Monochromatic light rays parallel to x-axis strike a convex lens AB. If the lens oscillates such that AB tilts upto a small angle $\mathrm{\theta }$ (in radian) on either side of y-axis, then find the distance between extreme positions of oscillating image (f = focal length of the lens) :

When the lens is tilted by , the image is formed at the intersection (Q) of focal plane of lens in tilted position and x-axis.

As the lens oscillates. The image shifts on x-axis in between P and Q.
$\therefore$ Distance between two extreme position of the image = PQ = $\frac{\mathrm{f}}{\cos \mathrm{\theta }}-\mathrm{f}=\mathrm{f}(\sec \mathrm{\theta }-1)$