(For Normal Vision):
When there is a denser medium between a real object and an observer (in air), the object appears to be shifted towards the observer .
If an object is placed at O, the ray of light coming from O gets refracted at different interfaces 1, 2 & 3 etc. to produce different images I1 I2, I3 etc. respectively. Then I3 is the final image.
The apparent shift of the image due to medium 1
$ \displaystyle \Delta t_1 = t_1(1-\frac{1}{n_1}) $
Similarly the apparent shift due to the medium 2, 3 etc can be given as
$ \displaystyle \Delta t_2 = t_1(1-\frac{1}{n_2}) $ ,
$ \displaystyle \Delta t_3 = t_1(1-\frac{1}{n_3}) $ …….etc
The net (total) apparent shift of the image
$ \displaystyle \Delta t = t_1(1-\frac{1}{n_1}) + t_2(1-\frac{1}{n_2}) + t_3(1-\frac{1}{n_3}) …. $
Apparent depth = Real depth − apparent shift
Apparent Depth = $ \displaystyle = \frac{t_1}{n_1} + \frac{t_2}{n_2} + \frac{t_3}{n_3} + …. $
Exercise : Find the apparent depth and apparent shift of an object placed at the bottom of the a vessel containing a glass slab of thickness 10 cm and water of height 50 cm. as shown in the figure.
