Solution to Question 23
1. Analyzing the Lens System
The lens is fixed in the wall of an aquarium. This forms an optical system separating two media: Air (refractive index $1$) and Water (refractive index $n$). The general formula for refraction through a thin lens separating two media $n_1$ and $n_2$ is: $$\frac{n_2}{v} – \frac{n_1}{u} = P$$ where $P$ is the optical power of the system. Importantly, the magnitude of this optical power $P$ remains constant regardless of the direction of light propagation, provided the lens geometry remains the same.
2. Case 1: Object in Air
Object is in air ($n_1 = 1$) at distance $a$. Image is in water ($n_2 = n$) at distance $b$. According to sign convention (light travels left to right): $$u = -a, \quad v = +b$$ Substituting into the refraction formula: $$\frac{n}{b} – \frac{1}{-a} = P$$ $$P = \frac{n}{b} + \frac{1}{a} \quad \text{— (i)}$$
3. Case 2: Object in Aquarium (Water)
Now the object is in water ($n_1 = n$) at a distance $a_1$. We need to find the image distance $v’$ in air ($n_2 = 1$). Sign convention: $$u = -a_1, \quad v = v’$$ Substituting into the refraction formula: $$\frac{1}{v’} – \frac{n}{-a_1} = P$$ $$\frac{1}{v’} + \frac{n}{a_1} = P \quad \text{— (ii)}$$
4. Solving for Image Position
Equating the values of $P$ from (i) and (ii): $$\frac{1}{v’} + \frac{n}{a_1} = \frac{n}{b} + \frac{1}{a}$$ Rearranging to solve for $v’$: $$\frac{1}{v’} = \frac{n}{b} + \frac{1}{a} – \frac{n}{a_1}$$ Taking the common denominator ($aba_1$): $$\frac{1}{v’} = \frac{n(a \cdot a_1) + b(a_1) – n(a \cdot b)}{a b a_1}$$ $$\frac{1}{v’} = \frac{n a a_1 + a_1 b – n a b}{a b a_1}$$ $$\frac{1}{v’} = \frac{a_1 b + na(a_1 – b)}{a b a_1}$$ Inverting to find $v’$: $$v’ = \frac{a b a_1}{a_1 b + na(a_1 – b)}$$