The distance between an object and the screen is 75 cm. When a convex lens of focal length 12 cm is placed in the between theobject and the screen, magnification of real formed can be (magnitude only)

Options

A. 2
B. 4
C. 8
D. 16

Answer

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

### Step 1: Understand the given information
- The distance between the object and the screen is 75 cm.
- The focal length of the convex lens (f) is 12 cm.

### Step 2: Define the variables
- Let the distance of the lens from the object be \( x \).
- Therefore, the distance of the lens from the screen will be \( 75 - x \).

### Step 3: Use the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \( f \) = focal length of the lens (12 cm)
- \( u \) = object distance (which will be negative in our case, so \( u = -x \))
- \( v \) = image distance (which will be positive, so \( v = 75 - x \))

Substituting these values into the lens formula:
\[ \frac{1}{12} = \frac{1}{75 - x} - \frac{1}{-x} \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ \frac{1}{12} = \frac{1}{75 - x} + \frac{1}{x} \]
Multiplying through by \( 12x(75 - x) \) to eliminate the denominators:
\[ x(75 - x) = 12(75 - x) + 12x \]
This simplifies to:
\[ x(75 - x) = 900 \]

### Step 5: Formulate the quadratic equation
Expanding and rearranging the equation gives:
\[ x^2 - 75x + 900 = 0 \]

### Step 6: Solve the quadratic equation
We can factor this quadratic equation:
\[ (x - 60)(x - 15) = 0 \]
Thus, the solutions for \( x \) are:
\[ x = 60 \quad \text{or} \quad x = 15 \]

### Step 7: Calculate magnification for both cases
The magnification \( m \) is given by:
\[ m = \frac{v}{u} \]

#### Case 1: \( x = 60 \)
- \( u = -60 \)
- \( v = 75 - 60 = 15 \)
\[ m = \frac{15}{-60} = -\frac{1}{4} \]

#### Case 2: \( x = 15 \)
- \( u = -15 \)
- \( v = 75 - 15 = 60 \)
\[ m = \frac{60}{-15} = -4 \]

### Step 8: Find the magnitude of magnification
The magnification can be \( \frac{1}{4} \) or \( 4 \). Since the question asks for the magnitude only, the possible magnifications are:
- Magnitude of \( m = \frac{1}{4} \) is \( \frac{1}{4} \)
- Magnitude of \( m = -4 \) is \( 4 \)

### Final Answer
The magnitude of the magnification of the real image formed can be \( 4 \).

---