The image of a needle placed `45 cm` from a lens is formed on a screen placed `90 cm` on the other side of lens. Find displacement of image if object is moved `5 cm` away from lens.

Options

A. `10 cm`, towards the lens
B. `15 cm,` away from the lens
C. `15 cm`, towards the lens
D. `10 cm`, away from the lens

Explanation

To solve the problem step by step, we will use the lens formula and the information provided in the question.

### Step 1: Identify the given values
- Initial object distance (U) = -45 cm (negative because it is on the same side as the object)
- Initial image distance (V) = +90 cm (positive because it is on the opposite side of the lens)
- Object is moved 5 cm away, so the new object distance (U') = -45 cm - 5 cm = -50 cm

### Step 2: Use the lens formula to find the focal length (F)
The lens formula is given by:
\[ \frac{1}{F} = \frac{1}{V} - \frac{1}{U} \]
Substituting the initial values:
\[ \frac{1}{F} = \frac{1}{90} - \frac{1}{-45} \]
This simplifies to:
\[ \frac{1}{F} = \frac{1}{90} + \frac{1}{45} \]
Finding a common denominator (which is 90):
\[ \frac{1}{F} = \frac{1}{90} + \frac{2}{90} = \frac{3}{90} \]
Thus,
\[ F = \frac{90}{3} = 30 \text{ cm} \]

### Step 3: Calculate the new image distance (V') when the object is moved to U' = -50 cm
Using the lens formula again with the new object distance:
\[ \frac{1}{F} = \frac{1}{V'} - \frac{1}{U'} \]
Substituting the values:
\[ \frac{1}{30} = \frac{1}{V'} - \frac{1}{-50} \]
This simplifies to:
\[ \frac{1}{30} = \frac{1}{V'} + \frac{1}{50} \]
Rearranging gives:
\[ \frac{1}{V'} = \frac{1}{30} - \frac{1}{50} \]
Finding a common denominator (which is 150):
\[ \frac{1}{V'} = \frac{5}{150} - \frac{3}{150} = \frac{2}{150} \]
Thus,
\[ V' = \frac{150}{2} = 75 \text{ cm} \]

### Step 4: Calculate the displacement of the image
The displacement of the image is the difference between the initial image distance and the new image distance:
\[ \text{Displacement} = V - V' = 90 \text{ cm} - 75 \text{ cm} = 15 \text{ cm} \]

### Conclusion
The displacement of the image when the object is moved 5 cm away from the lens is **15 cm towards the lens**.

---