JEE Main · 2012 · 29 Apr 2012 · Offline · Geometrical Optics

Without the plate the lens forms the image with $u=-240$ cm and $v=12$ cm, so $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{12}+\frac{1}{240}=\frac{7}{80}$. A glass plate of thickness $t=1$ cm and index $\mu =1.50$ shifts the image toward the lens by $t\left(1-\frac{1}{\mu }\right)=1\left(1-\frac{1}{1.5}\right)=\frac{1}{3}$ cm, so to keep the sharp image on the film the lens must now form it at $v^{\prime }=12-\frac{1}{3}=\frac{35}{3}$ cm. Using the same focal length, $\frac{1}{u^{\prime }}=\frac{1}{v^{\prime }}-\frac{1}{f}=\frac{3}{35}-\frac{7}{80}$, which gives $u^{\prime }=-560$ cm. The object must be moved to $5.6$ m from the lens.