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Optics

Combination of Thin Lenses in Contact and Separated by a Distance

The combination of thin lenses involves placing multiple lenses either in contact or separated by a finite distance along a common principal axis to modify the overall focal length and optical power. This technique is fundamental in optical instrument design to minimize aberrations and achieve desired magnification.

UP TGT ScienceUP PGT PhysicsStudy Notes + Practice

Scope and prerequisites

TGT core

Derivation and application of equivalent focal length and power for two thin lenses in contact, sign convention, and magnification.

PGT extension

Equivalent focal length of two thin lenses separated by a distance d, cardinal points, system matrix representation, and combination of multiple lenses.

Prerequisites

Refraction through spherical surfaces and Lens Maker's Formula
Sign convention in geometrical optics
Real and virtual images formed by convex and concave lenses

Equivalent Focal Length of Lenses in Contact

When two thin lenses of focal lengths f₁ and f₂ are kept in close contact, the combination acts as a single equivalent lens.
The reciprocal of the equivalent focal length (F) is equal to the algebraic sum of the reciprocals of individual focal lengths.
The total optical power P of the combination is the algebraic sum of individual powers: $P = P_{1} + P_{2}$.

Magnification of Lenses in Contact

The total linear magnification produced by a system of thin lenses in contact is equal to the product of linear magnifications of individual lenses.
$m = m_{1} × m_{2} × $... × m_n
For an object placed at a distance u, the final image distance v satisfies 1/v - 1/$u = 1/F$.

Two Thin Lenses Separated by a Distance

When two thin lenses of focal lengths f₁ and f₂ are separated by a distance d along their common principal axis, the equivalent focal length F is altered.
The separation d reduces the effective converging or diverging power compared to lenses in direct contact.
The position of the equivalent principal planes shifts depending on the separation distance d.

Power and Matrix Approach for Lens Systems

Optical power is measured in dioptres (D) when focal length is expressed in meters.
In advanced PGT scope, translation and refraction matrices are used to compute the overall system matrix for separated lenses.
A combination of lenses can eliminate chromatic and spherical aberrations to a considerable extent (e.g., achromatic doublets).

Derivations

Equivalent Focal Length of Two Thin Lenses in Contact

Conditions: Two thin lenses of focal lengths f₁ and f₂ placed coaxially in contact.

Let an object O be placed at a distance u from the combination on the principal axis.
The first lens alone would form an image at distance v₁ such that: \frac{1}{v_1} - \frac{1}{u} = \frac{1}{f_1}.
This image acts as a virtual object for the second lens, forming the final image at distance v such that: \frac{1}{v} - \frac{1}{$v_1 = \frac{1}{f_2}$.
Adding the two equations eliminates v₁: \frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2}.
Since the combination is replaced by a single equivalent lens of focal length F: \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}.

Result: \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}

Equivalent Focal Length of Two Lenses Separated by Distance d

Conditions: Two thin lenses of focal lengths f₁ and f₂ separated by distance d < f₁.

Consider a ray parallel to the principal axis incident on the first lens of focal length f₁.
It converges towards a point at distance f₁ behind the first lens in the absence of the second lens.
The second lens intercepts this ray at a distance d from the first lens, so the object distance for the second lens is u₂ = + (f₁ - d).
Applying the lens formula for the second lens: \frac{1}{v} - \frac{1}{f_1 - d} = \frac{1}{f_2}.
Solving for the equivalent focal length F from the optical center of the first lens gives the standard relation.

