A point object is placed on the principal axis of a convex lens (f = 15 cm) at a distance of 30 cm from it. A glass plate `(mu= 1.50)` of thickness 1 cm is placed on the other side of the lens perpendicular to the axis. Locate the image of the point object.

Explanation

To solve the problem step-by-step, we will follow these steps:

### Step 1: Identify the given parameters
- Focal length of the convex lens, \( f = 15 \, \text{cm} \)
- Object distance from the lens, \( u = -30 \, \text{cm} \) (negative as per the sign convention)
- Refractive index of the glass plate, \( \mu = 1.5 \)
- Thickness of the glass plate, \( t = 1 \, \text{cm} \)

### Step 2: Use the lens formula to find the image distance without the glass plate
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Rearranging this gives:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]

Substituting the values:
\[ \frac{1}{v} = \frac{1}{15} + \frac{1}{-30} \]
Calculating the right side:
\[ \frac{1}{v} = \frac{1}{15} - \frac{1}{30} \]
Finding a common denominator (30):
\[ \frac{1}{v} = \frac{2}{30} - \frac{1}{30} = \frac{1}{30} \]
Thus,
\[ v = 30 \, \text{cm} \]

### Step 3: Calculate the shift due to the glass plate
The shift \( S \) caused by the glass plate is given by:
\[ S = \left(1 - \frac{\mu_{\text{surrounding}}}{\mu_{\text{glass}}}\right) \cdot t \]
Here, \( \mu_{\text{surrounding}} = 1 \) (for air), and substituting the values:
\[ S = \left(1 - \frac{1}{1.5}\right) \cdot 1 \]
Calculating the shift:
\[ S = \left(1 - \frac{2}{3}\right) \cdot 1 = \frac{1}{3} \, \text{cm} \approx 0.33 \, \text{cm} \]

### Step 4: Determine the final image distance
The final image distance \( v' \) after accounting for the shift is:
\[ v' = v + S = 30 \, \text{cm} + 0.33 \, \text{cm} = 30.33 \, \text{cm} \]

### Conclusion
The image of the point object is located at a distance of \( 30.33 \, \text{cm} \) to the right of the lens.

---