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Animated Solution for Physics - Optics: The refractive index of a converging lens is $1.4$ . What will be the focal length of this lens if it is placed in a medium of same refractive index ? (Assume the radii of curvature of the faces of lens are $R_{1}$ and $R_{2}$ respectively)

Q: Does the shape of the lens ($R_1$ and $R_2$) matter in this case?

No! As long as $\mu_1 = \mu_2$, the term $(\frac{\mu_2}{\mu_1} - 1)$ becomes zero, making the entire expression for $\frac{1}{f}$ zero, regardless of the values of $R_1$ and $R_2$.

Q: What if the medium had a higher refractive index than the lens?

Then the term $(\frac{\mu_2}{\mu_1} - 1)$ would be negative. This means a converging (convex) lens would start behaving like a diverging (concave) lens!

Q: Is an infinite focal length the same as zero power?

Yes! Power is the reciprocal of focal length ($P = \frac{1}{f}$). Since $\frac{1}{f} = 0$, the optical power of the lens is zero.

Select Answer:

Visualized Solution

Lens Maker's Formula

$\frac{1}{f} = (\frac{μ _{2}}{μ _{1}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
$μ_{2}$ = Refractive index of lens
$μ_{1}$ = Refractive index of medium

Applying the Condition

Given: $μ_{2} = 1.4$
Medium has same refractive index: $μ_{1} = 1.4$
$\Rightarrow μ_{1} = μ_{2}$

Substitution

$\frac{1}{f} = (\frac{1.4}{1.4} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
$\frac{1}{f} = (1 - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

Final Calculation

$\frac{1}{f} = 0 \times (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
$\frac{1}{f} = 0$
$f = \infty$

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The Sigma Insight: Lens

Solution Diagram

The Magic of Disappearing Lenses

Imagine you have a glass lens, and you drop it into a special liquid. Suddenly, the lens vanishes! You can't see it anymore. Is it magic? No, it's physics!

This phenomenon happens when the refractive index of the lens matches the refractive index of the surrounding medium. Let's dive into the mathematics behind this cool trick using the Lens Maker's Formula.

The Master Equation

The focal length

f

of a lens placed in a medium is given by:

\frac{1}{f} = (\frac{μ _{2}}{μ _{1}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})

Here,

μ_{2}

is the refractive index of the lens material, and

μ_{1}

is the refractive index of the surrounding medium.

R_{1}

and

R_{2}

are the radii of curvature of the lens surfaces.

Analyzing the Setup

In our problem, we are given a converging lens with a refractive index of

1.4

. The twist is that it's placed in a medium that also has a refractive index of

1.4

So, we have:

μ_{2} = 1.4

μ_{1} = 1.4

This means

μ_{1} = μ_{2}

Final Calculation

Let's substitute this into our master equation:

\frac{1}{f} = (\frac{1.4}{1.4} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})

\frac{1}{f} = (1 - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})

\frac{1}{f} = 0

If the reciprocal of the focal length is zero, what is the focal length itself?

f = \infty

The focal length becomes infinite.

What does an infinite focal length mean physically? It means the lens has lost its converging power. Parallel light rays entering the lens will not converge at a point; they will just pass straight through, completely undisturbed. The lens behaves just like a flat, transparent glass plate. This is why the lens becomes invisible in the medium!

Common Traps & FAQs

Why does the lens disappear when placed in a medium of the same refractive index?

Because light doesn't bend (refract) when it passes from one medium to another with the same refractive index. Without refraction, our eyes can't detect the boundary of the lens.

Does the shape of the lens (

R_{1}

and

R_{2}

) matter in this case?

No! As long as

μ_{1} = μ_{2}

, the term

(\frac{μ _{2}}{μ _{1}} - 1)

becomes zero, making the entire expression for

\frac{1}{f}

zero, regardless of the values of

R_{1}

and

R_{2}

What if the medium had a higher refractive index than the lens?

Then the term

(\frac{μ _{2}}{μ _{1}} - 1)

would be negative. This means a converging (convex) lens would start behaving like a diverging (concave) lens!

Is an infinite focal length the same as zero power?

Yes! Power is the reciprocal of focal length (

P = \frac{1}{f}

). Since

\frac{1}{f} = 0

, the optical power of the lens is zero.

Can we use this principle in real life?

Absolutely! This principle is used in index-matching fluids to make glass objects invisible, which is useful in certain optical experiments and even magic tricks.

Animated Solution for Physics - Optics: The refractive index of a converging lens is $1.4$ . What will be the focal length of this lens if it is placed in a medium of same refractive index ? (Assume the radii of curvature of the faces of lens are $R_{1}$ and $R_{2}$ respectively)

Select Answer:

Visualized Solution

Lens Maker's Formula

Applying the Condition

Substitution

Final Calculation

The Sigma Insight: Lens

The Magic of Disappearing Lenses

The Master Equation

Analyzing the Setup

Final Calculation

Similar Questions

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A thin lens of refractive index 1.5 has a focal length of 15 cm in air. When the lens is placed in a medium of refractive index 4/3, its focal length will become ...... cm.

A point object in air is in front of the curved surface of a plano-convex lens. The radius of curvature of the curved surface is $30 cm$ and the refractive index of the lens material is $1.5$ , then the focal length of the lens (in cm) is ……… .

Select Answer:

Visualized Solution

Lens Maker's Formula

Applying the Condition

Substitution

Final Calculation

The Sigma Insight: Lens

The Magic of Disappearing Lenses

The Master Equation

Analyzing the Setup

Final Calculation

Similar Questions

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