Result: \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}

Formula sheet

Formula / law	Equation	Variables and units	Conditions
Equivalent Focal Length (In Contact)	$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} + \dots + \frac{1}{f_n}$$	F = equivalent focal length, f_i = focal length of individual lens meters (m)	Thin lenses in intimate contact
Equivalent Power (In Contact)	$$P = P_1 + P_2 + \dots + P_n$$	P = total power, P_i = power of individual lens Dioptres (m^-1)	Thin lenses in contact
Equivalent Focal Length (Separated by d)	$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$$	d = separation distance between lenses meters (m)	Coaxial thin lenses separated by distance d
Magnification of Lens Combination	$$m = m_1 \times m_2 \times \dots \times m_n$$	m = total magnification, m_i = magnification of i-th lens Dimensionless	Lenses in contact
Position of Equivalent Principal Focus (Separated Lenses)	$$F = \frac{f_1 f_2}{f_1 + f_2 - d}$$	F = focal length of system meters (m)	d < f₁
Power of Separated Lenses	$$P = P_1 + P_2 - d P_1 P_2$$	P, P₁, P₂ = powers, d = separation Dioptres (D)	Coaxial thin lenses
Lens Maker's Formula (Component basis)	$$\frac{1}{f} = (n - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$	n = refractive index, R₁, R₂ = radii of curvature m^-1 for 1/f	Thin spherical lens
Combined Focal Length with Liquid Immersion	$$\frac{1}{F_L} = (n_{rel} - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$	n_rel = refractive index of lens w.r.t liquid m^-1	Lens immersed in medium

Graphs and experiments

Ray Diagram for Two Convex Lenses in Contact

Draw: Two overlapping convex lenses on a principal axis showing a single incident ray refracting through both lenses to a common focus.

Interpret: Illustrates how the combined bending of light is greater than that of either individual lens, shortening the focal length.

Exam use: Tested in conceptual multiple-choice questions regarding focal length changes.

Ray Diagram for Separated Lenses

Draw: Two convex lenses separated by distance d, showing parallel incident rays bending at the first lens, traveling across distance d, and undergoing further refraction at the second lens.

Interpret: Demonstrates the shift in focal point and the formation of equivalent optical center.

Exam use: Frequently asked in PGT numericals regarding equivalent focal length.

Power vs Separation Distance Graph

Draw: A linear or non-linear decay curve of equivalent power P against separation distance d for two convex lenses.

Interpret: Shows that as separation d increases, the total power of two converging lenses decreases.

Exam use: Conceptual interpretation of combined lens behavior.

Solved numericals

TGT core

Two thin lenses of focal lengths +20 cm and -30 cm are placed in contact. Find the focal length and power of the combination.

Identify given values: f₁ = +20 c$m = +0.2 m$, f₂ = -30 c$m = -0.3 m$.
Apply lens combination formula: 1/$F = 1/f_{1} + 1/f_{2} = 1/20 - 1/30$.
Calculate common denominator: 1/$F = (3 - 2) / 60 = 1/60 cm^{-1}$.
Find $F = +60 cm = +0.6 m$.
Calculate power $P = 1/F(in meters) = 1 / 0.6 = +1.67 D$.

Answer: Focal lengt$h = +60 cm$, Powe$r = +1.67 D$

PGT extension

Two thin convex lenses of focal lengths 30 cm and 20 cm are separated by a distance of 10 cm. Find the equivalent focal length of the combination.

Identify given values: f₁ = 30 cm, f₂ = 20 cm, d = 10 cm.
Apply the formula for separated lenses: 1/$F = 1/f_{1} + 1/f_{2} - d / (f_{1}f_{2})$.
Substitute values: 1/$F = 1/30 + 1/20 - 10 / (30 × 20)$.
Simplify terms: 1/$F = 1/30 + 1/20 - 10/600 = 1/30 + 1/20 - 1/60$.
Find common denominator (60): 1/$F = (2 + 3 - 1) / 60 = 4 / 60 = 1 / 15 cm^{-1}$.
Calculate $F = +15 cm$.

Answer: Focal lengt$h = +15 cm$

PGT extension

A convex lens of focal length 25 cm is placed in contact with a concave lens of focal length 20 cm. What is the nature and focal length of the combination?

Identify given values: f₁ = +25 cm (convex), f₂ = -20 cm (concave).
Use combination formula: 1/$F = 1/f_{1} + 1/f_{2} = 1/25 - 1/20$.
Calculate common denominator (100): 1/$F = (4 - 5) / 100 = -1 / 100 cm^{-1}$.
Determine $F = -100 cm = -1 m$.
Interpret sign: Negative focal length indicates the combination behaves as a diverging (concave) lens.

Answer: Diverging lens with focal length -100 cm (-1 m)

Assumptions and limitations

[object Object]

Common misconceptions

Common error: Adding two convex lenses increases the focal length of the system.

Correct: Adding two convex lenses decreases the equivalent focal length.

Since 1/$F = 1/f_{1} + 1/f_{2}$, the reciprocal of the equivalent focal length increases, meaning F decreases and power increases.

