A thin lens made of glass (refractive index = 1.5) of focal length f = 16 cm in immersed in a liquid

Q: Why does the focal length change when we put the lens in a liquid?

The bending of light (refraction) depends on the difference in optical density between the lens and its surroundings. In a liquid, this difference is smaller than in air, so the light bends less, resulting in a longer focal length.

Q: What happened to the $R_1$ and $R_2$ terms? We don't have their values!

That's the beauty of taking a ratio! Since the physical shape of the lens doesn't change when you submerge it, the entire geometric term $(1/R_1 - 1/R_2)$ is identical in both equations and perfectly cancels out.

Q: Why didn't we use the $f = 16\text{ cm}$ value given in the question?

The question specifically asks for the ratio $f_1/f$. Because we are solving for a relative multiplier rather than an absolute focal length, the initial value of $16\text{ cm}$ is actually redundant information!

Q: I got stuck on the decimal calculation $\frac{0.5}{0.08/1.42}$. Is there a trick?

Yes! Always try to eliminate decimals early. Flip the denominator to get $\frac{0.5 \times 1.42}{0.08}$, which becomes $\frac{0.71}{0.08}$. Multiply top and bottom by 100 to get a clean fraction: $\frac{71}{8}$.

Q: What would happen if the liquid had a refractive index of 1.6 instead of 1.42?

Fascinating question! If $n_l > n_g$, the term $(n_g/n_l - 1)$ becomes negative. This means the focal length becomes negative, and our convex converging lens would magically transform into a diverging concave lens!

Question

A thin lens made of glass (refractive index $= 1.5$) of focal length $f = 16	ext{ cm}$ in immersed in a liquid of refractive index $1.42$. If its focal length in liquid is $f_1$, then the ratio $f_1 / f$ is closest to the integer

Options:

(A) 1

(B) 5

(C) 9

(D) 17

Sigma Percentile · Accepted Answer

Have you ever tried opening your eyes underwater? Everything looks blurry, right? That's not because your eyes suddenly stopped working, but because the physics of light bending—refraction—changes when the surrounding medium changes.

This exact phenomenon is the heart of our problem today. We have a glass lens, and we are going to dunk it into a liquid. Our mission? To find out exactly how much its focal length changes.

### The Master Equation

To solve this, we need the ultimate tool for lenses: the **Lens Maker's Formula**. This beautiful equation connects the physical shape of a lens to its optical power.

$$ \frac{1}{f} = \left(\frac{n_{	ext{lens}}}{n_{	ext{medium}}} - 1ight) \left(\frac{1}{R_1} - \frac{1}{R_2}ight) $$

Here, $n_{	ext{lens}}$ is the refractive index of the glass, and $n_{	ext{medium}}$ is the refractive index of whatever is surrounding it. The terms $R_1$ and $R_2$ represent the physical curvature of the lens surfaces.

The crucial insight here is that dunking the lens in a liquid doesn't physically deform it. The radii of curvature $R_1$ and $R_2$ remain absolutely constant!

### Analyzing the Setup

Let's write down the equation for our two specific scenarios. First, when the lens is chilling in the air:

$$ \frac{1}{f} = \left(\frac{n_g}{n_a} - 1ight) \left(\frac{1}{R_1} - \frac{1}{R_2}ight) $$

Next, we submerge the lens into the liquid. The focal length changes to a new value, $f_1$:

$$ \frac{1}{f_1} = \left(\frac{n_g}{n_l} - 1ight) \left(\frac{1}{R_1} - \frac{1}{R_2}ight) $$

We now have a system of two equations. But we don't know the values of $R_1$ and $R_2$. How do we proceed?

### The Power of Ratios

In physics, when you have messy constants that you don't need, you divide! By taking the ratio of the two equations, the entire geometric term $\left(\frac{1}{R_1} - \frac{1}{R_2}ight)$ completely cancels out.

Dividing the air equation by the liquid equation, we get:

$$ \frac{f_1}{f} = \frac{\left(\frac{n_g}{n_a} - 1ight)}{\left(\frac{n_g}{n_l} - 1ight)} $$

This is a massive victory. We have reduced a complex optical problem into pure, simple arithmetic.

### Final Calculation

Now, let's carefully substitute our known values: $n_g = 1.5$, $n_a = 1$, and $n_l = 1.42$.

$$ \frac{f_1}{f} = \frac{\left(\frac{1.5}{1} - 1ight)}{\left(\frac{1.5}{1.42} - 1ight)} $$

Let's simplify the numerator first. $1.5 - 1$ is simply $0.5$. For the denominator, we take the LCM:

$$ \frac{f_1}{f} = \frac{0.5}{\left(\frac{1.5 - 1.42}{1.42}ight)} = \frac{0.5}{\left(\frac{0.08}{1.42}ight)} $$

Flipping the fraction in the denominator up, we get:

$$ \frac{f_1}{f} = \frac{0.5 	imes 1.42}{0.08} $$

Multiplying $0.5$ by $1.42$ gives $0.71$. So we have:

$$ \frac{f_1}{f} = \frac{0.71}{0.08} $$

To make this easier, let's multiply the top and bottom by 100 to clear the decimals:

$$ \frac{f_1}{f} = \frac{71}{8} $$

Finally, we perform the division. $71 \div 8$ gives $8.875$.

The question asks for the closest integer. Since $8.875$ is much closer to $9$ than to $8$, our final answer is **9**.

And there you have it! By trusting the Lens Maker's formula and keeping our algebra clean, we conquered the problem. Keep practicing, and soon these optical maneuvers will become second nature to you!

Animated Solution for Physics - Optics: A thin lens made of glass (refractive index $= 1.5$ ) of focal length $f = 16 cm$ in immersed in a liquid of refractive index $1.42$ . If its focal length in liquid is $f_{1}$ , then the ratio $f_{1} / f$ is closest to the integer

Select Answer:

Visualized Solution

Lens in Air

Lens Maker's Formula

Focal Length in Air

Lens Immersed in Liquid

Focal Length in Liquid

Ratio of Focal Lengths

Substituting Values

Simplifying the Expression

Final Calculation

Closest Integer

The Way Forward

The Sigma Insight: Lens

The Master Equation

Analyzing the Setup

The Power of Ratios

Final Calculation

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Select Answer:

Visualized Solution

Lens in Air

Lens Maker's Formula

Focal Length in Air

Lens Immersed in Liquid

Focal Length in Liquid

Ratio of Focal Lengths

Substituting Values

Simplifying the Expression

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