A converging lens with a focal length of 12cm produces a 3cm light virtual image of a 1cm hgh object. When entry in the table below is correct?

Options

A. `{:("Image distance",,"Location of image") ,(8cm,,"Same side lens as object"):}`
B. `{:("Image distance",,"Location of image") ,(8cm,,"Opposite side of lens from object"):}`
C. `{:("Image distance",,"Location of image") ,(12cm,,"Opposite side of lens from object"):}`
D. `{:("Image distance",,"Location of image") ,(24cm,,"Same side of lens as object"):}`

Answer

To solve the problem step by step, we will use the lens formula and the magnification formula for a converging lens.

### Step 1: Understand the given data
- Focal length of the lens (f) = 12 cm
- Height of the object (h_o) = 1 cm
- Height of the image (h_i) = 3 cm (virtual image)
- Since the image is virtual, it is erect and located on the same side as the object.

### Step 2: Calculate the magnification (m)
The magnification (m) is given by the formula:
\[ m = \frac{h_i}{h_o} \]
Substituting the values:
\[ m = \frac{3 \text{ cm}}{1 \text{ cm}} = 3 \]
Since the image is virtual and erect, the magnification is positive:
\[ m = +3 \]

### Step 3: Relate magnification to object distance (u) and image distance (v)
The magnification can also be expressed in terms of object distance (u) and image distance (v):
\[ m = -\frac{v}{u} \]
From the previous step, we know:
\[ 3 = -\frac{v}{u} \]
This implies:
\[ v = -3u \]

### Step 4: Use the lens formula
The lens formula relates the focal length (f), object distance (u), and image distance (v):
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting \( f = 12 \text{ cm} \):
\[ \frac{1}{12} = \frac{1}{v} - \frac{1}{u} \]

### Step 5: Substitute \( v \) in terms of \( u \)
From Step 3, we have \( v = -3u \). Substitute this into the lens formula:
\[ \frac{1}{12} = \frac{1}{-3u} - \frac{1}{u} \]
Finding a common denominator:
\[ \frac{1}{12} = -\frac{1}{3u} + \frac{1}{u} \]
\[ \frac{1}{12} = -\frac{1}{3u} + \frac{3}{3u} \]
\[ \frac{1}{12} = \frac{2}{3u} \]

### Step 6: Solve for \( u \)
Cross-multiplying gives:
\[ 2 \cdot 12 = 3u \]
\[ 24 = 3u \]
\[ u = \frac{24}{3} = 8 \text{ cm} \]

### Step 7: Find \( v \)
Now substitute \( u \) back into the equation for \( v \):
\[ v = -3u = -3 \times 8 = -24 \text{ cm} \]

### Step 8: Conclusion
- The image distance \( v = -24 \text{ cm} \) indicates that the image is virtual and located on the same side as the object.
- The object distance \( u = 8 \text{ cm} \) is between the focal point and the lens.

### Step 9: Check the options
Based on the calculations:
- The correct entry for the image distance in the table is **-24 cm** (option D).