Common error: The separation distance d is added to individual focal lengths.

Correct: The separation distance is subtracted in the denominator term (-d / f₁f₂).

Separating converging lenses reduces the overall convergence of the ray bundle compared to direct contact.

Quick revision

For lenses in contact, 1/$F = 1/f_{1} + 1/f_{2}$ and $P = P_{1} + P_{2}$.
Equivalent Focal Length (In Contact): \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} + \dots + \frac{1}{f_n}
Equivalent Power (In Contact): $P = P_1 + P_2 + \dots + P_n$

Must remember

The reciprocal of the equivalent focal length of thin lenses in intimate contact is the algebraic sum of the reciprocals of individual focal lengths: $1/F = 1/f_1 + 1/f_2$.
Total power of lenses in contact is given by the algebraic sum of individual powers: $P = P_1 + P_2$.
When two thin lenses are separated by a distance $d$, the equivalent focal length formula becomes $1/F = 1/f_1 + 1/f_2 - d/(f_1 f_2)$.
The total linear magnification for lenses in contact is the product of individual magnifications: $m = m_1 \times m_2$.
Optical power is measured in dioptres ($D$) when the focal length is expressed in meters.

Glossary

Equivalent Focal Length ($F$)The focal length of a single hypothetical lens that produces the same optical effect as a combination of lenses.

Optical Power ($P$)The degree of convergence or divergence of light rays, defined as the reciprocal of focal length in meters ($P = 1/f$).

Separation Distance ($d$)The distance along the common principal axis between two coaxial thin lenses.

Achromatic DoubletA combination of two lenses designed to eliminate or reduce chromatic aberration.

Lenses in Contact vs Lenses Separated by Distance

Feature	Lenses in Contact ($d = 0$)	Lenses Separated by Distance ($d$)
Equivalent Focal Length Formula	$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$	$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$
Equivalent Power Formula	$P = P_1 + P_2$	$P = P_1 + P_2 - d P_1 P_2$
Effect of Separation on Power	Constant for a given contact setup	Decreases as separation $d$ increases for converging lenses

Practice Quiz

Choose an option to reveal the explanation.

1. What is the equivalent focal length of two thin lenses of focal lengths $f_1$ and $f_2$ kept in close contact?

Explanation: For lenses in contact, $1/F = 1/f_1 + 1/f_2$, which simplifies to $F = \frac{f_1 f_2}{f_1 + f_2}$.

2. If two lenses of powers $+2\text{ D}$ and $-3\text{ D}$ are placed in contact, what is the power of the combination?

Explanation: Total power $P = P_1 + P_2 = (+2\text{ D}) + (-3\text{ D}) = -1\text{ D}$.

3. How does the total linear magnification $m$ of a system of thin lenses in contact relate to individual magnifications?

Explanation: The total linear magnification produced by a system of thin lenses in contact is equal to the product of linear magnifications of individual lenses: $m = m_1 \times m_2$.

4. When two thin lenses of focal lengths $f_1$ and $f_2$ are separated by a distance $d$, the formula for the equivalent focal length $F$ is:

Explanation: The correct formula for two coaxial thin lenses separated by a distance $d$ is $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$.

5. What happens to the equivalent focal length of two convex lenses when they are separated by a small distance $d$ compared to when they are in contact?

Explanation: Since separation subtracts a term from $1/F$, $1/F$ decreases, which means the equivalent focal length $F$ increases.

6. What is the unit of optical power of a lens when its focal length is measured in meters?

Explanation: Optical power is measured in dioptres ($D$), which are equivalent to $\text{m}^{-1}$ when focal length is in meters.

7. If a convex lens of focal length $20\text{ cm}$ is placed in contact with a concave lens of focal length $25\text{ cm}$, what is the focal length of the combination?

Explanation: $f_1 = +20\text{ cm}$, $f_2 = -25\text{ cm}$. $1/F = 1/20 + 1/(-25) = 1/100$, so $F = +100\text{ cm}$.

8. Which of the following describes the power $P$ of two separated lenses with individual powers $P_1$ and $P_2$ separated by distance $d$?

Explanation: The equivalent power for separated lenses is given by $P = P_1 + P_2 - d P_1 P_2$.

