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 of where to place an object in front of two lenses (one with a power of +10 D and the other with a power of -5 D) to obtain a virtual image with a magnification of 2, we can follow these steps:

### Step 1: Calculate the Net Power of the Lens Combination
The power of the first lens (P1) is +10 D and the power of the second lens (P2) is -5 D. When two lenses are placed in contact, the net power (P_net) is given by the sum of the individual powers:
\[ P_{net} = P_1 + P_2 \]
\[ P_{net} = +10 D - 5 D = +5 D \]

### Step 2: Calculate the Focal Length of the Combination
The focal length (f) of a lens is related to its power (P) by the formula:
\[ f = \frac{1}{P} \]
For the net power of +5 D:
\[ f_{net} = \frac{1}{5} = 0.2 \text{ m} = 20 \text{ cm} \]

### Step 3: Use the Magnification Formula
The magnification (m) is given by the formula:
\[ m = \frac{v}{u} \]
where \( v \) is the image distance and \( u \) is the object distance. Given that the magnification is +2 (indicating a virtual image), we have:
\[ m = 2 = \frac{v}{u} \]
From this, we can express \( v \) in terms of \( u \):
\[ v = 2u \]

### Step 4: Apply the Lens Formula
The lens formula relates the object distance (u), image distance (v), and focal length (f):
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting \( f = 20 \text{ cm} \) and \( v = 2u \):
\[ \frac{1}{20} = \frac{1}{2u} - \frac{1}{u} \]

### Step 5: Simplify the Equation
To simplify the equation:
\[ \frac{1}{20} = \frac{1}{2u} - \frac{2}{2u} \]
\[ \frac{1}{20} = -\frac{1}{2u} \]
Taking the reciprocal gives:
\[ -2u = 20 \]
Thus, solving for \( u \):
\[ u = -10 \text{ cm} \]

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