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 can follow these steps:

### Step 1: Understand the initial conditions
We know that the image is formed at a distance of 30 cm to the right of the lens. Let's denote:
- \( V_1 = 30 \, \text{cm} \) (image distance)
- \( U_1 \) (object distance, which we need to find)
- \( F \) (focal length of the lens)

### Step 2: Use the lens formula
The lens formula is given by:
\[ \frac{1}{F} = \frac{1}{V} + \frac{1}{U} \]
Substituting the known values:
\[ \frac{1}{F} = \frac{1}{30} + \frac{1}{U_1} \quad \text{(1)} \]

### Step 3: Analyze the second situation
When the lens is moved 5 cm to the right, the new object distance becomes:
\[ U_2 = U_1 + 5 \]
The screen is moved 5 cm to the left, so the new image distance becomes:
\[ V_2 = 30 - 5 = 25 \, \text{cm} \]

### Step 4: Apply the lens formula again
Using the lens formula for the new situation:
\[ \frac{1}{F} = \frac{1}{25} + \frac{1}{U_2} \quad \text{(2)} \]

### Step 5: Substitute \( U_2 \) into equation (2)
Substituting \( U_2 \) into equation (2):
\[ \frac{1}{F} = \frac{1}{25} + \frac{1}{U_1 + 5} \]

### Step 6: Set equations (1) and (2) equal
Since both equations equal \( \frac{1}{F} \), we can set them equal to each other:
\[ \frac{1}{30} + \frac{1}{U_1} = \frac{1}{25} + \frac{1}{U_1 + 5} \]

### Step 7: Cross-multiply to eliminate fractions
Cross-multiplying gives:
\[ \left(\frac{1}{30} - \frac{1}{25}\right) = \left(\frac{1}{U_1 + 5} - \frac{1}{U_1}\right) \]

### Step 8: Simplify the left side
Calculating the left side:
\[ \frac{1}{30} - \frac{1}{25} = \frac{25 - 30}{750} = -\frac{5}{750} = -\frac{1}{150} \]

### Step 9: Simplify the right side
The right side can be simplified as:
\[ \frac{U_1 - (U_1 + 5)}{U_1(U_1 + 5)} = \frac{-5}{U_1(U_1 + 5)} \]

### Step 10: Set the two sides equal
Setting both sides equal:
\[ -\frac{1}{150} = -\frac{5}{U_1(U_1 + 5)} \]

### Step 11: Cross-multiply to solve for \( U_1 \)
Cross-multiplying gives:
\[ U_1(U_1 + 5) = 750 \]
This simplifies to:
\[ U_1^2 + 5U_1 - 750 = 0 \]

### Step 12: Solve the quadratic equation
Using the quadratic formula \( U = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 5, c = -750 \):
\[ U_1 = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-750)}}{2 \cdot 1} \]
\[ U_1 = \frac{-5 \pm \sqrt{25 + 3000}}{2} \]
\[ U_1 = \frac{-5 \pm \sqrt{3025}}{2} \]
\[ U_1 = \frac{-5 \pm 55}{2} \]
Calculating the two possible values:
1. \( U_1 = \frac{50}{2} = 25 \)
2. \( U_1 = \frac{-60}{2} = -30 \) (not valid since distance cannot be negative)

Thus, \( U_1 = 25 \, \text{cm} \).

### Step 13: Find the focal length \( F \)
Substituting \( U_1 \) back into equation (1):
\[ \frac{1}{F} = \frac{1}{30} + \frac{1}{25} \]
Finding a common denominator (150):
\[ \frac{1}{F} = \frac{5}{150} + \frac{6}{150} = \frac{11}{150} \]
Thus,
\[ F = \frac{150}{11} \approx 13.64 \, \text{cm} \]

### Final Answer
The focal length of the lens is approximately \( 13.64 \, \text{cm} \).