A 3.0 cm object is placed 8.0 cm in front of a mirror.The virtual image is 4.0 cm further from the mirror when the mirror is concave than when it is planar. Determine the image height in the concave mirror.

Options

A. `0.5`
B. `4.5`
C. `1.5`
D. `3.0`

Answer

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

### Step 1: Identify the given data
- Height of the object (h_o) = 3.0 cm
- Object distance for the concave mirror (u) = -8.0 cm (the negative sign is due to the convention that distances measured against the direction of incident light are negative)
- The virtual image is 4.0 cm further from the mirror when it is concave than when it is planar.

### Step 2: Determine the image distance for the planar mirror
For a planar mirror, the image distance (v) is equal to the object distance (u) in magnitude but positive. Therefore:
- For a planar mirror, v = +8.0 cm.

### Step 3: Calculate the image distance for the concave mirror
Since the virtual image is 4.0 cm further from the mirror in the case of the concave mirror, we can find the image distance (v) for the concave mirror:
- v_concave = v_planar + 4.0 cm = 8.0 cm + 4.0 cm = 12.0 cm.
- However, since the image is virtual in the concave mirror, we take this value as negative: v_concave = -12.0 cm.

### Step 4: Use the magnification formula
The magnification (m) for mirrors is given by:
\[ m = -\frac{v}{u} \]
Substituting the values we have:
\[ m = -\frac{-12.0 \, \text{cm}}{-8.0 \, \text{cm}} = \frac{12.0}{8.0} = 1.5 \]

### Step 5: Calculate the height of the image
The magnification is also related to the heights of the object and the image:
\[ m = \frac{h_i}{h_o} \]
Where:
- \( h_i \) = height of the image
- \( h_o \) = height of the object

Rearranging the formula gives:
\[ h_i = m \cdot h_o \]
Substituting the known values:
\[ h_i = 1.5 \cdot 3.0 \, \text{cm} = 4.5 \, \text{cm} \]

### Final Answer
The height of the image in the concave mirror is **4.5 cm**.

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