91Ó°ÊÓ

$\frac{{\mathbf{n}}_{\mathbf{a}}}{\mathbf{s}}\mathbf{+}\frac{{\mathbf{n}}_{\mathbf{b}}}{\mathbf{s}\mathbf{\text{'}}}\mathbf{=}\mathbf{0}$

a) There is a special case for the spherical refracting surface which is the plane surface between two materials where $R=∞$. In this case, use the below equation for a plane refracting surface in the form

$\frac{{\mathrm{n}}_{\mathrm{a}}}{\mathrm{s}}+\frac{{\mathrm{n}}_{\mathrm{b}}}{\mathrm{s}\text{'}}=0$

Where is the refractive index of the water and equals 1.333,$n_b$ is the refractive index of air and equals 1.0. The distances is the distance from the air to the fish and equals 7cm.$s\text{'}$ is the apparent depth. So, solve equation for$s\text{'}$

$s\text{'}=-\left(\frac{n_b}{n_a}\right)s$ (1)

Put the values in above equation,

$s\text{'}=-\left(\frac{n_b}{n_a}\right)s=-\left(\frac{1.0}{1.333}\right)\left(7cm\right)=-5.25cm$

Hence, the apparent depth is -5.25cm.

(b) In this case, the object distances changes. The depth of the water is$d_{water}=20cm$ while the depth of the fish is $d_{fish}=7cm$. This depth is from the surface. So, the distance

from the mirror to the fish is

$d=d_{water}-d_{fish}=20cm-7cm=13cm$

Thus, the distance between the mirror and the image below the mirror is

$s=d+d_{water}=13cm+20cm=33cm$

Put the values in the equation (1),

$s\text{'}=-\left(\frac{n_b}{n_a}\right)s=-\left(\frac{1.0}{1.333}\right)\left(33cm\right)=-24.8cm$

Hence, the apparent depth is -24.8cm.