Monochromatic light rays parallel to x-axis strike a convex lens AB of refractive index `0.5`. If the lens oscillates such that AB tilts upto a small angle `theta` (in radian) on either side of y-axis, then find the distance between extreme positions of oscillating image:

Options

A. `2f(sectheta-1)`
B. `fsec^(2)theta`
C. `f(sectheta-1)`
D. the image will not move

Explanation

When the lens is tited by `theta`, the image is formed at the intersection (Q) of focal plane of lens in tited position and x-axis.
As the lens ocillates. The image shifts on x-axis in between P and Q
`therefore` Distance between two exterme position of the image `=PQ=(f)/(costheta)-f=f(sectheta-1)`