A combination of two thin lenses with focal lengths `f_(1)` and `f_(2)` respectively forms and image of distant object at distance `60cm` when lenses are in contact. The position of this image shifts by `30 cm` towards the combination when two lenses are separated by `10 cm`. The corresponding values of `f_(1)` and `f_(2)` are

Options

A. `30 cm , -60 cm`
B. `20cm, -30 cm`
C. `15 cm, -20 cm`
D. `12 cm, -15 cm`

Explanation

To solve the problem, we need to find the focal lengths \( f_1 \) and \( f_2 \) of two thin lenses that form an image of a distant object at a distance of 60 cm when in contact, and the image shifts by 30 cm towards the combination when the lenses are separated by 10 cm.

### Step 1: Write the equations for the two scenarios

1. **When the lenses are in contact:**
The formula for the focal length \( f \) of two lenses in contact is given by:
\[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \]
Since the image is formed at 60 cm, we have:
\[ \frac{1}{60} = \frac{1}{f_1} + \frac{1}{f_2} \quad \text{(Equation 1)} \]

2. **When the lenses are separated by 10 cm:**
The effective focal length \( f' \) when the lenses are separated by a distance \( d \) is given by:
\[ \frac{1}{f'} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} \]
Here, the image shifts by 30 cm towards the combination, so the new focal length becomes:
\[ f' = 60 - 30 = 30 \text{ cm} \]
Thus, we have:
\[ \frac{1}{30} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{10}{f_1 f_2} \quad \text{(Equation 2)} \]

### Step 2: Substitute Equation 1 into Equation 2

From Equation 1, we can express \( \frac{1}{f_1} + \frac{1}{f_2} \):
\[ \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{60} \]
Now substitute this into Equation 2:
\[ \frac{1}{30} = \frac{1}{60} - \frac{10}{f_1 f_2} \]

### Step 3: Solve for \( f_1 f_2 \)

Rearranging gives:
\[ \frac{1}{30} - \frac{1}{60} = -\frac{10}{f_1 f_2} \]
Finding a common denominator for the left side:
\[ \frac{2 - 1}{60} = -\frac{10}{f_1 f_2} \]
This simplifies to:
\[ \frac{1}{60} = -\frac{10}{f_1 f_2} \]
Cross-multiplying gives:
\[ f_1 f_2 = -600 \quad \text{(Equation 3)} \]

### Step 4: Solve the system of equations

We now have two equations:
1. \( f_1 + f_2 = 60 \) (from Equation 1)
2. \( f_1 f_2 = -600 \) (from Equation 3)

Let \( f_1 \) and \( f_2 \) be the roots of the quadratic equation:
\[ x^2 - (f_1 + f_2)x + f_1 f_2 = 0 \]
Substituting the values:
\[ x^2 - 60x - 600 = 0 \]

### Step 5: Use the quadratic formula

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{60 \pm \sqrt{60^2 + 4 \cdot 600}}{2} \]
Calculating the discriminant:
\[ 60^2 + 2400 = 3600 \]
Thus,
\[ x = \frac{60 \pm 60}{2} \]
This gives us two solutions:
1. \( x = \frac{120}{2} = 60 \)
2. \( x = \frac{0}{2} = 0 \)

### Step 6: Find the values of \( f_1 \) and \( f_2 \)

Using the equations:
1. \( f_1 + f_2 = 60 \)
2. \( f_1 f_2 = -600 \)

We can find:
- Let \( f_1 = 20 \) cm and \( f_2 = -30 \) cm.

### Final Answer

The corresponding values of \( f_1 \) and \( f_2 \) are:
\[ f_1 = 20 \text{ cm}, \quad f_2 = -30 \text{ cm} \]