What is the Lens Maker's Formula?

The formula is 1/f = (μ - 1)(1/R1 - 1/R2), where f is focal length, μ is refractive index, and R1, R2 are radii of curvature.

How do you calculate the focal length of a lens combination?

For lenses in contact, the effective focal length F is given by 1/F = 1/f1 + 1/f2.

How does the liquid act as a lens?

The liquid filled in the concave cavity forms a plano-convex or plano-concave liquid lens depending on the boundary surfaces.

Why is sign convention important in optics?

Sign convention determines whether a surface is converging or diverging; misapplying it leads to incorrect focal lengths.

What is the refractive index of the glass in this problem?

The refractive index of the glass lens is given as 1.5.

What is the refractive index of the liquid?

The refractive index of the liquid is given as 1.3.

Is the final combination converging or diverging?

Since the final focal length is positive (600/11 cm), the combination is converging.

How do we treat the interface between liquid and glass?

We treat them as two separate lenses in contact: one glass lens and one liquid lens.

What happens if the liquid index matches the glass index?

If indices match, the interface effectively disappears, and the system behaves as a single lens of that index.

A concave-convex lens of refractive index 1.5 and the radii of curvature of its surfaces are 30 cm and 20 cm, respectively. The concave surface is upwards and is filled with a liquid of refractive index 1.3. The focal length of the liquid-glass combination will be

Q: What is a concave-convex lens?

It is a lens where one surface is concave and the other is convex. In this problem, both radii are on the same side, making it a meniscus lens.

A concave-convex lens of refractive index 1.5 and the radii of curvature of its surfaces are 30 cm and 20 cm, respectively. The concave surface is upwards and is filled with a liquid of refractive index 1.3. The focal length of the liquid-glass combination will be | JEE Main Mathematics

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QMultiple Choice

A concave-convex lens of refractive index $1.5$ and the radii of curvature of its surfaces are $30$ cm and $20$ cm, respectively. The concave surface is upwards and is filled with a liquid of refractive index $1.3$. The focal length of the liquid-glass combination will be:

Options: A) $700/11$ cm, B) $600/11$ cm, C) $800/11$ cm, D) $500/11$ cm

✅ Correct Answer

600/11 cm

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Solution Steps

Analyze the Glass Lens

For the concave-convex glass lens: $n_g = 1.5$. Radii of curvature are $R_1 = +30$ cm and $R_2 = +20$ cm (both on the same side for a meniscus lens).

Using Lens Maker’s Formula: $$\frac{1}{f_g} = (n_g – 1)\left(\frac{1}{R_1} – \frac{1}{R_2}\right)$$

$$\frac{1}{f_g} = (1.5 – 1)\left(\frac{1}{30} – \frac{1}{20}\right) = 0.5 \left(\frac{2-3}{60}\right) = 0.5 \left(-\frac{1}{60}\right) = -\frac{1}{120}$$

Analyze the Liquid Lens

The liquid ($n_l = 1.3$) fills the upper concave surface ($R = 30$ cm). The top surface of the liquid is flat (planar), so $R_{top} = \infty$.

For the liquid lens: $R_1 = \infty$ and $R_2 = +30$ cm.

$$\frac{1}{f_l} = (n_l – 1)\left(\frac{1}{R_1} – \frac{1}{R_2}\right) = (1.3 – 1)\left(\frac{1}{\infty} – \frac{1}{30}\right)$$

$$\frac{1}{f_l} = 0.3 \left(0 – \frac{1}{30}\right) = -\frac{0.3}{30} = -\frac{1}{100}$$

Check Sign Convention for Orientation

The problem states the concave surface ($R=30$) is upwards and filled. If light comes from below, the glass lens focal length is calculated. If the system acts together, we sum the powers.

Wait, if $R_1=30$ and $R_2=20$, and light goes through liquid then glass: Liquid lens ($R_1=\infty, R_2=30$) then Glass lens ($R_1=30, R_2=20$).

Calculate Total Power

Equivalent focal length $F$ is: $$\frac{1}{F} = \frac{1}{f_l} + \frac{1}{f_g}$$

Note: We must be careful with signs based on the shape. Usually, for such combinations: $$\frac{1}{F} = \frac{n_l – 1}{R_{top} – R_{int}} + \frac{n_g – 1}{R_{int} – R_{bot}}$$

Using values: $\frac{1}{F} = \frac{1.3-1}{\infty – (-30)} + \frac{1.5-1}{-30 – (-20)}$ is wrong. Let’s use standard $1/f_1 + 1/f_2$.

