91Ó°ÊÓ

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,$n=\frac{4}{3}$

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 $r=∞$, the image distance relation can be given using the above equation as:

$$\begin{gathered} \frac{1}{p}+\frac{1.6}{i}=0 \\ \frac{1}{p}=-\frac{1.6}{i} \\ i=-1.6p \\ =-1.6×8.0 \end{gathered}$$

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

Now for the second surface, the object is at distance

$$\begin{gathered} p\text{'}=3+\frac{64}{5} \\ =\frac{79}{5}\text{cm} \end{gathered}$$

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

$$\begin{gathered} \frac{1}{p\text{'}}+\frac{4}{3i\text{'}}=0 \\ i\text{'}=-\frac{79}{6} \\ ≈-13.2 \end{gathered}$$

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 $r=∞$, the image distance relation can be given using the above equation as:

$$\begin{gathered} \frac{4}{3p}=-\frac{8}{5i} \\ i=-1.2p \\ =-1.2×6.8\text{cm} \\ =-8.16\text{cm} \end{gathered}$$

Now for the second surface,object is at distance,

role="math" localid="1663065349982" $$\begin{gathered} p\text{'}=3\text{cm}+8.16\text{cm} \\ =11.16\text{cm} \end{gathered}$$

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

$$\begin{gathered} \frac{8}{5p\text{'}}+\frac{1}{i\text{'}}=0 \\ i\text{'}=-\frac{5}{8}×p\text{'} \\ =-\frac{5}{8}×11.16\text{cm} \\ =-7.0\text{cm} \end{gathered}$$

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$15\text{cm}$