For position of real object at `x_(1)` and `x_(2) (x_(2)gtx_(1))` magnification is equal to 2. find out `(x_(1))/(x_(2))` if focal length of converging lens `f=20cm`

Options

A. `(1)/(2)`
B. `(1)/(4)`
C. `2`
D. `4`

Answer

To solve the problem, we need to find the ratio \( \frac{x_1}{x_2} \) given that the magnification \( m \) is equal to 2 for two different positions of a real object in front of a converging lens with a focal length \( f = 20 \, \text{cm} \).

### Step-by-Step Solution:

1. **Understanding Magnification**:
The magnification \( m \) for a lens is given by the formula:
\[ m = \frac{h'}{h} = -\frac{v}{u} \]
where \( h' \) is the height of the image, \( h \) is the height of the object, \( v \) is the image distance, and \( u \) is the object distance. Since the problem states that the magnification is 2, we can write:
\[ m = -2 = -\frac{v}{u} \]
This implies \( v = 2u \).

2. **Using the Lens Formula**:
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting \( v = 2u \) into the lens formula:
\[ \frac{1}{20} = \frac{1}{2u} - \frac{1}{u} \]
Simplifying the right side:
\[ \frac{1}{20} = \frac{1 - 2}{2u} = -\frac{1}{2u} \]
Therefore:
\[ 2u = -20 \implies u = -10 \, \text{cm} \]
Since \( u \) is negative for real objects, we take the absolute value:
\[ x_1 = 10 \, \text{cm} \]

3. **Finding \( x_2 \)**:
Now we need to find the second position \( x_2 \) where the magnification is still 2. For this position, we can again use the lens formula:
\[ m = -\frac{v}{u} = -2 \implies v = -2u \]
Substituting into the lens formula:
\[ \frac{1}{20} = \frac{1}{-2u} - \frac{1}{u} \]
Simplifying gives:
\[ \frac{1}{20} = -\frac{1}{2u} - \frac{1}{u} = -\frac{3}{2u} \]
Therefore:
\[ 2u = -60 \implies u = -30 \, \text{cm} \]
Taking the absolute value, we have:
\[ x_2 = 30 \, \text{cm} \]

4. **Finding the Ratio**:
Now, we can find the ratio \( \frac{x_1}{x_2} \):
\[ \frac{x_1}{x_2} = \frac{10}{30} = \frac{1}{3} \]

### Final Answer:
\[ \frac{x_1}{x_2} = \frac{1}{3} \]