Question

how deep is the goldfish? a tank whose bottom is a mirror is filled with water to a depth of 19.2 cm. a small fish floats motionless 8.00 cm under the surface of the water. part b what is the apparent depth of the reflection of the fish in the bottom of the tank when viewed at normal incidence? express your answer in centimeters. use 1.33 for the index of refraction of water. view available hint(s) submit previous answers correct

Explanation:

Step1: Find the real depth of the reflection

The fish is 8.00 cm below the surface, and the tank's depth is 19.2 cm. The distance from the fish to the bottom (mirror) is \(19.2 - 8.00 = 11.2\) cm. So the reflection of the fish in the mirror is 11.2 cm below the bottom, meaning the real depth of the reflection from the surface is \(19.2 + 11.2 = 30.4\) cm.

Step2: Calculate the apparent depth

The formula for apparent depth \(d_{apparent}\) when real depth is \(d_{real}\) and refractive index of water is \(n\) is \(d_{apparent}=\frac{d_{real}}{n}\). Here, \(d_{real} = 30.4\) cm and \(n = 1.33\). So \(d_{apparent}=\frac{30.4}{1.33}\approx22.86\) cm? Wait, no, wait. Wait, the reflection: first, the fish is at 8 cm below surface. The mirror is at 19.2 cm below surface. So the image of the fish in the mirror (reflection) is at a distance from the mirror equal to the distance from the fish to the mirror. Distance from fish to mirror: \(19.2 - 8.00 = 11.2\) cm. So the reflection (image) is 11.2 cm below the mirror, so its real depth from the surface is \(19.2 + 11.2 = 30.4\) cm. But wait, no—wait, when we look at the reflection, the light from the reflection (image) comes up through water? Wait, no. Wait, the image is in the mirror (bottom), so the light from the image (reflection) has to go from the image (which is a virtual image in the mirror, but then the light from that image travels through water to the surface. Wait, maybe I messed up. Let's re - think.

The fish is at depth \(y_1 = 8.00\) cm below the surface. The bottom (mirror) is at \(y = 19.2\) cm below the surface. The distance from the fish to the mirror is \(19.2 - 8.00=11.2\) cm. So the image of the fish in the mirror (reflection) is at a distance \(11.2\) cm below the mirror, so the depth of the reflected image (real depth) from the surface is \(19.2 + 11.2 = 30.4\) cm. Now, when we view this reflected image from above the water, the apparent depth of an object in water is given by \(d_{apparent}=\frac{d_{real}}{n}\), where \(n\) is the refractive index of water. But wait, is the reflected image a real object? No, the reflected image is a virtual image, but the light rays from the reflected image (which is a virtual image in the mirror) will travel from the mirror (bottom) up through the water. Wait, no—actually, the light from the fish goes down to the mirror, reflects, and then comes back up through the water. So the path of the light is: fish -> mirror (distance \(19.2 - 8.00 = 11.2\) cm), then the reflected ray goes back up to the surface, traveling through water a distance of \(19.2+11.2 = 30.4\) cm? No, wait, the reflected ray from the mirror: the image of the fish in the mirror is at a depth of \(19.2+(19.2 - 8.00)=30.4\) cm below the surface (since the mirror is at 19.2 cm, and the image is 11.2 cm below the mirror). Now, when we look at this image (reflection), the light from the image (which is a virtual image) has to travel from the image (depth 30.4 cm) up through the water to the surface. But the apparent depth formula is for an object in water: \(d_{apparent}=\frac{d_{real}}{n}\), where \(d_{real}\) is the real depth of the object (or image, in this case) in water. So the real depth of the reflected image is 30.4 cm, and the apparent depth is \(\frac{30.4}{1.33}\)? Wait, no, that can't be. Wait, maybe I made a mistake in the real depth of the reflection.

Wait, let's start over. The fish is at \(y = 8.00\) cm below the surface. The mirror is at \(y = 19.2\) cm below the surface. The distance between the fish and the mirror is \(19.2 - 8.00=11.2\) cm. The image of the fish in the mirror (reflection) is located at a distance \(11.2\) cm below the mirror (because the mirror forms an image as far behind it as the object is in front). So the depth of the reflected image (from the surface) is \(19.2+11.2 = 30.4\) cm. Now, this reflected image is a virtual image, but the light rays coming from this image (to reach our eye) have to travel through the water (from depth 30.4 cm to the surface, a distance of 30.4 cm) and then into air. Wait, no—when we look at the reflection, the light from the reflected image (which is in the mirror) is actually the light that was reflected from the fish to the mirror and then back. But when we consider the apparent depth of the reflected image, we need to consider that the light from the reflected image (which is a virtual image in the mirror) is now a real object? No, the reflected image is a virtual image, but the light rays from that virtual image (as they travel towards the surface) are in water. Wait, maybe the correct approach is:

The reflected image (of the fish in the mirror) has a real - depth (distance from the surface) of \(d = 19.2+(19.2 - 8.00)=30.4\) cm. But when we look at this image from above the water, the apparent depth \(d_{app}\) is given by \(d_{app}=\frac{d}{n}\), where \(n\) is the refractive index of water. Wait, but let's check the formula for apparent depth. The apparent depth of an object in a medium with refractive index \(n\) (relative to air, \(n_{air}=1\)) is \(d_{app}=\frac{d_{real}}{n}\), where \(d_{real}\) is the real depth of the object.

