JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Geometrical Optics

Here the lens first forms a sharp image with the source at $u=-10$ cm and the screen at $v=+10$ cm, so from $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{10}-\frac{1}{-10}$ we get $f=5$ cm. Now slip in the glass block of thickness $t=1.5$ cm and refractive index $\mu =1.5$ right against the source. A slab shifts the object towards the lens by $t\left(1-\frac{1}{\mu }\right)=1.5\left(1-\frac{1}{1.5}\right)=0.5$ cm, so the effective object distance becomes $u^{\prime }=-9.5$ cm. Feeding this back into the lens formula, $\frac{1}{v^{\prime }}=\frac{1}{5}-\frac{1}{9.5}=\frac{4.5}{47.5}$, which gives $v^{\prime }\approx 10.55$ cm. Since the image used to sit at $10$ cm, the screen must be pushed out by $v^{\prime }-v\approx 0.55$ cm, and because $v^{\prime }$ has increased, that shift is away from the lens.