A convex lens is put $10\ \text{cm}$ from a light source and it makes a sharp image on a screen, kept $10\ \text{cm}$ from the lens. Now a glass block (refractive index $1.5$) of $1.5\ \text{cm}$ 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
Answer: (B) $0.55\ \text{cm}$ away from the lens
From $u = -10\ \text{cm}$, $v = +10\ \text{cm}$: $\dfrac1f = \dfrac1{10} + \dfrac1{10} \Rightarrow f = 5\ \text{cm}$.
The glass block shifts the object towards the lens by
$$t\left(1 - \frac1\mu\right) = 1.5\left(1 - \frac{1}{1.5}\right) = 0.5\ \text{cm}$$
so now $u = -9.5\ \text{cm}$:
$$\frac1v = \frac15 - \frac{1}{9.5} = \frac{4.5}{47.5} \Rightarrow v \approx 10.55\ \text{cm}$$
The screen must be moved about $0.55\ \text{cm}$ away from the lens.
Solution by Sreeraj P, M.Sc Physics