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 analyze the optical system consisting of a thin convex lens and a plane mirror.

### Step 1: Identify the parameters
- Focal length of the convex lens (F) = +30 cm (positive because it is a convex lens)
- Distance of the object (U) = -15 cm (negative because the object is placed in front of the lens)
- Distance of the plane mirror from the lens = 15 cm

### Step 2: Calculate the image formed by the lens (I1)
Using the lens formula:
\[ \frac{1}{F} = \frac{1}{V_1} - \frac{1}{U} \]
Substituting the values:
\[ \frac{1}{30} = \frac{1}{V_1} - \frac{1}{-15} \]
This simplifies to:
\[ \frac{1}{30} = \frac{1}{V_1} + \frac{1}{15} \]
Finding a common denominator (which is 30):
\[ \frac{1}{30} = \frac{2}{30} + \frac{1}{V_1} \]
Rearranging gives:
\[ \frac{1}{V_1} = \frac{1}{30} - \frac{2}{30} = -\frac{1}{30} \]
Thus:
\[ V_1 = -30 \text{ cm} \]
This means the image I1 is formed 30 cm on the same side as the object.

### Step 3: Determine the distance from the lens to the mirror
The mirror is placed 15 cm behind the lens. Therefore, the distance from the image I1 to the mirror is:
\[ \text{Distance from lens to I1} = 30 \text{ cm} (image distance) + 15 \text{ cm} (distance to mirror) = 45 \text{ cm} \]

### Step 4: Calculate the image formed by the mirror (I2)
For a plane mirror, the image distance (V2) is equal to the object distance (U2) from the mirror:
\[ V_2 = -U_2 \]
Here, U2 = 45 cm (the distance from the mirror to I1, which is treated as the object for the mirror):
\[ V_2 = -(-45) = 45 \text{ cm} \]
This means the image I2 is formed 45 cm behind the mirror.

### Step 5: Calculate the distance from the lens to the new image I2
The distance from the lens to the mirror is 15 cm, and the distance from the mirror to the image I2 is 45 cm. Therefore, the distance from the lens to I2 is:
\[ \text{Distance from lens to I2} = 15 \text{ cm} + 45 \text{ cm} = 60 \text{ cm} \]

### Step 6: Calculate the final image formed by the lens (I3)
Now, we will find the final image I3 formed by the lens using the lens formula again:
\[ \frac{1}{F} = \frac{1}{V_3} - \frac{1}{U_3} \]
Where U3 = -60 cm (the object distance for the lens is negative since the rays are coming from the right):
\[ \frac{1}{30} = \frac{1}{V_3} - \frac{1}{-60} \]
This simplifies to:
\[ \frac{1}{30} = \frac{1}{V_3} + \frac{1}{60} \]
Finding a common denominator (which is 60):
\[ \frac{1}{30} = \frac{2}{60} + \frac{1}{V_3} \]
Rearranging gives:
\[ \frac{1}{V_3} = \frac{1}{30} - \frac{2}{60} = \frac{2}{60} - \frac{2}{60} = 0 \]
Thus:
\[ V_3 = 60 \text{ cm} \]

### Step 7: Calculate the distance of the final image from the object
The final distance from the object is:
\[ \text{Distance from object to final image} = V_3 - U = 60 \text{ cm} - (-15 \text{ cm}) = 60 + 15 = 75 \text{ cm} \]

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