A point object is placed on the axis in front of a convex lens of focal length 30 cm. The distance of the object from the lens is 45 cm. If the lens is moved away from object by 1 mm and also the object is moved apart from the line joining the object and optical centre by 1 mm the displacement of the image has magnitude-

Options

A. `sqrt(29)mm`
B. `sqrt(34)mm`
C. `2sqrt(5)mm`
D. `sqrt(44)mm`

Answer

To solve the problem, we will use the lens formula and the concept of image displacement due to the movement of the lens and the object.

### Given:
- Focal length of the convex lens, \( f = 30 \, \text{cm} \)
- Initial distance of the object from the lens, \( u = -45 \, \text{cm} \) (object distance is taken as negative in lens formula)
- Movement of the lens away from the object, \( d_L = 1 \, \text{mm} = 0.1 \, \text{cm} \)
- Movement of the object away from the optical axis, \( d_O = 1 \, \text{mm} = 0.1 \, \text{cm} \)

### Step 1: Find the initial image distance using the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]
Substituting the values:
\[ \frac{1}{v} = \frac{1}{30} + \frac{1}{-45} \]
Finding a common denominator (which is 90):
\[ \frac{1}{v} = \frac{3}{90} - \frac{2}{90} = \frac{1}{90} \]
Thus,
\[ v = 90 \, \text{cm} \]

### Step 2: Calculate the new object distance after moving the lens
The lens is moved away from the object by 1 mm. Therefore, the new object distance \( u' \) is:
\[ u' = -45 - 0.1 = -45.1 \, \text{cm} \]

### Step 3: Calculate the new image distance using the lens formula again
Using the lens formula again for the new object distance:
\[ \frac{1}{v'} = \frac{1}{f} + \frac{1}{u'} \]
Substituting the new values:
\[ \frac{1}{v'} = \frac{1}{30} + \frac{1}{-45.1} \]
Finding a common denominator (approximately 1353):
\[ \frac{1}{v'} = \frac{45.1}{1353} - \frac{30}{1353} = \frac{15.1}{1353} \]
Thus,
\[ v' \approx \frac{1353}{15.1} \approx 89.7 \, \text{cm} \]

### Step 4: Calculate the displacement of the image
The displacement of the image \( \Delta v \) is given by:
\[ \Delta v = v' - v = 89.7 - 90 = -0.3 \, \text{cm} \]

### Step 5: Consider the displacement due to the object moving away from the optical axis
Since the object is moved away from the optical axis by 1 mm, this will also affect the image position. For small angles, the displacement of the image due to the object moving laterally can be approximated using the magnification formula:
\[ \text{Magnification} (m) = \frac{h'}{h} = -\frac{v}{u} \]
For small displacements, the lateral displacement of the image \( \Delta y \) can be approximated as:
\[ \Delta y = m \cdot d_O \]
Calculating \( m \):
\[ m = -\frac{90}{-45} = 2 \]
Thus,
\[ \Delta y = 2 \cdot 0.1 = 0.2 \, \text{cm} \]

### Step 6: Total displacement of the image
The total displacement of the image \( \Delta \) is:
\[ \Delta = \Delta v + \Delta y = -0.3 + 0.2 = -0.1 \, \text{cm} \]
The magnitude of the displacement is:
\[ |\Delta| = 0.1 \, \text{cm} = 1 \, \text{mm} \]

### Final Answer:
The displacement of the image has a magnitude of **1 mm**.

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