A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens. Now a glass block (refractive index 1.5) of 1.5 cm thickness is placed in between the light source and the lens. To get the sharp image again, the screen is shifted by a distance $$d$$. Then $$d$$ is:

$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{10} - \left(-\frac{1}{10}\right) = \frac{2}{10} \implies f = 5\text{ cm}$$

$$\Delta x = t\left(1 - \frac{1}{\mu}\right) = 1.5 \left(1 - \frac{1}{1.5}\right) = 1.5 \left(1 - \frac{2}{3}\right) = 0.5\text{ cm (in direction of incident light)}$$

New object distance from the lens: $$u' = -(10 - \Delta x) = -(10 - 0.5) = -9.5\text{ cm}$$

$$\frac{1}{v'} = \frac{1}{f} + \frac{1}{u'} = \frac{1}{5} - \frac{1}{9.5} = \frac{1}{5} - \frac{2}{19} = \frac{19 - 10}{95} = \frac{9}{95}$$

$$v' = \frac{95}{9} = 10.55\text{ cm}$$

$$d = v' - v = 10.55 - 10 = 0.55\text{ cm (away from the lens)}$$