Susan Nall

2022-01-05

A tank whose bottom is a mirror is filled with water to a depth of 20.0 cm. A small fish floats motionless 7.0 cm 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?

esfloravaou

Step 1
a) There is a special case for the spherical refrecting surface which is the plane surface between two materials where $R=\mathrm{\infty }$. In this case, we use equation (34.13) for a plane refracting surface in the form $\frac{{n}_{a}}{s}+\frac{{n}_{b}}{{s}^{\prime }}=0$ (1)
Where ${n}_{a}$ 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' is the apparent depth. So, let us solve equation (1) for s'
${s}^{\prime }=-\left(\frac{{n}_{b}}{{n}_{a}}\right)s$ (2)
Now, plug the values for ${n}_{b},{n}_{a}$ and s into equation (2) to get s'
${s}^{\prime }=-\left(\frac{{n}_{b}}{{n}_{a}}\right)s=-\left(\frac{1.0}{1.333}\right)\left(7cm\right)=-5.25cm$
Hence, the apparent depth is 5.25cm.
b) In this case, the object 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
$d={d}_{water}-{d}_{fish}=20cm-7cm=13cm$
Now, we can get the distance between the mirror and the image below the mirror by
$s=d+{d}_{water}=13cm+20cm=33cm$
Now, plug the values for ${n}_{b},{n}_{a}$ and s into equation (2) to get s'
${s}^{\prime }=-\left(\frac{{n}_{b}}{{n}_{a}}\right)s=-\left(\frac{1.0}{1.333}\right)\left(33cm\right)=-24.8cm$
Hence, the apparent depth is 24.8cm.

Jillian Edgerton

The apparent depth of (a) the fish is 5.3 cm and (b) the image of the fish is 24.8 cm.
Explanation:
According to the following equation:
$\frac{{n}_{w}}{s}+\frac{{n}_{a}}{{s}^{\prime }}=\frac{{n}_{a}-{n}_{w}}{{R}_{c}}$
where nw and na is the refractive indices of water (1.33) and air (1.00) respectively; s is the depth of the fish below the surface of the water; s' is the apparent depth of the fish from normal incidence and Rc is the radius of curvature of the mirror at the bottom of the tank.
Note that the bottom of the tank is assumed to be a flat mirror, therefore the radius of curvature is very large $\left(R⇒\mathrm{\infty }\right)$.
Therefore, the above equation can be expressed as:
$\frac{{n}_{w}}{s}+\frac{{n}_{a}}{{s}^{\prime }}=0$
Now we can solve for the apparent depth of the fish.
a) ${s}^{\prime }=-\left(\frac{{n}_{a}}{{n}_{w}}\right)xs$ (Make s' subject of the formula from the above equation)
${s}^{\prime }=\left(\frac{1.00}{1.33}\right)x7cm$
$\therefore {s}^{\prime }=5.3cm$.
b) The motionless fish floats 13 cm above the mirror, therefore the image of the fish will be situated at $13+20=33cm$ away from the real fish.
Therefore, $s=33cm$
${s}^{\prime }=-\left(\frac{{n}_{a}}{{n}_{w}}\right)xs$
${s}^{\prime }=\left(\frac{1.00}{1.33}\right)x33cm$
${s}^{\prime }=24.8cm$.
NB: Here, it is assumed that the water is pure, as impurities may alter the refractive index of water.

karton

Step 1
Given data:
- The height of the water is ${h}_{w}=20.0cm$.
- The real depth is ${h}_{r}=7.0cm$.
Part (a)
The apparent depth of the fish can be calculated as:
$\begin{array}{}{h}_{app}=\frac{{n}_{air}}{{n}_{water}}{h}_{r}\\ {h}_{app}=\frac{\left(1\right)}{\left(1.33\right)}\left(7cm\right)\\ {h}_{app}=5.26cm\\ \text{Thus, the apparent depth of the fish is 5.26cm.}\\ \text{Step 2}\\ \text{Part (b)}\\ \text{In this case, the real depth will be:}\\ {h}_{r1}=2{h}_{w}-{h}_{r}\\ {h}_{r1}=2\left(20cm\right)-\left(7cm\right)\\ {h}_{r1}=33cm\\ \text{The apparent depth of the image of the fish can be calculated as:}\\ {h}_{app1}=\frac{{n}_{air}}{{n}_{water}}{h}_{r1}\\ {h}_{app1}=\frac{\left(1\right)}{\left(1.33\right)}\left(33cm\right)\\ {h}_{app1}=24.8cm\end{array}$
Thus, the apparent depth of the image of the fish is 24.8cm.

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