Two lenses of power `+ 10 D and - 5 D` are placed in contact,
(i) Calculate the focal length of the combination
(ii) where should an object be held from the combination so as to obtain a virtual image of magnification `2` ?

Explanation

To solve the given problem step by step, we will break it down into two parts as mentioned in the question.

### Part (i): Calculate the focal length of the combination

1. **Identify the powers of the lenses**:
- The power of the first lens (P1) = +10 D (diopters)
- The power of the second lens (P2) = -5 D (diopters)

2. **Calculate the net power (P_net)**:
- The net power of the combination of lenses in contact is given by the sum of their individual powers:
\[ P_{net} = P_1 + P_2 = +10 D - 5 D = +5 D \]

3. **Calculate the focal length (f)**:
- The focal length of a lens is related to its power by the formula:
\[ f = \frac{1}{P_{net}} \]
- Substituting the value of net power:
\[ f = \frac{1}{5 D} = 0.2 \text{ m} = 20 \text{ cm} \]

### Part (ii): Where should an object be held to obtain a virtual image of magnification 2?

1. **Identify the magnification (m)**:
- The magnification is given as +2 (since the image is virtual and erect).

2. **Use the magnification formula**:
- The magnification (m) for a lens is given by:
\[ m = \frac{v}{u} = \frac{f}{f + u} \]
- Here, \(v\) is the image distance, \(u\) is the object distance, and \(f\) is the focal length we calculated.

3. **Set up the equation**:
- From the magnification formula, we can express it as:
\[ 2 = \frac{20}{20 + u} \]

4. **Cross-multiply and solve for u**:
- Cross-multiplying gives:
\[ 2(20 + u) = 20 \]
- Expanding this:
\[ 40 + 2u = 20 \]
- Rearranging gives:
\[ 2u = 20 - 40 \]
\[ 2u = -20 \]
\[ u = -10 \text{ cm} \]

### Final Answers:
- (i) The focal length of the combination is **20 cm**.
- (ii) The object should be held at **10 cm** in front of the lens (indicated by the negative sign).