9. Adding two convex lenses in contact has what effect on the overall system?

Explanation: Adding convex lenses increases individual powers, so the combined power increases ($P = P_1 + P_2$) and the equivalent focal length decreases ($1/F = 1/f_1 + 1/f_2$).

10. What is the primary purpose of combining lenses, such as in an achromatic doublet?

Explanation: A combination of lenses, such as achromatic doublets, is specifically designed to minimize chromatic and spherical aberrations.

11. For two thin lenses in contact, if $f_1 = +10\text{ cm}$ and $f_2 = +15\text{ cm}$, the equivalent focal length is:

Explanation: $F = \frac{f_1 f_2}{f_1 + f_2} = \frac{10 \times 15}{10 + 15} = \frac{150}{25} = +6\text{ cm}$.

12. In advanced PGT scope, what mathematical tool is often used to compute the overall system matrix for separated lenses?

Explanation: Matrix methods (ABCD matrices involving translation and refraction matrices) are used in advanced optics to analyze ray paths through lens systems.

13. If the separation $d$ between two lenses equals the sum of their focal lengths ($d = f_1 + f_2$), what is the equivalent focal length of the system?

Explanation: Using $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{f_1 + f_2}{f_1 f_2} = \frac{f_1 + f_2}{f_1 f_2} - \frac{f_1 + f_2}{f_1 f_2} = 0$, so $F = \infty$.

14. What is the misconception regarding adding two convex lenses?

Explanation: A common misconception is that adding two convex lenses increases focal length, whereas it actually decreases the equivalent focal length and increases power.

15. How does immersion of a lens system in a liquid affect its focal length compared to air?

Explanation: The focal length of a lens depends on the relative refractive index of the lens material with respect to the surrounding medium, as per Lens Maker's Formula.

PYQ Patterns

These are pattern-based practice questions. Authentic year labels are only used when verified from a source.

1. Two thin lenses of power $+5\text{ D}$ and $-2\text{ D}$ are placed in close contact. The focal length of the combination is:

Explanation: Total power $P = +5 - 2 = +3\text{ D}$. Equivalent focal length $F = 1/P = 1/3\text{ m} \approx +0.33\text{ m}$.

2. A convex lens of focal length $30\text{ cm}$ is placed in contact with a concave lens of focal length $50\text{ cm}$. The power of the resulting combination is:

Explanation: $P_1 = 1/0.3 = +3.33\text{ D}$, $P_2 = 1/(-0.5) = -2\text{ D}$. $P = 3.33 - 2 = +1.33\text{ D}$? Wait: $1/F = 1/30 + 1/(-50) = (5 - 3)/150 = 2/150 = 1/75\text{ cm} = 1/0.75\text{ m}$. Thus $P = 1/0.75 = +1.33\text{ D}$. Let's re-verify: $P_1 = +3.33$, $P_2 = -2.0$, sum is $+1.33\text{ D}$. Wait, option 1 is $+1.33\text{ D}$ and option 2 is $-1.33\text{ D}$. Convex is +30, concave is -50, so equivalent is positive. Correct index is 0.

3. Two thin lenses with focal lengths $f_1$ and $f_2$ are separated by a distance $d$. The condition for the equivalent focal length to be independent of wavelength (achromatic combination) typically involves:

Explanation: For two lenses of the same material separated by $d$, the condition for achromatism is $d = \frac{f_1 + f_2}{2}$.

4. If two identical convex lenses each of focal length $f$ are placed in contact, the equivalent focal length of the combination is:

Explanation: $1/F = 1/f + 1/f = 2/f$, hence $F = f/2$.

5. When a convex lens and a concave lens of equal focal length are placed in contact, the equivalent focal length and power are:

Explanation: If $f_1 = f$ and $f_2 = -f$, then $1/F = 1/f - 1/f = 0$, so $F = \infty$ and $P = 0$.

6. An object is placed at a distance $u$ in front of a system of thin lenses in contact having equivalent focal length $F$. The image distance $v$ satisfies:

Explanation: The standard thin lens equation applies to the equivalent lens: $\frac{1}{v} - \frac{1}{u} = \frac{1}{F}$.

Timed Topic Test

Attempt these questions under a ten-minute limit using a separate timer.

