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 find the position of the object with respect to the concave mirror, we can follow these steps:

### Step 1: Understand the given information
We have a concave mirror with a focal length (f) of -0.15 m (negative because it's a concave mirror). The image produced is virtual and twice the size of the object.

### Step 2: Define the magnification
The magnification (m) is defined as the ratio of the height of the image (h_i) to the height of the object (h_o):
\[ m = \frac{h_i}{h_o} \]
Given that the image is twice the size of the object:
\[ h_i = 2h_o \]
Thus, the magnification can be expressed as:
\[ m = \frac{2h_o}{h_o} = 2 \]

### Step 3: Relate magnification to object and image distances
The magnification is also related to the object distance (u) and image distance (v) by the formula:
\[ m = -\frac{v}{u} \]
Substituting the value of magnification:
\[ 2 = -\frac{v}{u} \]
This implies:
\[ v = -2u \]

### Step 4: Use the mirror formula
The mirror formula relates the object distance (u), image distance (v), and focal length (f):
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the known values:
- f = -0.15 m
- v = -2u

This gives us:
\[ \frac{1}{-0.15} = \frac{1}{-2u} + \frac{1}{u} \]

### Step 5: Simplify the equation
To simplify, we can find a common denominator for the right side:
\[ \frac{1}{-0.15} = \frac{1}{-2u} + \frac{2}{2u} \]
\[ \frac{1}{-0.15} = \frac{1}{2u} \]
Now, we can cross-multiply:
\[ 2u = -0.15 \]
Thus:
\[ u = -0.075 \, \text{m} \]

### Step 6: Convert to centimeters
Converting meters to centimeters:
\[ u = -0.075 \times 100 = -7.5 \, \text{cm} \]

### Conclusion
The position of the object with respect to the mirror is -7.5 cm.

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