A biconvex lens is used to project a slide on screen. The slide is `2 cm` high and placed at `10 cm` from the lens. The image is `18 cm` high. What is the focal length of the lens?

Options

A. `9 cm`
B. `18 cm`
C. `4.5 cm`
D. `20 cm`

Explanation

To find the focal length of the biconvex lens used to project a slide, we can follow these steps:

### Step 1: Identify Given Values
- Height of the object (slide), \( H_o = 2 \, \text{cm} \)
- Distance of the object from the lens, \( U = -10 \, \text{cm} \) (the object distance is taken as negative in lens formula)
- Height of the image, \( H_i = 18 \, \text{cm} \)

### Step 2: Calculate Magnification
The magnification \( m \) of the lens is given by the formula:
\[ m = \frac{H_i}{H_o} = \frac{-V}{U} \]
Where \( V \) is the image distance.

Substituting the known values:
\[ m = \frac{18}{2} = 9 \]
Since the image is real and inverted, we can write:
\[ 9 = \frac{-V}{-10} \]
Thus, we can solve for \( V \):
\[ V = 9 \times 10 = 90 \, \text{cm} \]

### Step 3: Use the Lens Formula
The lens formula relates the object distance \( U \), image distance \( V \), and focal length \( F \):
\[ \frac{1}{F} = \frac{1}{V} - \frac{1}{U} \]

Substituting the values we found:
\[ \frac{1}{F} = \frac{1}{90} - \frac{1}{-10} \]
This simplifies to:
\[ \frac{1}{F} = \frac{1}{90} + \frac{1}{10} \]
Converting \( \frac{1}{10} \) to a common denominator:
\[ \frac{1}{10} = \frac{9}{90} \]
Thus:
\[ \frac{1}{F} = \frac{1}{90} + \frac{9}{90} = \frac{10}{90} \]

### Step 4: Calculate the Focal Length
Now, we can find \( F \):
\[ F = \frac{90}{10} = 9 \, \text{cm} \]

### Final Answer
The focal length of the lens is \( 9 \, \text{cm} \).

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