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?
Answer & explanation
Correct answer: option 4
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.