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Q109P

Expert-verifiedFound in: Page 1044

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 ${{\mathit{d}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{8}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{\text{cm}}}{\mathbf{}}{\mathbf{}}{\mathbf{,}}{{\mathit{d}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{3}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{\text{cm}}}{\mathbf{}}{\mathbf{,}}{{\mathit{d}}}_{{\mathbf{3}}}{\mathbf{=}}{\mathbf{6}}{\mathbf{.}}{\mathbf{8}}{\mathbf{}}{\mathbf{\text{cm}}}$****. (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?**

- The apparent distance of the watcherfrom the fish is $20\text{cm}$.
- The apparent distance of fishfrom the observer is $15\text{cm}$.

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

**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)**

We have, for spherical refracting surface,

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

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:

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

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

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 $20\text{cm}$.

(b)

We have, for spherical refracting surface,

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

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

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

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

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

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:

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

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 $15\text{cm}$

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