A convex lens focuses a distant object on a screen placed 10 cm away from it. A glass plate (n = 1.5) of thickness 1.5 is inserted between the lens and the screen. Where should the object be placed so that its image is again focused on the screen?

190 cm

The situation when the glass plate is inserted between the lens and the screen, is shown in figure. The lens forms the image of object O at point $I_1$ but the glass plate intercepts the rays and forms the final image at I on the screen. The shift in the position of image after insertion of glass plate

$I_1I=t(1-\frac{1}{n})=(1.5cm)(1-\frac{1}{1.5})=0.5cm$

Thus, the lens forms the image at a distance of 9.5 cm from itself. Using

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, we get

$\frac{1}{u}=\frac{1}{v}-\frac{1}{f}=\frac{1}{9.5}-\frac{1}{10}$ or u = +190 cm.

i.e. the object should be virtual and to be taken at a distance of 190 cm. from the lens right of it.