An object is placed at a distance of 10cm form a co-axial combination of two lenses A and B in contact. The combination forms a real image three times the size of the object. If lens B is concave with a focal length of 30cm. The nature and focal length of lens A is

Options

A. convex,12cm
B. concave,12cm
C. convex,6cm
D. convex,18cm

Explanation

To solve the problem step by step, we will analyze the information given and apply the relevant formulas from optics.

### Step 1: Identify the given parameters
- Object distance (U) = -10 cm (the negative sign indicates that the object is placed on the same side as the incoming light).
- Magnification (M) = 3 (the image is three times the size of the object).
- Focal length of lens B (f_B) = -30 cm (since lens B is concave, its focal length is negative).

### Step 2: Use the magnification formula
The magnification formula is given by:
\[ M = -\frac{V}{U} \]
Where:
- V = image distance
- U = object distance

Substituting the known values:
\[ 3 = -\frac{V}{-10} \]
This simplifies to:
\[ 3 = \frac{V}{10} \]
Thus, we can find V:
\[ V = 3 \times 10 = 30 \text{ cm} \]

### Step 3: Apply the lens formula for the combination of lenses
For a combination of two lenses in contact, the effective focal length (F) can be calculated using the formula:
\[ \frac{1}{F} = \frac{1}{f_A} + \frac{1}{f_B} \]
Where:
- f_A = focal length of lens A (which we need to find)
- f_B = focal length of lens B = -30 cm

### Step 4: Use the lens formula to find the focal length of lens A
From the lens formula, we can rearrange it to find f_A:
\[ \frac{1}{F} = \frac{1}{f_A} + \frac{1}{-30} \]
We also know that the image distance V = 30 cm and the object distance U = -10 cm. We can find the effective focal length F using the lens formula:
\[ \frac{1}{F} = \frac{1}{V} - \frac{1}{U} \]
Substituting the values:
\[ \frac{1}{F} = \frac{1}{30} - \frac{1}{-10} \]
This simplifies to:
\[ \frac{1}{F} = \frac{1}{30} + \frac{1}{10} \]
Finding a common denominator (30):
\[ \frac{1}{F} = \frac{1}{30} + \frac{3}{30} = \frac{4}{30} = \frac{2}{15} \]
Thus:
\[ F = \frac{15}{2} = 7.5 \text{ cm} \]

### Step 5: Substitute back to find f_A
Now we can substitute F back into the equation:
\[ \frac{1}{\frac{15}{2}} = \frac{1}{f_A} - \frac{1}{30} \]
This simplifies to:
\[ \frac{2}{15} = \frac{1}{f_A} + \frac{1}{30} \]
Rearranging gives:
\[ \frac{1}{f_A} = \frac{2}{15} - \frac{1}{30} \]
Finding a common denominator (30):
\[ \frac{1}{f_A} = \frac{4}{30} - \frac{1}{30} = \frac{3}{30} = \frac{1}{10} \]
Thus:
\[ f_A = 10 \text{ cm} \]

### Step 6: Determine the nature of lens A
Since the focal length f_A is positive, lens A is a **convex lens**.

### Final Answer
The nature of lens A is **convex** and its focal length is **10 cm**.

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