A convex lens (of focal length `20 cm)` and a concave mirror, having their principal axes along the same lines , are Kept `80cm` is to from eath other . The concave mirror is to the right of the convex lens. When an object is Kept at a distance of `30 cm` to the left of the convex lens,its image remains at the same position even if the concave mirror is removed . The maximum distance of the object for which the concave mirror , by itself would produce a virtual would be:

Options

A. `30 cm`
B. `25cm`
C. `10cm`
D. `20cm`

Answer

To solve the problem step by step, we will follow the concepts of optics, specifically the lens and mirror formulas.

### Step 1: Understand the setup
We have a convex lens with a focal length \( f_L = 20 \, \text{cm} \) and a concave mirror. The distance between the lens and the mirror is \( 80 \, \text{cm} \). An object is placed \( 30 \, \text{cm} \) to the left of the convex lens.

### Step 2: Determine the object distance for the lens
The object distance \( u_L \) for the convex lens is given as:
\[ u_L = -30 \, \text{cm} \quad (\text{negative as per sign convention}) \]

### Step 3: Use the lens formula to find the image distance
The lens formula is given by:
\[ \frac{1}{f_L} = \frac{1}{v_L} - \frac{1}{u_L} \]
Rearranging gives:
\[ \frac{1}{v_L} = \frac{1}{f_L} + \frac{1}{u_L} \]
Substituting the values:
\[ \frac{1}{v_L} = \frac{1}{20} + \frac{1}{-30} \]
Finding a common denominator (60):
\[ \frac{1}{v_L} = \frac{3}{60} - \frac{2}{60} = \frac{1}{60} \]
Thus,
\[ v_L = 60 \, \text{cm} \]

### Step 4: Analyze the position of the image
The image formed by the convex lens is \( 60 \, \text{cm} \) to the right of the lens. Since the distance between the lens and the mirror is \( 80 \, \text{cm} \), the image is located \( 80 - 60 = 20 \, \text{cm} \) to the left of the concave mirror.

### Step 5: Determine the object distance for the concave mirror
The image formed by the lens acts as the object for the concave mirror. The object distance \( u_M \) for the concave mirror is:
\[ u_M = -20 \, \text{cm} \quad (\text{negative as per sign convention}) \]

### Step 6: Relate the object distance to the focal length of the concave mirror
For the concave mirror, the mirror formula is:
\[ \frac{1}{f_M} = \frac{1}{v_M} + \frac{1}{u_M} \]
Since the image formed by the lens coincides with the image formed by the mirror, we can say that the image distance \( v_M \) for the concave mirror is equal to the distance of the image from the mirror, which is \( 20 \, \text{cm} \) (to the left of the mirror). Thus:
\[ v_M = -20 \, \text{cm} \quad (\text{negative for virtual image}) \]

### Step 7: Substitute the values into the mirror formula
Substituting the values into the mirror formula:
\[ \frac{1}{f_M} = \frac{1}{-20} + \frac{1}{-20} \]
This simplifies to:
\[ \frac{1}{f_M} = -\frac{1}{20} - \frac{1}{20} = -\frac{2}{20} = -\frac{1}{10} \]
Thus,
\[ f_M = -10 \, \text{cm} \]

### Step 8: Determine the maximum distance for a virtual image
A concave mirror produces a virtual image when the object is placed between the focus and the pole. The focal length of the concave mirror is \( 10 \, \text{cm} \), so the maximum distance of the object from the mirror for which it can produce a virtual image is:
\[ \text{Maximum object distance} = f_M = 10 \, \text{cm} \]

### Final Answer
The maximum distance of the object for which the concave mirror, by itself, would produce a virtual image is:
\[ \boxed{10 \, \text{cm}} \]