A square card of side length 1 mm is being seen through a magnifying lens of focal length 10 cm. The card is placed at a distance of 9 cm from the lens. The appaent area of the card thorugh the lens is

Options

A. `1 cm^(2)`
B. `0.81 cm^(2)`
C. `0.27 cm^(2)`
D. `0.60 cm^(2)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the area of the square card.
The area \( A \) of a square is given by the formula:
\[ A = \text{side} \times \text{side} \]
Given that the side length of the card is 1 mm, we can calculate the area:
\[ A = 1 \, \text{mm} \times 1 \, \text{mm} = 1 \, \text{mm}^2 \]

### Step 2: Identify the given values.
- Focal length of the lens \( f = 10 \, \text{cm} \)
- Object distance \( u = -9 \, \text{cm} \) (the negative sign indicates that the object is on the same side as the incoming light)

### Step 3: Use the lens formula to find the image distance \( v \).
The lens formula is:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]
Substituting the known values:
\[ \frac{1}{v} = \frac{1}{10} + \frac{1}{-9} \]
Calculating the right side:
\[ \frac{1}{v} = \frac{1}{10} - \frac{1}{9} \]
Finding a common denominator (which is 90):
\[ \frac{1}{v} = \frac{9}{90} - \frac{10}{90} = \frac{-1}{90} \]
Thus,
\[ v = -90 \, \text{cm} \]

### Step 4: Calculate the magnification \( m \).
Magnification is given by the formula:
\[ m = -\frac{v}{u} \]
Substituting the values we found:
\[ m = -\frac{-90}{-9} = 10 \]

### Step 5: Calculate the apparent area of the card through the lens.
The apparent area \( A' \) through the lens can be calculated using the magnification:
\[ A' = m^2 \times A \]
Substituting the magnification and the area of the card:
\[ A' = 10^2 \times 1 \, \text{mm}^2 = 100 \, \text{mm}^2 \]

### Step 6: Convert the area to cm².
Since \( 1 \, \text{cm}^2 = 100 \, \text{mm}^2 \):
\[ A' = 1 \, \text{cm}^2 \]

Thus, the apparent area of the card through the lens is \( 1 \, \text{cm}^2 \).

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