Consider an extended object immersed in water contained in a plane trough. When seen from close to the edge of the trough the object looks distorted because

Options

A. the apparent depth of the points close to the edge are nearer the surface of the water compared to the points away from the edge.
B. the angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air.
C. some of the points of the object far awa from the edge may not be visible because total internal reflection.
D. water in a trough acts as a lens an magnifies the object.

Answer

A, B, C

Explanation

Consider a plane trough (a long and shallow water reservoir with plane bottom) with AD a water surface. Now, observe point like objects `P_1` and `P_2`, at equal depth AB = CD = h obliquely a point Q, nearer to edge CD. Here, for incidence ray `P_1Q`, angle of incidence is `theta_1`. and angle o refraction is `theta._1`. Similarly, for incident ray `vac(P_2Q)` angle of incidence is `theta_2`, and angle of refraction is `theta._2` . Here because of refraction at surface AD of water, we get `theta._1 gt theta_1` and `theta._2 gt theta_2`. Here `theat_2 gt theta_1` and so we get `theta._2 gt theat._1` . Because of this virtual depths of virtual positions `P._1` and `P._2` (of real objects, respectively `P_1` and `P_2`.) are obtained as `d_1` and `d_2` where `d_1 lt d_2` as seen from the figure)
Thus virtual depth is found to be less nearer the edge of trough and farther the edge of trough, it is found to be more.
Now, here if object is an extended object, extended from `P_1` to `P_2` then due to same depth, object is actually planar but when it is observed from near the edge of water surface in a trough image of planar object appears to be distorted. Thus, option (A) is correct.
Now, as per the statement if we consider object in air at `R_1`, near the water surface then its real image is formed at point `P_1`. Here `angleR_1QM = theta._1` and `angleP_1QN = theta_1` where `theta_1 lt theta._1`
`implies` Option (B) is also correct.
Now, in above figure if we shift point-like object towards end B then angle of incidence `theta` goes on increasing. At one stage when `theta ge C` (where C = critical angle), object is not seen from air due to total internal reflection. Thus, option (C) is also correct.