An object placed in front of a concave mirror of focal length 0.15 m produces a virtual image, which is twice the size of the object. The position of the object with respect to the mirror is

Options

A. `-5.5cm`
B. `-6.5cm`
C. `-7.5cm`
D. `-8.5cm`

Answer

To solve the problem step by step, we will use the mirror formula and the magnification formula.

### Step 1: Understand the Given Information
- Focal length of the concave mirror, \( f = -0.15 \, m \) (negative because it's a concave mirror).
- The image produced is virtual and twice the size of the object, which means the magnification \( M = 2 \).

### Step 2: Write the Magnification Formula
The magnification \( M \) for mirrors is given by the formula:
\[ M = -\frac{V}{U} \]
Where:
- \( V \) is the image distance,
- \( U \) is the object distance.

Since the image is virtual and twice the size of the object:
\[ 2 = -\frac{V}{U} \]
From this, we can express \( V \) in terms of \( U \):
\[ V = -2U \]

### Step 3: Use the Mirror Formula
The mirror formula is given by:
\[ \frac{1}{V} + \frac{1}{U} = \frac{1}{f} \]
Substituting \( V = -2U \) into the mirror formula:
\[ \frac{1}{-2U} + \frac{1}{U} = \frac{1}{-0.15} \]

### Step 4: Simplify the Equation
To simplify the left side:
\[ \frac{1}{-2U} + \frac{1}{U} = \frac{-1 + 2}{2U} = \frac{1}{2U} \]
So, we have:
\[ \frac{1}{2U} = \frac{1}{-0.15} \]

### Step 5: Cross Multiply to Solve for \( U \)
Cross-multiplying gives:
\[ 1 \cdot (-0.15) = 2U \cdot 1 \]
Thus:
\[ -0.15 = 2U \]
Now, solving for \( U \):
\[ U = \frac{-0.15}{2} = -0.075 \, m \]

### Step 6: Convert to Centimeters
To convert meters to centimeters:
\[ U = -0.075 \, m = -7.5 \, cm \]

### Final Answer
The position of the object with respect to the mirror is:
\[ U = -7.5 \, cm \]