Actually, if concave side up: Liquid lens is plano-convex ($R_1=\infty, R_2=30$). Power $P_l = \frac{1.3-1}{30} = \frac{0.3}{30} = \frac{1}{100}$.

Glass lens power $P_g = (1.5-1)(\frac{1}{30} – \frac{1}{20}) = -\frac{1}{120}$.

Final Calculation

$$\frac{1}{F} = \frac{1}{100} + \frac{1}{120} = \frac{6 + 5}{600} = \frac{11}{600}$$

$$F = \frac{600}{11} \text{ cm}$$

Final Answer: 600/11 cm

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Theory

1. Lens Maker’s Formula

The Lens Maker’s formula is used to construct a lens of desired focal length using a material of known refractive index. It relates the focal length ($f$), the refractive index of the material relative to the medium ($\mu$), and the radii of curvature ($R_1$ and $R_2$) of the two spherical surfaces. The formula is expressed as $1/f = (\mu – 1)(1/R_1 – 1/R_2)$. Proper sign convention is paramount: distances measured in the direction of incident light are positive, while those against it are negative. For a meniscus lens (like the concave-convex one in the problem), both radii often have the same sign.

2. Combination of Lenses

When two thin lenses are placed in contact, the effective power of the combination is the algebraic sum of their individual powers. Mathematically, $P_{eq} = P_1 + P_2$. In terms of focal length, this translates to $1/F = 1/f_1 + 1/f_2$. This principle is not restricted to glass lenses; it applies to any transparent media with curved boundaries, such as a liquid layer on a glass lens. In JEE problems, the interface between a liquid and a lens is often treated as a shared surface, creating a “liquid lens” whose focal length depends on the refractive index of the liquid and the curvature of the glass surface it rests upon.

3. Liquid Lenses in Optics

A liquid lens is formed when a transparent liquid is contained in a cavity or placed on a curved surface. The shape of the liquid boundary (meniscus) determines its optical properties. For instance, a liquid poured into a concave glass surface forms a plano-convex liquid lens. The focal length is calculated using the standard lens maker’s formula, where one surface is the curved glass interface and the other is usually a flat horizontal plane ($R = \infty$) due to gravity. The combined system’s focal length allows for interesting applications in variable-focus lenses and is a recurring theme in advanced physics examinations.

4. Refractive Index and Optical Path

Refractive index ($\mu$) is a measure of how much light slows down in a medium compared to a vacuum. It dictates the bending of light at interfaces according to Snell’s Law. In lens combinations, the relative refractive index at each boundary determines the convergence or divergence of light rays. The “effective” focal length of a combination depends heavily on these indices. If a lens is immersed in a liquid of the same refractive index, it becomes invisible and its power becomes zero. Understanding how light propagates through multiple media with different indices is key to solving complex optical system problems.

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FAQs

Why did we take one radius as infinity for the liquid?

The liquid is filled in an upward-facing cavity, so its top surface remains flat due to gravity, making its radius of curvature infinite.

What is a concave-convex lens?

It is a lens where both centers of curvature lie on the same side. It is also known as a meniscus lens.

How do we determine the sign of R1 and R2?

Using the Cartesian sign convention, if the surface is convex towards the incident light, R is positive; if concave, R is negative.

Does the order of lenses matter?

For thin lenses in contact, the order does not change the equivalent focal length.

What if the liquid had a higher refractive index than the glass?

The liquid lens would have more power, potentially changing the nature of the combination from converging to diverging or vice versa.

What is the unit of power in optics?

The SI unit is Dioptre ($D$), which is equal to $1/f$ where $f$ is in meters.

Is the combination converging?

Yes, because the resulting focal length is positive ($+600/11$ cm).

Can we use this for thick lenses?

No, the $1/F = 1/f1 + 1/f2$ formula is only valid for thin lenses in contact.

Why is $n-1$ used in the formula?

It represents the relative refractive index of the lens material compared to air ($\mu_{lens}/\mu_{air} – 1$).

What happens if the radii are equal?

If $R_1 = R_2$, the power of the lens becomes zero, and it acts like a plane glass plate.

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