A point object is situated on the principal axis of a convex lens (...

For the convex lens, the object distance is $u = -30 \text{ cm}$ and the focal length is $f = +15 \text{ cm}$.

Using the lens formula:

$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$

$\dfrac{1}{v} - \dfrac{1}{-30} = \dfrac{1}{15}$

$\dfrac{1}{v} = \dfrac{1}{15} - \dfrac{1}{30} = \dfrac{1}{30}$

$v = 30 \text{ cm}$

The lens forms an image at a distance of $30 \text{ cm}$ on the opposite side.

When a glass plate is placed in the path of the converging rays, it produces a shift in the position of the image in the direction of the incident light. The shift is given by:

\Delta x = t \left(1 -\dfrac{1}{\mu}\right)

Substituting $t = 1 \text{ cm}$ and $\mu = 1.5 =$\dfrac{3}{2}:

\Delta x = 1 \left(1 -\dfrac{2}{3}\right) = \dfrac{1}{3} \text{ cm} \approx$0.33 \text{ cm}$

Since the shift is in the direction of the incident light (away from the lens), the distance of the final image from the lens is:

$v' = v + \Delta x = 30 + 0.33 = 30.33 \text{ cm}$