P_1 \cdot P_2}{100} = \frac{27}{5} \quad \text{(Equation 2)} \]

### Step 3: Substitute Equation 1 into Equation 2
From Equation 1, we know \( P_1 + P_2 = 9 \). Substitute this into Equation 2:
\[ 9 - \frac{20 \cdot P_1 \cdot P_2}{100} = \frac{27}{5} \]

### Step 4: Simplify the equation
Rearranging gives:
\[ \frac{20 \cdot P_1 \cdot P_2}{100} = 9 - \frac{27}{5} \]
Calculating the right side:
\[ 9 = \frac{45}{5} \quad \Rightarrow \quad 9 - \frac{27}{5} = \frac{45 - 27}{5} = \frac{18}{5} \]
Thus, we have:
\[ \frac{20 \cdot P_1 \cdot P_2}{100} = \frac{18}{5} \]
Multiplying both sides by 100 gives:
\[ 20 \cdot P_1 \cdot P_2 = \frac{1800}{5} = 360 \]
Dividing by 20:
\[ P_1 \cdot P_2 = 18 \quad \text{(Equation 3)} \]

### Step 5: Solve the system of equations
Now we have two equations:
1. \( P_1 + P_2 = 9 \) (Equation 1)
2. \( P_1 \cdot P_2 = 18 \) (Equation 3)

Let’s express \( P_2 \) in terms of \( P_1 \):
\[ P_2 = 9 - P_1 \]
Substituting into Equation 3:
\[ P_1(9 - P_1) = 18 \]
Expanding:
\[ 9P_1 - P_1^2 = 18 \]
Rearranging gives:
\[ P_1^2 - 9P_1 + 18 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( P_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -9, c = 18 \):
\[ P_1 = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot 18}}{2 \cdot 1} = \frac{9 \pm \sqrt{81 - 72}}{2} = \frac{9 \pm \sqrt{9}}{2} = \frac{9 \pm 3}{2} \]
This gives:
\[ P_1 = \frac{12}{2} = 6 \quad \text{or} \quad P_1 = \frac{6}{2} = 3 \]

### Step 7: Find \( P_2 \)
If \( P_1 = 6 \), then \( P_2 = 9 - 6 = 3 \).
If \( P_1 = 3 \), then \( P_2 = 9 - 3 = 6 \).

Thus, the individual powers are:
\[ P_1 = 3 \, \text{D}, \quad P_2 = 6 \, \text{D} \]

### Final Answer:
The individual powers of the lenses are \( 3 \, \text{D} \) and \( 6 \, \text{D} \).

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