An object and a screen are fixed at a distance d apart. When a lens of focal length f is moved between the object and the screen, sharp images of the object are formed on the screen for two positions of the lens.
The magnifications produced at these two positions are `M_(1) "and" M_(2)`-

Options

A. d gt 2f
B. dgt4f
C. `M_(1),M_(2)= 1`
D. `|M_(1)|- |M_(2)| = 1`

Explanation

To solve the problem step by step, we will analyze the situation involving a convex lens placed between a fixed object and screen. The goal is to find the relationship between the magnifications produced at two positions of the lens.

### Step 1: Define the Variables
Let:
- \( d \) = distance between the object and the screen
- \( f \) = focal length of the lens
- \( x \) = distance from the object to the lens
- \( v \) = distance from the lens to the screen

From the setup, we can express \( v \) as:
\[ v = d - x \]

### Step 2: Apply the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting \( u = -x \) (the negative sign indicates the object distance is taken as negative in lens formula) and \( v = d - x \):
\[ \frac{1}{f} = \frac{1}{d - x} - \frac{1}{-x} \]
This simplifies to:
\[ \frac{1}{f} = \frac{1}{d - x} + \frac{1}{x} \]

### Step 3: Rearranging the Equation
Combining the fractions on the right side:
\[ \frac{1}{f} = \frac{x + (d - x)}{x(d - x)} \]
This simplifies to:
\[ \frac{1}{f} = \frac{d}{x(d - x)} \]

### Step 4: Cross-Multiplying
Cross-multiplying gives:
\[ f \cdot d = x(d - x) \]
Rearranging this leads to:
\[ x^2 - dx + fd = 0 \]

### Step 5: Solving the Quadratic Equation
This is a quadratic equation in \( x \). The solutions can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -d, c = fd \):
\[ x = \frac{d \pm \sqrt{d^2 - 4fd}}{2} \]

### Step 6: Conditions for Real Solutions
For \( x \) to have real solutions, the discriminant must be non-negative:
\[ d^2 - 4fd \geq 0 \]
This implies:
\[ d \geq 4f \]

### Step 7: Finding the Magnifications
For the two positions of the lens, the magnifications \( M_1 \) and \( M_2 \) can be expressed as:
- For the first position:
\[ M_1 = \frac{v}{u} = \frac{d - x}{x} \]
- For the second position (where the object and screen positions are swapped):
\[ M_2 = \frac{v}{u} = \frac{x}{d - x} \]

### Step 8: Relationship Between Magnifications
Multiplying the two magnifications together:
\[ M_1 \cdot M_2 = \left(\frac{d - x}{x}\right) \cdot \left(\frac{x}{d - x}\right) = 1 \]

### Conclusion
Thus, we conclude that the product of the magnifications at the two positions of the lens is equal to 1:
\[ M_1 \cdot M_2 = 1 \]