A real image of an object is formed at a distance of `20 cm` from a lens. On putting another lens in contact with it, the image is shifted `10 cm` towards the combination, Determine the power of the second lens.

Options

A. 2D
B. 5D
C. 6D
D. 10 D

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the given information
We have a real image formed by the first lens at a distance of \( V_1 = 20 \, \text{cm} \). When a second lens is placed in contact with the first lens, the image shifts \( 10 \, \text{cm} \) towards the combination, resulting in a new image distance of \( V_2 = 20 \, \text{cm} - 10 \, \text{cm} = 10 \, \text{cm} \).

### Step 2: Use the lens formula for the first lens
The lens formula is given by:
\[ \frac{1}{F} = \frac{1}{V} - \frac{1}{U} \]
For the first lens, we can express it as:
\[ \frac{1}{F_1} = \frac{1}{V_1} - \frac{1}{U} \]
Substituting \( V_1 = 20 \, \text{cm} \):
\[ \frac{1}{F_1} = \frac{1}{20} - \frac{1}{U} \quad \text{(Equation 1)} \]

### Step 3: Use the lens formula for the combination of lenses
For the combination of the two lenses, we have:
\[ \frac{1}{F} = \frac{1}{F_1} + \frac{1}{F_2} \]
Using the new image distance \( V_2 = 10 \, \text{cm} \):
\[ \frac{1}{F} = \frac{1}{V_2} - \frac{1}{U} \]
Substituting \( V_2 = 10 \, \text{cm} \):
\[ \frac{1}{F} = \frac{1}{10} - \frac{1}{U} \quad \text{(Equation 2)} \]

### Step 4: Substitute Equation 1 into Equation 2
From Equation 1, we know \( \frac{1}{F_1} \):
\[ \frac{1}{F} = \frac{1}{10} - \left( \frac{1}{20} - \frac{1}{U} \right) \]
This simplifies to:
\[ \frac{1}{F} = \frac{1}{10} - \frac{1}{20} + \frac{1}{U} \]

### Step 5: Simplify the equation
Now, we can simplify the left-hand side:
\[ \frac{1}{10} - \frac{1}{20} = \frac{2 - 1}{20} = \frac{1}{20} \]
So we have:
\[ \frac{1}{F} = \frac{1}{20} + \frac{1}{U} \]
Now, we can cancel \( \frac{1}{U} \) from both sides:
\[ \frac{1}{F} - \frac{1}{U} = \frac{1}{20} \]

### Step 6: Solve for \( F_2 \)
Now, we know that \( F = F_1 + F_2 \):
\[ \frac{1}{F_2} = \frac{1}{F} - \frac{1}{F_1} \]
Substituting \( F_1 = 20 \, \text{cm} \):
\[ \frac{1}{F_2} = \frac{1}{F} - \frac{1}{20} \]
Since we found \( F = 20 \, \text{cm} \):
\[ \frac{1}{F_2} = \frac{1}{10} - \frac{1}{20} \]
This gives:
\[ \frac{1}{F_2} = \frac{2 - 1}{20} = \frac{1}{20} \]
Thus, \( F_2 = 20 \, \text{cm} \).

### Step 7: Calculate the power of the second lens
The power \( P \) of a lens is given by:
\[ P = \frac{1}{F} \quad \text{(in meters)} \]
Converting \( F_2 \) to meters:
\[ F_2 = 20 \, \text{cm} = 0.2 \, \text{m} \]
So,
\[ P_2 = \frac{1}{0.2} = 5 \, \text{diopters} \]

### Final Answer
The power of the second lens is \( 5 \, \text{diopters} \).

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