An object `5 cm` in length is held `25 cm` away from a converging lens of focal length `10 cm`. Then the position and height of the image is

Options

A. 50/3 cm , 3.33 cm
B. -50/3 cm , -3.33 cm
C. 50/3 cm , -3.33 cm
D. -50/3 cm , 3.33 cm

Explanation

To solve the problem step by step, we will follow the lens formula and magnification formula to find the position and height of the image formed by a converging lens.

### Step 1: Identify Given Values
- Object height (H1) = 5 cm
- Object distance (U) = -25 cm (negative because the object is on the same side as the incoming light)
- Focal length (F) = +10 cm (positive for a converging lens)

### Step 2: Use the Lens Formula
The lens formula is given by:
\[ \frac{1}{V} - \frac{1}{U} = \frac{1}{F} \]
Rearranging gives:
\[ \frac{1}{V} = \frac{1}{F} + \frac{1}{U} \]

### Step 3: Substitute the Values
Substituting the values of F and U into the equation:
\[ \frac{1}{V} = \frac{1}{10} + \frac{1}{-25} \]
Calculating the right side:
\[ \frac{1}{V} = \frac{1}{10} - \frac{1}{25} \]

### Step 4: Find a Common Denominator
The common denominator for 10 and 25 is 50:
\[ \frac{1}{V} = \frac{5}{50} - \frac{2}{50} = \frac{3}{50} \]

### Step 5: Calculate V
Taking the reciprocal gives:
\[ V = \frac{50}{3} \approx 16.67 \text{ cm} \]
This indicates that the image is formed at approximately 16.67 cm on the opposite side of the lens.

### Step 6: Calculate Magnification
Magnification (m) is given by:
\[ m = \frac{H2}{H1} = -\frac{V}{U} \]
Substituting the values of V and U:
\[ m = -\frac{50/3}{-25} = \frac{50/3}{25} = \frac{2}{3} \]

### Step 7: Find the Height of the Image (H2)
Using the magnification formula:
\[ H2 = m \times H1 = \frac{2}{3} \times 5 \text{ cm} = \frac{10}{3} \text{ cm} \approx 3.33 \text{ cm} \]
The positive value indicates that the image is upright.

### Step 8: Determine the Nature of the Image
- The image is real (since V is positive).
- The image is inverted (due to the negative sign in the magnification).
- The image is diminished (since the height of the image is less than the height of the object).

### Final Results
- Position of the image (V) = 16.67 cm
- Height of the image (H2) = 3.33 cm
- Nature of the image: Real, inverted, and diminished.