A thin converging lens of focal length f = 25 cm forms the image of an object on a screen placed at a distance of 75 cm from the lens. Now the screen in moved closer to the lens by a distance of 25 cm. The distance through which the object has to be shifted so that its image on the screen is sharp again is

Options

A. 37.5 cm
B. 16.25 cm
C. 12.5 cm
D. 13.5 cm

Answer

To solve the problem step by step, we will use the lens formula and the given information about the distances involved.

### Step 1: Understand the initial conditions
We have a converging lens with a focal length \( f = 25 \, \text{cm} \). The initial image distance \( v_1 \) is given as \( 75 \, \text{cm} \).

### Step 2: Apply the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \( f \) is the focal length,
- \( v \) is the image distance,
- \( u \) is the object distance.

### Step 3: Calculate the initial object distance
Using the lens formula for the initial conditions:
\[ \frac{1}{25} = \frac{1}{75} - \frac{1}{u_1} \]
Rearranging gives:
\[ \frac{1}{u_1} = \frac{1}{75} - \frac{1}{25} \]
Finding a common denominator (which is 75):
\[ \frac{1}{u_1} = \frac{1 - 3}{75} = \frac{-2}{75} \]
Thus,
\[ u_1 = -\frac{75}{2} = -37.5 \, \text{cm} \]
(Note: The negative sign indicates that the object is on the same side as the incoming light.)

### Step 4: Determine the new image distance
When the screen is moved closer by \( 25 \, \text{cm} \), the new image distance \( v_2 \) becomes:
\[ v_2 = 75 - 25 = 50 \, \text{cm} \]

### Step 5: Calculate the new object distance
Now, we apply the lens formula again for the new image distance:
\[ \frac{1}{25} = \frac{1}{50} - \frac{1}{u_2} \]
Rearranging gives:
\[ \frac{1}{u_2} = \frac{1}{50} - \frac{1}{25} \]
Finding a common denominator (which is 50):
\[ \frac{1}{u_2} = \frac{1 - 2}{50} = \frac{-1}{50} \]
Thus,
\[ u_2 = -50 \, \text{cm} \]

### Step 6: Calculate the distance the object has to shift
The distance the object has to shift is given by:
\[ \Delta u = u_1 - u_2 = (-37.5) - (-50) = 12.5 \, \text{cm} \]

### Final Answer
The distance through which the object has to be shifted to get a sharp image again is \( 12.5 \, \text{cm} \).

---