how do you find the total normal displacement if an oblect is placed say beneath a glass slab over which is kept a beaker of water when viewed from air?

i can solve such a numerical by direct application of the derieved formulae but fail to understand it physically...

"My conception":--

(1) a ray travels from the object to the glass water interface normally, and thus undeviated.

(2) an inclined ray travels from the object to the glass - water interface, adn suffers its "first" refraction, which when produced backwards intersects with the undeviated ray to form an img..

(3) the img now acts like a virtual object for the water- air interface, whose refracted ray when produced backwards cuts the undeviated ray to form the final img whose position helps to find the normal displacement..

 

to find the normal disp of the 1st img (say ur eyes are in water), do u use the absolute R.I or the relative R.I of glass w.r.t water???

 

after finding out the disp. due to glass, does the apparent depth due to glass become the real depth for displacement due to water??

 

can it ever be that the 1st image not be in glass but in water??

 

please explain wherevr iv gone wrong..and clear my doubt!!!

 

thnx a million!!