An object to the left of a lens is imaged by the lens on a screen `30cm` to the right of the lens. When the lens is moved ` 5cm` to the right, the screen must be moved `5cm` to the left to refocus the images. The focal length of lens is :-

Options

A. `10cm`
B. `15cm`
C. `20cm`
D. `30cm`

Explanation

To find the focal length of the lens, we will use the lens formula and the information provided in the question. Let's break down the solution step by step.

### Step 1: Understand the given information
- The initial image distance \( V \) is \( 30 \, \text{cm} \) (to the right of the lens).
- When the lens is moved \( 5 \, \text{cm} \) to the right, the new image distance \( V' \) becomes \( 30 - 5 = 25 \, \text{cm} \).
- The screen is then moved \( 5 \, \text{cm} \) to the left, making the new image distance \( V'' = 25 - 5 = 20 \, \text{cm} \).

### Step 2: Apply the lens formula
The lens formula is given by:
\[ \frac{1}{F} = \frac{1}{V} - \frac{1}{U} \]
Where:
- \( F \) is the focal length of the lens.
- \( V \) is the image distance.
- \( U \) is the object distance (which we need to find).

### Step 3: Set up the equations
1. For the initial position of the lens:
\[ \frac{1}{30} = \frac{1}{F} - \frac{1}{U} \quad \text{(1)} \]

2. For the new position of the lens:
\[ \frac{1}{20} = \frac{1}{F} - \frac{1}{(U + 5)} \quad \text{(2)} \]

### Step 4: Solve the equations
From equation (1):
\[ \frac{1}{F} = \frac{1}{30} + \frac{1}{U} \quad \text{(3)} \]

From equation (2):
\[ \frac{1}{F} = \frac{1}{20} + \frac{1}{(U + 5)} \quad \text{(4)} \]

### Step 5: Set equations (3) and (4) equal to each other
\[ \frac{1}{30} + \frac{1}{U} = \frac{1}{20} + \frac{1}{(U + 5)} \]

### Step 6: Cross-multiply and simplify
Cross-multiplying gives:
\[ \left(\frac{1}{30} - \frac{1}{20}\right) = \left(\frac{1}{(U + 5)} - \frac{1}{U}\right) \]

Calculating the left side:
\[ \frac{1}{30} - \frac{1}{20} = \frac{2 - 3}{60} = -\frac{1}{60} \]

Calculating the right side:
\[ \frac{U - (U + 5)}{U(U + 5)} = \frac{-5}{U(U + 5)} \]

Equating both sides:
\[ -\frac{1}{60} = -\frac{5}{U(U + 5)} \]

### Step 7: Cross-multiply again
\[ U(U + 5) = 300 \]
This simplifies to:
\[ U^2 + 5U - 300 = 0 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula:
\[ U = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 1, b = 5, c = -300 \):
\[ U = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-300)}}{2 \cdot 1} \]
\[ U = \frac{-5 \pm \sqrt{25 + 1200}}{2} \]
\[ U = \frac{-5 \pm \sqrt{1225}}{2} \]
\[ U = \frac{-5 \pm 35}{2} \]

Calculating the roots:
1. \( U = \frac{30}{2} = 15 \, \text{cm} \) (valid)
2. \( U = \frac{-40}{2} = -20 \, \text{cm} \) (not valid since distance cannot be negative)

### Step 9: Substitute \( U \) back to find \( F \)
Using \( U = 15 \, \text{cm} \) in equation (1):
\[ \frac{1}{F} = \frac{1}{30} + \frac{1}{15} \]
\[ \frac{1}{F} = \frac{1 + 2}{30} = \frac{3}{30} = \frac{1}{10} \]
Thus,
\[ F = 10 \, \text{cm} \]

### Final Answer
The focal length of the lens is \( \boxed{10 \, \text{cm}} \).