Two lenses of power +10 D and - 5 D are placed in contact. Where should an object be held from the lens, so as to obtain a virtual image of magnification 2 ?

Options

A. 5 cm
B. `-5 cm`
C. 10 cm
D. `-10 cm`

Answer

To solve the problem, we need to find the position of the object (u) from the combined lens system to obtain a virtual image with a magnification of 2. Here’s a step-by-step solution:

### Step 1: Calculate the Power of the Combined Lens
The power of a lens is given in diopters (D). When two lenses are in contact, their powers add up:
\[ P = P_1 + P_2 \]
Where:
- \( P_1 = +10 \, D \) (convex lens)
- \( P_2 = -5 \, D \) (concave lens)

Calculating the combined power:
\[ P = 10 - 5 = 5 \, D \]

### Step 2: Calculate the Focal Length of the Combined Lens
The focal length (f) of a lens can be calculated from its power using the formula:
\[ P = \frac{1}{f} \]
Thus,
\[ f = \frac{1}{P} = \frac{1}{5} \, m = 0.2 \, m = 20 \, cm \]

### Step 3: Use the Magnification Formula
The magnification (m) is given by:
\[ m = \frac{V}{U} \]
Where:
- \( V \) is the image distance
- \( U \) is the object distance

Given that the magnification is 2:
\[ 2 = \frac{V}{U} \]
This implies:
\[ V = 2U \]

### Step 4: Apply the Lens Formula
The lens formula relates the focal length, object distance, and image distance:
\[ \frac{1}{f} = \frac{1}{V} - \frac{1}{U} \]

Substituting the known values:
\[ \frac{1}{20} = \frac{1}{2U} - \frac{1}{U} \]

### Step 5: Simplify the Equation
Finding a common denominator:
\[ \frac{1}{20} = \frac{1}{2U} - \frac{2}{2U} \]
\[ \frac{1}{20} = \frac{-1}{2U} \]

### Step 6: Solve for U
Cross-multiplying gives:
\[ -1 \cdot 20 = 2U \]
Thus:
\[ U = -10 \, cm \]

### Conclusion
The object should be placed 10 cm in front of the lens system (the negative sign indicates that the object is on the same side as the incoming light).