Q16E - University Physics with Modern Physics

Question

34.16 A tank whose bottom is a mirror is filled with water to a depth of 20.0cm. A small fish floats motionless 7.0cm under the surface of the water. (a) What is the apparent depth of the fish when viewed at normal incidence? (b) What is the apparent depth of the image of the fish when viewed at normal incidence?

Step-by-Step Solution

Answer

(a) The apparent depth of the fish is 5.25cm.

(b) The apparent depth of the image of the fish is 24.8cm.

Step 1: Formula for Spherical refracting surface with plane surface between two optical materials

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

Step 2: Determine the apparent depth of the fish

a) There is a special case for the spherical refracting surface which is the plane surface between two materials where $R=\infty$. 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 distance s 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.

Step 3: Determine the apparent depth of the image of the fish

(b) In this case, the object distance s 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.