But wait, another way: the fish is at \(y_1 = 8.00\) cm. The mirror is at \(y = 19.2\) cm. The image of the fish in the mirror is at \(y_2=19.2+(19.2 - 8.00)=30.4\) cm (real depth). Now, the light from this image (reflection) travels from \(y_2 = 30.4\) cm to the surface (\(y = 0\)) through water. The apparent depth is the depth that the image appears to be, which is given by \(d_{app}=\frac{d_{real}}{n}\). So \(d_{app}=\frac{30.4}{1.33}\approx22.86\) cm? Wait, but that seems high. Wait, maybe I messed up the real depth of the reflection.

Wait, no—wait, the fish is at 8 cm below surface. The mirror is at 19.2 cm below surface. The distance from the fish to the mirror is \(19.2 - 8 = 11.2\) cm. So the image of the fish in the mirror is 11.2 cm below the mirror, so its depth from the surface is \(19.2+11.2 = 30.4\) cm. Now, when we look at this image, the light from the image has to go up through 30.4 cm of water. The apparent depth is \(\frac{30.4}{1.33}\approx22.86\)? Wait, but let's check with the formula. The apparent depth formula: when an object is in a medium with refractive index \(n\) (water, \(n = 1.33\)) and real depth \(d\), the apparent depth \(d_{app}=\frac{d}{n}\).

Wait, but maybe the error is in the real depth of the reflection. Let's think again. The fish is at 8 cm below surface. The mirror is at 19.2 cm below surface. The image of the fish in the mirror is at a distance from the mirror equal to the distance from the fish to the mirror. So distance from fish to mirror: \(19.2 - 8 = 11.2\) cm. So image is 11.2 cm below mirror, so depth from surface: \(19.2+11.2 = 30.4\) cm. Now, the light from this image (reflection) comes from 30.4 cm below the surface, through water, to the surface. So the apparent depth is \(\frac{30.4}{1.33}\approx22.9\) cm? Wait, but let's calculate \(\frac{30.4}{1.33}\): \(30.4\div1.33\approx22.86\), which is approximately 22.9 cm. Wait, but maybe I made a mistake in the real depth. Wait, no—wait, the reflection is a mirror image, so the real depth of the reflected image is \(19.2+(19.2 - 8.00)=30.4\) cm. Then, using the apparent depth formula \(d_{app}=\frac{d_{real}}{n}\), with \(d_{real}=30.4\) cm and \(n = 1.33\), we get \(d_{app}=\frac{30.4}{1.33}\approx22.9\) cm. Wait, but let's check another way.

Alternatively, the fish is at 8 cm below surface. The mirror is at 19.2 cm below surface. The image of the fish in the mirror is at \(19.2+(19.2 - 8)=30.4\) cm below surface (real depth). The apparent depth of this image is \(\frac{30.4}{1.33}\approx22.9\) cm.

Wait, but let's do the calculation again: \(30.4\div1.33\). Let's compute \(1.33\times22 = 29.26\), \(30.4 - 29.26 = 1.14\), \(1.14\div1.33\approx0.857\), so total is \(22 + 0.857\approx22.86\approx22.9\) cm.

Wait, but maybe I messed up the real depth. Let's think about the light path. The fish emits light, some of it goes to the mirror (bottom), reflects, and comes back up. The image of the fish in the mirror is a virtual image located at a distance from the mirror equal to the distance from the fish to the mirror. So fish is 8 cm below surface, mirror is 19.2 cm below surface, so distance from fish to mirror is \(19.2 - 8 = 11.2\) cm. So the image is 11.2 cm below the mirror, so its depth from the surface is \(19.2+11.2 = 30.4\) cm. Now, when we look at this image, the light from the image (which is a virtual image) has to travel from 30.4 cm below the surface, through water, to the surface. The apparent depth is given by the formula for apparent depth in a medium: \(d_{app}=\frac{d_{real}}{n}\), where \(n\) is the refractive index of the medium (water, \(n = 1.33\)) relative to air (\(n_{air}=1\)). So this formula applies here because the object (the reflected image) is in water, and we are viewing it from air. So the calculation is correct.

Answer:

\(\boxed{22.9}\) (or more precisely, around 22.86 which can be rounded to 22.9 or 22.8)