An optical system consists of a thin convex lens of focal length `30 cm` and a plane mirror placed `15 cm` behind the lens. An object is placed `15 cm` in front of the lens. The distance of the final image from the object is

Options

A. `60 cm`
B. `30 cm`
C. `75 cm`
D. `45 cm`

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Identify the given data
- Focal length of the convex lens (F) = +30 cm (positive for convex lens)
- Distance of the plane mirror from the lens = 15 cm
- Object distance from the lens (U) = -15 cm (negative as per sign convention)

### Step 2: Use the lens formula to find the position of the first image
The lens formula is given by:
\[ \frac{1}{F} = \frac{1}{V} - \frac{1}{U} \]
Substituting the known values:
\[ \frac{1}{30} = \frac{1}{V_1} - \frac{1}{-15} \]
This simplifies to:
\[ \frac{1}{30} = \frac{1}{V_1} + \frac{1}{15} \]
To solve for \(V_1\), we first find a common denominator:
\[ \frac{1}{30} = \frac{1}{V_1} + \frac{2}{30} \]
Rearranging gives:
\[ \frac{1}{V_1} = \frac{1}{30} - \frac{2}{30} = -\frac{1}{30} \]
Thus, we find:
\[ V_1 = -30 \text{ cm} \]
This means the first image is formed 30 cm to the left of the lens.

### Step 3: Determine the object distance for the mirror
The distance from the lens to the mirror is 15 cm. Since the first image is 30 cm to the left of the lens, the distance from the image to the mirror (acting as a new object for the mirror) is:
\[ \text{Distance from image to mirror} = 30 \text{ cm} + 15 \text{ cm} = 45 \text{ cm} \]
Thus, the object distance for the mirror (U2) is:
\[ U_2 = -45 \text{ cm} \]

### Step 4: Use the mirror formula to find the position of the second image
The mirror formula is similar to the lens formula:
\[ \frac{1}{F} = \frac{1}{V} - \frac{1}{U} \]
For a plane mirror, the focal length (F) is considered infinite, thus:
\[ 0 = \frac{1}{V_2} - \frac{1}{-45} \]
This simplifies to:
\[ \frac{1}{V_2} = \frac{1}{45} \]
Thus, we find:
\[ V_2 = +45 \text{ cm} \]
This means the second image is formed 45 cm behind the mirror.

### Step 5: Determine the object distance for the lens again
The second image (acting as an object for the lens) is now 45 cm behind the mirror. Since the mirror is 15 cm behind the lens, the total distance from the lens to the new object is:
\[ \text{Distance from lens to new object} = 15 \text{ cm} + 45 \text{ cm} = 60 \text{ cm} \]
Thus, the object distance for the lens (U3) is:
\[ U_3 = -60 \text{ cm} \]

### Step 6: Use the lens formula again to find the final image position
Using the lens formula again:
\[ \frac{1}{30} = \frac{1}{V_3} - \frac{1}{-60} \]
This simplifies to:
\[ \frac{1}{30} = \frac{1}{V_3} + \frac{1}{60} \]
Finding a common denominator gives:
\[ \frac{1}{30} = \frac{2}{60} + \frac{1}{V_3} \]
Rearranging gives:
\[ \frac{1}{V_3} = \frac{1}{30} - \frac{2}{60} = \frac{1}{30} - \frac{1}{30} = 0 \]
Thus, we find:
\[ V_3 = +60 \text{ cm} \]
This means the final image is formed 60 cm to the right of the lens.

### Step 7: Calculate the distance between the object and the final image
The object is located 15 cm in front of the lens, and the final image is located 60 cm behind the lens. Therefore, the distance between the object and the final image is:
\[ \text{Distance} = 60 \text{ cm} - (-15 \text{ cm}) = 60 + 15 = 75 \text{ cm} \]

### Final Answer
The distance of the final image from the object is **75 cm**.