A convex lens forms an image of an object on a screen 30 cm from the lens. When the lens is moved 90 cm towards the object, then the image is again formed on the screen. Then, the focal length of the lens is

Options

A. 13 cm
B. 24 cm
C. 33 cm
D. 40 cm

Explanation

To solve the problem step by step, we will use the lens formula and analyze the two scenarios given in the question.

### Step 1: Understand the given information
- In the first scenario, the image is formed 30 cm from the lens.
- In the second scenario, the lens is moved 90 cm towards the object, and the image is formed again on the same screen.

### Step 2: Define the variables
Let:
- \( v_1 = 30 \) cm (image distance in the first case)
- \( u_1 = -x \) (object distance in the first case, negative as per convention)
- \( v_2 = 120 \) cm (image distance in the second case, since it is now 30 cm + 90 cm)
- \( u_2 = -(x - 90) \) (object distance in the second case)

### Step 3: Write the lens formula for both cases
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

#### For the first case:
\[ \frac{1}{f} = \frac{1}{30} - \frac{1}{-x} \]
This simplifies to:
\[ \frac{1}{f} = \frac{1}{30} + \frac{1}{x} \quad \text{(Equation 1)} \]

#### For the second case:
\[ \frac{1}{f} = \frac{1}{120} - \frac{1}{-(x - 90)} \]
This simplifies to:
\[ \frac{1}{f} = \frac{1}{120} + \frac{1}{x - 90} \quad \text{(Equation 2)} \]

### Step 4: Set the two equations equal to each other
Since the focal length \( f \) is the same for both cases, we can set Equation 1 equal to Equation 2:
\[ \frac{1}{30} + \frac{1}{x} = \frac{1}{120} + \frac{1}{x - 90} \]

### Step 5: Solve for \( x \)
To solve this equation, we will first eliminate the fractions by finding a common denominator. The common denominator for 30, 120, and \( x(x - 90) \) can be used. However, for simplicity, we can cross-multiply and rearrange:
\[ \frac{1}{30} - \frac{1}{120} = \frac{1}{x - 90} - \frac{1}{x} \]

Calculating the left side:
\[ \frac{1}{30} - \frac{1}{120} = \frac{4 - 1}{120} = \frac{3}{120} = \frac{1}{40} \]

Now, equating the right side:
\[ \frac{1}{40} = \frac{1}{x - 90} - \frac{1}{x} \]

Finding a common denominator for the right side:
\[ \frac{1}{40} = \frac{x - (x - 90)}{x(x - 90)} = \frac{90}{x(x - 90)} \]

Cross-multiplying gives:
\[ 90 \cdot 40 = x(x - 90) \]
\[ 3600 = x^2 - 90x \]
\[ x^2 - 90x - 3600 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -90, c = -3600 \):
\[ x = \frac{90 \pm \sqrt{(-90)^2 - 4 \cdot 1 \cdot (-3600)}}{2 \cdot 1} \]
\[ x = \frac{90 \pm \sqrt{8100 + 14400}}{2} \]
\[ x = \frac{90 \pm \sqrt{22500}}{2} \]
\[ x = \frac{90 \pm 150}{2} \]
Calculating the two possible values:
1. \( x = \frac{240}{2} = 120 \)
2. \( x = \frac{-60}{2} = -30 \) (not valid since distance cannot be negative)

Thus, \( x = 120 \) cm.

### Step 7: Substitute \( x \) back to find \( f \)
Now substitute \( x = 120 \) cm back into either Equation 1 or Equation 2 to find \( f \):
Using Equation 1:
\[ \frac{1}{f} = \frac{1}{30} + \frac{1}{120} \]
Finding a common denominator (120):
\[ \frac{1}{f} = \frac{4 + 1}{120} = \frac{5}{120} = \frac{1}{24} \]
Thus, \( f = 24 \) cm.

### Final Answer
The focal length of the lens is \( \boxed{24 \text{ cm}} \).