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 the information provided in the question.

### Step 1: Understand the situation
We have a convex lens that forms an image at a distance of 30 cm from the lens. When the lens is moved 90 cm towards the object, the image is still formed on the screen. We need to find the focal length of the lens.

### Step 2: Set up the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
where:
- \( f \) = focal length of the lens
- \( v \) = image distance from the lens
- \( u \) = object distance from the lens

### Step 3: Analyze the first condition
In the first scenario, the image is formed at 30 cm from the lens, so:
- \( v = 30 \) cm
- Let \( u \) be the object distance (which we need to find).

Using the lens formula:
\[ \frac{1}{f} = \frac{1}{30} - \frac{1}{u} \quad \text{(Equation 1)} \]

### Step 4: Analyze the second condition
When the lens is moved 90 cm towards the object, the new object distance becomes \( u - 90 \) cm (since the lens is moving closer to the object). The new image distance is now \( v_1 = 30 + 90 = 120 \) cm (the image remains on the screen).

Using the lens formula again:
\[ \frac{1}{f} = \frac{1}{120} - \frac{1}{(u - 90)} \quad \text{(Equation 2)} \]

### Step 5: Equate the two equations
From Equation 1:
\[ \frac{1}{f} = \frac{1}{30} + \frac{1}{u} \]
From Equation 2:
\[ \frac{1}{f} = \frac{1}{120} + \frac{1}{(u - 90)} \]

Since both equations equal \( \frac{1}{f} \), we can set them equal to each other:
\[ \frac{1}{30} + \frac{1}{u} = \frac{1}{120} + \frac{1}{(u - 90)} \]

### Step 6: Solve for \( u \)
Cross-multiplying to eliminate the fractions:
\[ \frac{1}{u} - \frac{1}{(u - 90)} = \frac{1}{120} - \frac{1}{30} \]

Finding a common denominator for the left side:
\[ \frac{(u - 90) - u}{u(u - 90)} = \frac{1}{120} - \frac{4}{120} \]
\[ \frac{-90}{u(u - 90)} = -\frac{3}{120} \]

Cross-multiplying gives:
\[ 90 \times 120 = 3u(u - 90) \]
\[ 10800 = 3u^2 - 270u \]
\[ 3u^2 - 270u - 10800 = 0 \]
Dividing the entire equation by 3:
\[ u^2 - 90u - 3600 = 0 \]

### Step 7: Use the quadratic formula
Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ u = \frac{90 \pm \sqrt{(-90)^2 - 4 \cdot 1 \cdot (-3600)}}{2 \cdot 1} \]
\[ u = \frac{90 \pm \sqrt{8100 + 14400}}{2} \]
\[ u = \frac{90 \pm \sqrt{22500}}{2} \]
\[ u = \frac{90 \pm 150}{2} \]
Calculating the two possible values:
1. \( u = \frac{240}{2} = 120 \) cm (valid)
2. \( u = \frac{-60}{2} = -30 \) cm (not valid)

### Step 8: Substitute \( u \) back to find \( f \)
Now substituting \( u = 120 \) cm back into Equation 1:
\[ \frac{1}{f} = \frac{1}{30} + \frac{1}{120} \]
Finding a common denominator (120):
\[ \frac{1}{f} = \frac{4}{120} + \frac{1}{120} = \frac{5}{120} \]
Thus:
\[ f = \frac{120}{5} = 24 \text{ cm} \]

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

---