# Refraction Through Number of media , Total apparent shift of the image

### (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.