In Fig. 34-54, a fish watcher at point P watches a fish...

In Fig. $34-54$, a fish watcher at point $P$ watches a fish through a glass wall of a fish tank. The watcher is level with the fish; the index of refraction of the glass is $8 / 5$, and that of the water is $4 / 3$. The distances are $d_{1}=8.0 \mathrm{~cm}, d_{2}=3.0 \mathrm{~cm}$, and $d_{3}=$ $6.8 \mathrm{~cm}$. (a) To the fish, how far away does the watcher appear to be? (Hint: The watcher is the object. Light from that object passes through the wall's outside surface, which acts as a refracting surface. Find the image produced by that surface. Then treat that image as an object whose light passes through the wall's inside surface, which acts as another refracting surface.) (b) To the watcher, how far away does the fish appear to be?

In Fig. $34-54$, a fish watcher at point $P$ watches a fish through a glass wall of a fish tank. The watcher is level with the fish; the index of refraction of the glass is $8 / 5$, and that of the water is $4 / 3$. The distances are $d_{1}=8.0 \mathrm{~cm}, d_{2}=3.0 \mathrm{~cm}$, and $d_{3}=$
$6.8 \mathrm{~cm}$. (a) To the fish, how far away does the watcher appear to be? (Hint: The watcher is the object. Light from that object passes
through the wall's outside surface, which acts as a refracting surface. Find the image produced by that surface. Then treat that image as an object whose light passes through the wall's inside surface, which acts as another refracting surface.) (b) To the watcher, how far away does the fish appear to be?

Key Concepts

Apparent Depth

Apparent depth refers to the phenomenon where an object submerged in a medium appears to be at a shallower depth than its actual physical depth. This effect occurs because of the refraction of light as it travels from one medium to another with a different refractive index, often leading to a compression of the perceived distance between the object and the observer.

Virtual Image Formation

In optics, a virtual image is formed when the refracted or reflected light rays diverge in such a way that they appear to originate from a location other than the object’s actual position. This concept is crucial when analyzing how objects are perceived through different media, as the apparent position can differ significantly from the true position due to refraction.

Refraction at Multiple Interfaces

When light encounters more than one interface between materials with differing refractive indices, it undergoes multiple bending events, each governed by Snell's Law. Analyzing such situations involves sequentially applying refraction principles at each boundary to accurately determine the final path of the light and the perceived locations of the objects involved.

Snell's Law

Snell's Law is the fundamental principle in optics that governs the relationship between the angles of incidence and refraction when light passes from one medium to another. It is mathematically expressed as n? sin(??) = n? sin(??), where n represents the refractive indices of the media and ? the corresponding angles. This law is essential for understanding how light bends at interfaces between different materials.

Refractive Index

The refractive index of a medium quantifies how much the speed of light is reduced relative to its speed in a vacuum. It is a critical parameter in determining how much light will bend when transitioning between different media. Variations in refractive index across materials lead directly to phenomena such as bending, focusing, and distortion of images viewed through these materials.

Transcript

00:01 For our question, we're given the distances d1, d2, and d3, as well as the index of refraction of the three different mediums, n -sub -a for air, n -sub -g for glass, and in sub -w for the water.

00:14 For part a of our question, we are asked to figure out how far away the watcher appears to be relative to the fish.

00:26 Okay, so to begin this, we're going to first, first of all, calculate the image distance of the image produced.

00:32 By the first surface of the wall.

00:34 The relationship between the object distance, image distance, and the radius of curvature is as follows.

00:40 So we'll indicate this as part a.

00:44 So in general, we have n sub 1 over p plus n sub 2 over i is equal to n sub 2 minus n sub 1 over the radius of curvature r.

01:03 So for us, in all cases, is r is equal to infinity because we have a flat wall.

01:09 And anything over infinity is equal to zero.

01:11 So the right side of the equation for all of us, or for all of our situations, it's going to be equal to zero.

01:17 So let's get more specific for our case, n sub a is the first index of refraction, because we're starting out in error, divided by p.

01:26 But we're told that p originally is the distance d sub 1.

01:32 And this is plus n sub 2, which happens to be the index of refraction of glass, or n sub g, divided by i.

01:41 And this is all equal to zero.

01:43 So we can rearrange this to solve for i, and we find that i is equal to n sub g divided by n sub a.

01:58 There's a negative sign here, negative n sub g over n sub a times d1.

02:03 Or in other words, i is equal to negative 12 .8 centimeters.

02:18 Okay, so now for the second surface, or now relative to the second service, the image produced by the first surface is calculated as such.

02:30 And we'll call this image p prime.

02:33 So now p prime for the second surface is equal to d2 minus i that we just found.

02:43 So plugging the values for d2 and i, we find that this is equal to positive 15 .8 centimeters.

02:55 Okay, now to find i prime, we still have the same equation, except for this time, n sub 1 is the glass.

03:05 So it's the n sub g.

03:07 Let's rewrite that g to make it look more legible.

03:11 N sub g divided by p prime, which we just found, plus the second index of refraction is water.

03:20 So it would be n sub w divided by i prime, which is what we're trying to find.

03:25 And again, the radius of curvature is still infinity, so that...

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