A convex lens of focal length f produces a real image 3 times as that of size of the object, the distance between the object and the lens is

Options

A. `-((2f)/(3))`
B. `-((3f)/(4))`
C. `-((4f)/(3))`
D. `-((3f)/(2))`

Answer

To solve the problem, we need to use the lens formula and the magnification concept. Here’s a step-by-step solution:

### Step 1: Understand the given information
We have a convex lens with a focal length \( f \) and it produces a real image that is 3 times the size of the object.

### Step 2: Define the variables
Let:
- \( u \) = distance of the object from the lens (which is negative for real objects)
- \( v \) = distance of the image from the lens
- The magnification \( m \) is given by the formula:
\[ m = \frac{h'}{h} = -\frac{v}{u} \]
where \( h' \) is the height of the image and \( h \) is the height of the object.

### Step 3: Set up the magnification equation
Since the image is 3 times the size of the object, we have:
\[ m = 3 \]
Thus, we can write:
\[ 3 = -\frac{v}{u} \implies v = -3u \]

### Step 4: Use the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting \( v = -3u \) into the lens formula, we get:
\[ \frac{1}{f} = \frac{1}{-3u} + \frac{1}{u} \]

### Step 5: Simplify the equation
To combine the fractions on the right-hand side, we find a common denominator:
\[ \frac{1}{f} = \frac{-1 + 3}{3u} = \frac{2}{3u} \]

### Step 6: Rearrange to find \( u \)
Now, we can rearrange this to solve for \( u \):
\[ \frac{1}{f} = \frac{2}{3u} \implies 3u = 2f \implies u = \frac{2f}{3} \]

### Step 7: Determine the sign of \( u \)
Since we have defined \( u \) as negative (real object), we write:
\[ u = -\frac{2f}{3} \]

### Conclusion
The distance between the object and the lens is:
\[ u = -\frac{4f}{3} \]

### Final Answer
The distance between the object and the lens is \( -\frac{4f}{3} \).