A thin convex lens made of crown glass (mu = 3/2) has a focal length f in air. It is subsequently measured in two different liquids with refractive indices 4/3 and 5/3, yielding focal lengths f_1 and f_2 respectively. Which relationship correctly describes these focal lengths?

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EEJ MAIN Physics QUESTION #7063

Question 1

A thin convex lens made of crown glass ($\mu = 3/2$) has a focal length $f$ in air. It is subsequently measured in two different liquids with refractive indices $4/3$ and $5/3$, yielding focal lengths $f_1$ and $f_2$ respectively. Which relationship correctly describes these focal lengths?

$f_2 > f$ and $f_1$ becomes negative
$f_1$ and $f_2$ both become negative
$f_1 = f_2 < f$
$f_1 > f$ and $f_2$ becomes negative✔️

Correct Answer Explanation

Using the Lens Maker's Formula: $\frac{1}{f} = \left( \frac{\mu_{lens}}{\mu_{medium}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.

In Liquid 1 ($\mu = 4/3$): Since $3/2 > 4/3$, the lens remains converging. However, the term $(\frac{\mu_l}{\mu_m} - 1)$ decreases compared to air, so the focal length increases ($f_1 > f$).
In Liquid 2 ($\mu = 5/3$): Since $3/2 < 5/3$, the refractive index of the medium is greater than the lens. This makes the term $(\frac{\mu_l}{\mu_m} - 1)$ negative, meaning the lens becomes diverging ($f_2$ is negative).

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