A point object O is placed on the principal axis of a convex lens of focal length 10 cm at 12 cm from the lens. When object is displaced 1mm along the principal axis magnitude of displacement of image is `x_(1)`. When the lens is displaced by 1mm perpendicular to the principal axis displacement of image is `x_(2)` in magnitude. find the value of `(x_(1))/(x_(2))`

Answer

To solve the problem, we need to analyze the displacement of the image when the object is displaced along the principal axis and when the lens is displaced perpendicular to the principal axis. We will use the lens formula and the concept of magnification.

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

### Step 2: Calculate the initial image distance
Given:
- Focal length \( f = 10 \, \text{cm} \)
- Object distance \( u = -12 \, \text{cm} \) (negative because the object is on the same side as the incoming light)

Using the lens formula:
\[ \frac{1}{10} = \frac{1}{v} - \frac{1}{-12} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{10} + \frac{1}{12} \]
Finding a common denominator (60):
\[ \frac{1}{v} = \frac{6}{60} + \frac{5}{60} = \frac{11}{60} \]
Thus:
\[ v = \frac{60}{11} \approx 5.45 \, \text{cm} \]

### Step 3: Displacement of the object
When the object is displaced by \( \Delta u = 1 \, \text{mm} = 0.1 \, \text{cm} \), the new object distance becomes:
\[ u' = -12 + 0.1 = -11.9 \, \text{cm} \]

### Step 4: Calculate the new image distance
Using the lens formula again:
\[ \frac{1}{f} = \frac{1}{v'} - \frac{1}{u'} \]
Substituting the new object distance:
\[ \frac{1}{10} = \frac{1}{v'} - \frac{1}{-11.9} \]
Rearranging gives:
\[ \frac{1}{v'} = \frac{1}{10} + \frac{1}{11.9} \]
Finding a common denominator (119):
\[ \frac{1}{v'} = \frac{11.9}{119} + \frac{10}{119} = \frac{21.9}{119} \]
Thus:
\[ v' = \frac{119}{21.9} \approx 5.43 \, \text{cm} \]

### Step 5: Calculate the displacement of the image
The displacement of the image when the object is displaced is:
\[ x_1 = v' - v \approx 5.43 - 5.45 = -0.02 \, \text{cm} = -0.2 \, \text{mm} \]

### Step 6: Displacement of the lens
When the lens is displaced by \( \Delta d = 1 \, \text{mm} \) perpendicular to the principal axis, the new image distance is calculated by considering the new position of the lens. The effective object distance remains the same, but the image distance will change due to the lens displacement.

### Step 7: Calculate the new image distance
The new image distance will be:
\[ v'' = v + \Delta d \cdot \frac{v}{f} = 5.45 + 1 \cdot \frac{5.45}{10} = 5.45 + 0.545 = 5.995 \, \text{cm} \]

### Step 8: Calculate the displacement of the image
The displacement of the image when the lens is displaced is:
\[ x_2 = v'' - v \approx 5.995 - 5.45 = 0.545 \, \text{cm} = 5.45 \, \text{mm} \]

### Step 9: Calculate the ratio \( \frac{x_1}{x_2} \)
Now, we can find the ratio:
\[ \frac{x_1}{x_2} = \frac{-0.2 \, \text{mm}}{5.45 \, \text{mm}} = \frac{-0.2}{5.45} \approx -0.0367 \]

### Final Result
The value of \( \frac{x_1}{x_2} \) is approximately \( -0.0367 \).

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