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 find the height of the image formed by a concave lens, we can use the lens formula and the magnification formula. Here are the steps to solve the problem:

### Step 1: Identify the given values
- Distance of the object (tree) from the lens, \( u = -18.0 \, m \) (the object distance is taken as negative in lens formula)
- Height of the object (tree), \( h_o = 2.0 \, m \)
- Focal length of the lens, \( f = -6.0 \, m \) (the focal length of a concave lens is negative)

### Step 2: Use the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \( v \) = image distance from the lens
- \( f \) = focal length
- \( u \) = object distance

Rearranging the formula to find \( v \):
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]

### Step 3: Substitute the values into the lens formula
Substituting the known values:
\[ \frac{1}{v} = \frac{1}{-6} + \frac{1}{-18} \]

### Step 4: Calculate \( \frac{1}{v} \)
Calculating the right side:
\[ \frac{1}{v} = -\frac{1}{6} - \frac{1}{18} \]
To add these fractions, find a common denominator (which is 18):
\[ \frac{1}{v} = -\frac{3}{18} - \frac{1}{18} = -\frac{4}{18} = -\frac{2}{9} \]

### Step 5: Find \( v \)
Taking the reciprocal to find \( v \):
\[ v = -\frac{9}{2} = -4.5 \, m \]
The negative sign indicates that the image is formed on the same side as the object (which is typical for a concave lens).

### Step 6: Calculate the magnification
The magnification \( m \) is given by:
\[ m = -\frac{v}{u} \]
Substituting the values:
\[ m = -\frac{-4.5}{-18} = \frac{4.5}{18} = \frac{1}{4} \]

### Step 7: Find the height of the image
The height of the image \( h_i \) can be calculated using the magnification:
\[ h_i = m \cdot h_o \]
Substituting the values:
\[ h_i = \frac{1}{4} \cdot 2.0 = 0.5 \, m \]

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

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