1. Calculate the equivalent power of two lenses of power $+4\text{ D}$ and $-1.5\text{ D}$ separated by a distance of $0.05\text{ m}$.

Explanation: $P = P_1 + P_2 - d P_1 P_2 = 4 + (-1.5) - (0.05)(4)(-1.5) = 2.5 - (-0.3) = 2.5 + 0.3 = 2.8\text{ D}$? Wait: $4 - 1.5 = 2.5$. $0.05 \times 4 \times (-1.5) = 0.05 \times (-6) = -0.3$. So $2.5 - (-0.3) = 2.8\text{ D}$. Let's re-verify options: A) 2.2, B) 2.7, C) 3.18, D) 1.85. Let's check with $P_1=4, P_2=-1.5, d=0.05$: $P = 4 - 1.5 - 0.05(4)(-1.5) = 2.5 + 0.3 = 2.8\text{ D}$. Wait, none is 2.8. Let's recalculate carefully: $P_1=4, P_2=2$ ($+2$)? Let's use $+4$ and $+2$ and $d=0.05$: $4 + 2 - 0.05(8) = 6 - 0.4 = 5.6$. Let's provide a clean question: $P_1 = +4\text{ D}$, $P_2 = +2\text{ D}$, $d = 0.05\text{ m}$. Then $P = 4 + 2 - (0.05)(4)(2) = 6 - 0.4 = 5.6\text{ D}$. Let's fix options for $P_1=4, P_2=2, d=0.05$: options ["5.6 D", "6.0 D", "5.2 D", "4.8 D"], correct_index 0.

2. Calculate the equivalent power when $P_1 = +4\text{ D}$, $P_2 = +2\text{ D}$, and separation $d = 0.05\text{ m}$.

Explanation: $P = P_1 + P_2 - d P_1 P_2 = 4 + 2 - (0.05)(4)(2) = 6 - 0.4 = 5.6\text{ D}$.

3. Two lenses of focal lengths $f_1 = +20\text{ cm}$ and $f_2 = -10\text{ cm}$ are placed in contact. The nature of the combination is:

Explanation: $1/F = 1/20 + 1/(-10) = 1/20 - 2/20 = -1/20\text{ cm}$. Since $F = -20\text{ cm}$ (negative), the combination behaves as a diverging lens.

4. What is the equivalent focal length of two lenses in contact if their powers are $+2.5\text{ D}$ and $+1.5\text{ D}$?

Explanation: Total power $P = 2.5 + 1.5 = +4.0\text{ D}$. Equivalent focal length $F = 1/P = 1/4 = +0.25\text{ m}$.

5. If three lenses of focal lengths $f_1$, $f_2$, and $f_3$ are kept in contact, the reciprocal of equivalent focal length is:

Explanation: For any number of thin lenses in intimate contact, the reciprocal of the equivalent focal length is the sum of reciprocals of individual focal lengths.

6. Which parameter determines the shift in principal planes when two lenses are separated by a distance $d$?

Explanation: The positions of the equivalent principal planes and focal points depend on the individual focal lengths $f_1$, $f_2$ and the separation distance $d$.

7. When two lenses are separated by a distance $d$, what happens to the convergence of parallel incident rays compared to contact?

Explanation: Separating converging lenses reduces the effective converging power of the system compared to when they are in direct contact.

8. A convex lens ($f = +15\text{ cm}$) and a concave lens ($f = -10\text{ cm}$) are separated by $5\text{ cm}$. What is the reciprocal of equivalent focal length $1/F$?

Explanation: Using $1/F = 1/f_1 + 1/f_2 - d/(f_1 f_2)$, with $f_1 = 15$, $f_2 = -10$, and $d = 5$: $1/15 + 1/(-10) - 5/(15(-10)) = 1/15 - 1/10 + 5/150$.

9. If the total magnification of a two-lens system is $+2$, and the first lens produces a magnification of $-1$, what is the magnification of the second lens?

Explanation: $m = m_1 \times m_2 \implies +2 = (-1) \times m_2 \implies m_2 = -2$.

10. What is the primary limitation of treating lens combinations using simple scalar addition of focal lengths?

Explanation: The simple formula $1/F = 1/f_1 + 1/f_2$ assumes thin lenses in direct contact ($d=0$). Thick lenses or separated lenses require extended formulas and matrix methods.