A tree is `18.0 m` away from `2.0 m` high from a concave lens. How high is the image formed by the given lens of focal length `6 m` ?

Options

A. `1.0 m`
B. `1.5 m`
C. `0.75 m`
D. `0.50 m`

Explanation

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

### Step 1: Identify the given parameters
- Distance of the object (tree), \( u = -18.0 \, m \) (negative because it is on the same side as the incoming light)
- Height of the object (tree), \( h_o = 2.0 \, m \)
- Focal length of the concave lens, \( f = -6.0 \, m \) (negative for concave lenses)

### Step 2: Use the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the known values:
\[ \frac{1}{-6} = \frac{1}{v} - \frac{1}{-18} \]
This simplifies to:
\[ \frac{1}{-6} = \frac{1}{v} + \frac{1}{18} \]

### Step 3: Rearranging to find \( v \)
To isolate \( \frac{1}{v} \), we can rearrange the equation:
\[ \frac{1}{v} = \frac{1}{-6} - \frac{1}{18} \]
Finding a common denominator (which is 18):
\[ \frac{1}{v} = \frac{-3}{18} - \frac{1}{18} = \frac{-4}{18} \]
Thus,
\[ \frac{1}{v} = -\frac{2}{9} \]
Taking the reciprocal gives:
\[ v = -\frac{9}{2} = -4.5 \, m \]

### Step 4: Calculate the magnification
The magnification \( m \) is given by:
\[ m = \frac{h_i}{h_o} = \frac{v}{u} \]
Substituting the values we have:
\[ m = \frac{-4.5}{-18} = \frac{1}{4} \]

### Step 5: Find the height of the image
Using the magnification to find the height of the image:
\[ h_i = m \cdot h_o = \frac{1}{4} \cdot 2.0 \, m = 0.5 \, m \]

### Final Answer
The height of the image formed by the concave lens is \( 0.5 \, m \).

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