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Expert-verified Found in: Page 1044 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # 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 ${\mathbf{8}}{\mathbf{/}}{\mathbf{5}}$, and that of the water is ${\mathbf{4}}{\mathbf{/}}{\mathbf{3}}$. The distances are ${{\mathbit{d}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{8}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{\text{cm}}}{\mathbf{}}{\mathbf{}}{\mathbf{,}}{{\mathbit{d}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{3}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{\text{cm}}}{\mathbf{}}{\mathbf{,}}{{\mathbit{d}}}_{{\mathbf{3}}}{\mathbf{=}}{\mathbf{6}}{\mathbf{.}}{\mathbf{8}}{\mathbf{}}{\mathbf{\text{cm}}}$. (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 walls 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 walls inside surface, which acts as another refracting surface.) (b) To the watcher, how far away does the fish appear to be? 1. The apparent distance of the watcherfrom the fish is $20\text{cm}$.
2. The apparent distance of fishfrom the observer is $15\text{cm}$.
See the step by step solution

## Step 1: The given data

1. The distances are ${d}_{1}=8.0\text{cm},{d}_{2}=3.0\text{cm},{d}_{3}=6.8\text{cm}$
2. Refractive index for glass, $n=\frac{8}{5}$
3. Refractive index for water, $n=\frac{4}{3}$

## Step 2: Understanding the concept of properties of the lens

When an object faces a convex refracting surface, the radius of curvature is positive, and when it faces a concave refracting surface, the radius of curvature is negative. We will use the relation for the spherical refracting surface to find the distance of the image, and for a flat surface, the radius of curvature is infinite.

Formula:

The lens maker equation for a spherical surface,

$\frac{{\mathbf{n}}_{\mathbf{1}}\mathbf{}}{\mathbf{p}}{\mathbf{+}}\frac{{\mathbf{n}}_{\mathbf{2}}\mathbf{}}{\mathbf{i}}{\mathbf{=}}\frac{{\mathbf{n}}_{\mathbf{1}}\mathbf{-}{\mathbf{n}}_{\mathbf{2}}}{\mathbf{r}}$ ...(i)

## Step 3: Calculation of the apparent distance of the watcher from fish

We have, for spherical refracting surface,

$\frac{{n}_{1}}{p}+\frac{{n}_{2}}{i}=\frac{\left({n}_{2}-{n}_{1}\right)}{r}$

With ${n}_{1}=1.0,{n}_{2}=1.6$, the lens equation can be given using equation (i) as follows:

$\frac{1}{p}+\frac{1.6}{i}=\frac{1.6-1}{r}$

For flat surface $r=\infty$, the image distance relation can be given using the above equation as:

$\frac{1}{p}+\frac{1.6}{i}=0\phantom{\rule{0ex}{0ex}}\frac{1}{p}=-\frac{1.6}{i}\phantom{\rule{0ex}{0ex}}i=-1.6p\phantom{\rule{0ex}{0ex}}=-1.6×8.0$

$=-12.8\text{cm or}-\frac{64}{5}\text{cm}$

Now for the second surface, the object is at distance

$p\text{'}=3+\frac{64}{5}\phantom{\rule{0ex}{0ex}}=\frac{79}{5}\text{cm}$

Again using the same formula of equation (i), the image distance for the calculated object distance can be given for $r=\infty$ case as follows:

$\frac{1}{p\text{'}}+\frac{4}{3i\text{'}}=0\phantom{\rule{0ex}{0ex}}i\text{'}=-\frac{79}{6}\phantom{\rule{0ex}{0ex}}\approx -13.2$

Thus, the observer is at distance $13.2+6.8=20\text{cm}$from the fish.

Hence, the apparent distance of the watcher from fish is $20\text{cm}$.

## Step 4: Calculation of the apparent distance of the fish from the observer

(b)

We have, for spherical refracting surface,

$\frac{{n}_{1}}{p}+\frac{{n}_{2}}{i}=\frac{\left({n}_{2}-{n}_{1}\right)}{r}$

With ${n}_{1}=\frac{4}{3},{n}_{2}=1.6$, the lens equation can be given using equation (i) as follows:

$\frac{4}{3p}+\frac{8}{5i}=\frac{\left({n}_{2}-{n}_{1}\right)}{r}$

For flat surface $r=\infty$, the image distance relation can be given using the above equation as:

$\frac{4}{3p}=-\frac{8}{5i}\phantom{\rule{0ex}{0ex}}i=-1.2p\phantom{\rule{0ex}{0ex}}=-1.2×6.8\text{cm}\phantom{\rule{0ex}{0ex}}=-8.16\text{cm}$

Now for the second surface,object is at distance,

role="math" localid="1663065349982" $p\text{'}=3\text{cm}+8.16\text{cm}\phantom{\rule{0ex}{0ex}}=11.16\text{cm}\phantom{\rule{0ex}{0ex}}$

Again using the same formula of equation (i), the image distance for the calculated object distance can be given for $r=\infty$ case as follows:

$\frac{8}{5p\text{'}}+\frac{1}{i\text{'}}=0\phantom{\rule{0ex}{0ex}}i\text{'}=-\frac{5}{8}×p\text{'}\phantom{\rule{0ex}{0ex}}=-\frac{5}{8}×11.16\text{cm}\phantom{\rule{0ex}{0ex}}=-7.0\text{cm}$

Thus, the final fish image is to the right of the air wall interface at $7.0\text{cm}$.

So, the distance of the fish from the observer is given as: $7.0\text{cm}+8.0\text{cm}=15\text{cm}$

Hence, the value of the apparent distance from the observer is $15\text{cm}$ ### Want to see more solutions like these? 