An upright object is placed at a distance of `40cm` in front of a convergent less of focal length `20cm`. A convergent mirror of focal length `10cm` is placed at a distance of `60cm` on the other side of the lens. The position and size of the final image will be:

Options

A. `20cm` from the convergent mirror, same size as the object
B. `40 cm` from the convergent mirror same size as the object
C. `40cm` from the convergent lens, twice the size of the object
D. `20cm` from the convergent mirror, twice the size of the object

Answer

To solve the problem step by step, we will analyze the situation involving the convex lens and the concave mirror.

### Step 1: Identify the Object and Lens Parameters
- The object is placed at a distance of \( 40 \, \text{cm} \) in front of a convex lens with a focal length of \( 20 \, \text{cm} \).

### Step 2: Use the Lens Formula to Find the Image Position
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \( f \) is the focal length of the lens (positive for a convex lens),
- \( v \) is the image distance from the lens,
- \( u \) is the object distance from the lens (negative in lens convention).

Here, \( f = 20 \, \text{cm} \) and \( u = -40 \, \text{cm} \).

Substituting the values into the lens formula:
\[ \frac{1}{20} = \frac{1}{v} - \frac{1}{-40} \]
\[ \frac{1}{20} = \frac{1}{v} + \frac{1}{40} \]
To solve for \( \frac{1}{v} \):
\[ \frac{1}{v} = \frac{1}{20} - \frac{1}{40} \]
Finding a common denominator (which is 40):
\[ \frac{1}{v} = \frac{2}{40} - \frac{1}{40} = \frac{1}{40} \]
Thus, \( v = 40 \, \text{cm} \).

### Step 3: Determine the Characteristics of the Image Formed by the Lens
- The image \( A'B' \) formed by the lens is located \( 40 \, \text{cm} \) on the opposite side of the lens.
- Since the object is at \( 2F \), the image is also at \( 2F \) and is inverted with the same size as the object. Therefore, the size of the image \( A'B' \) is \( 20 \, \text{cm} \) (same as the object).

### Step 4: Analyze the Position of the Concave Mirror
- The concave mirror is placed \( 60 \, \text{cm} \) from the lens. Since the image \( A'B' \) is \( 40 \, \text{cm} \) from the lens, the distance from the image to the mirror is:
\[ 60 \, \text{cm} - 40 \, \text{cm} = 20 \, \text{cm} \]
Thus, the image \( A'B' \) acts as the object for the concave mirror and is located \( 20 \, \text{cm} \) in front of it.

### Step 5: Use the Mirror Formula to Find the Final Image Position
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v'} + \frac{1}{u'} \]
Where:
- \( f \) is the focal length of the mirror (negative for a concave mirror),
- \( v' \) is the image distance from the mirror,
- \( u' \) is the object distance from the mirror (negative in mirror convention).

Here, \( f = -10 \, \text{cm} \) and \( u' = -20 \, \text{cm} \).

Substituting the values into the mirror formula:
\[ \frac{1}{-10} = \frac{1}{v'} - \frac{1}{20} \]
To solve for \( \frac{1}{v'} \):
\[ \frac{1}{v'} = \frac{1}{-10} + \frac{1}{20} \]
Finding a common denominator (which is 20):
\[ \frac{1}{v'} = \frac{-2}{20} + \frac{1}{20} = \frac{-1}{20} \]
Thus, \( v' = -20 \, \text{cm} \).

### Step 6: Determine the Characteristics of the Final Image
- The image \( A''B'' \) formed by the concave mirror is located \( 20 \, \text{cm} \) on the same side as the object (since \( v' \) is negative).
- The image is inverted and has the same size as the previous image \( A'B' \), which is \( 20 \, \text{cm} \).

### Final Result
- The final image \( A''B'' \) is located \( 20 \, \text{cm} \) from the mirror and has a size of \( 20 \, \text{cm} \).