91Ó°ÊÓ

Refraction from either a datum plane can create a picture. For dispersion with such a surface, the object and image distances are linked by

$s^{\mathrm{′}}=−\frac{n_2}{n_1}s$

The negative line shows that we would be dealing with a virtual image. $n_1$is the reflectance of the object's medium, $n_2$is the absorption coefficient of the writer's medium, and proximity The object distance, and the image distance are measured as from threshold.

• Diver's true height is: $h=150\mathrm{cm}$.
• The refractive index of water is $n_1=1.33$.
• The refractive index of air is $n_2=1.00$.

We've been tasked with determining the diver's apparent height in relation to you.

Equation combines the considerable distance of the dolphin's neck but also its picture here to viewer's eyes.

$s_{\text{head}}^{\mathrm{′}}=\left(\frac{n_2}{n_1}\right)s_{\text{head}}$

Equation compares the sort of barrier between it diver's legs to their picture in the eyes of the observer.

The difference between the visual distance of his face and his toes dictates the diver's apparent height. So

$h^{\mathrm{′}}=s_{\text{head}}^{\mathrm{′}}−s_{\text{toes}}^{\mathrm{′}}$

$=\left(\frac{n_2}{n_1}\right)s_{\text{head}}−\left(\frac{n_2}{n_1}\right)s_{\text{toes}}$

$=\left(s_{\text{head}}−s_{\text{toes}}\right)\frac{n_2}{n_1}$ ($h$=$s_{\text{head}}−s_{\text{toes}}$)

$h^{\mathrm{′}}=h\left(\frac{n_2}{n_1}\right)$

localid="1649161227990" $h^{\mathrm{′}}=\left(150\mathrm{cm}\right)\left(\frac{1.00}{1.33}\right)=133\text{cm}$