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Formulae for Focal Lengths in Geometrical Optics

The focal length of an optical system is determined by its radii of curvature, refractive index, and the surrounding medium according to rigorous geometrical optics principles. These formulations govern the convergence and divergence properties of thin and thick lenses as well as spherical mirrors.

UP TGT ScienceUP PGT PhysicsStudy Notes + Practice

Scope and prerequisites

TGT core

Lens Maker's Formula, mirror formula, magnification, simple combination of thin lenses in contact, sign convention.

PGT extension

Lenses separated by a distance, equivalent focal length, focal length of systems with varying refractive indices, displacement method for finding focal length.

Prerequisites

Laws of refraction and Snell's Law
Cartesian sign convention for spherical surfaces
Paraxial approximation

Lens Maker's Formula

Relates the focal length (f) of a thin lens to the relative refractive index (μ) and the radii of curvature (R₁ and R₂) of its two surfaces.
Derived by applying refraction successively at two spherical boundaries assuming a paraxial ray tracing approach.
Depends directly on the ratio of the refractive index of the lens material to that of the surrounding medium.

Spherical Mirror Focal Length

The focal length of a spherical mirror is strictly half of its radius of curvature under paraxial conditions: $f = R / 2$.
Independent of the surrounding medium because reflection does not depend on the speed of light in the medium.
Concave mirrors possess negative focal lengths while convex mirrors possess positive focal lengths in standard Cartesian sign conventions.

Combination of Thin Lenses

When two thin lenses of focal lengths f₁ and f₂ are placed in close contact, the reciprocal of the equivalent focal length is the algebraic sum of their individual reciprocals.
The total power of a combination of lenses in contact is given by the algebraic sum of the individual powers: $P = P_{1} + P_{2}$.
Useful for designing achromatic doublets and estimating cumulative optical convergence.

Displacement Method (U-V Method)

An experimental method to determine the focal length of a convex lens using an object-screen distance greater than 4f.
Yields two distinct positions of the lens for which a sharp image is formed on the fixed screen.
Provides an accurate estimation of focal length without directly measuring object and image distances from the optical center.

Derivations

Lens Maker's Formula Derivation

Conditions: Thin lens in a medium, paraxial rays, spherical surfaces of radii R₁ and R₂

Apply refraction formula for the first spherical surface separating media of refractive index n₁ and n₂: \frac{n_2}{v_1} - \frac{n_1}{u} = \frac{n_2 - n_1}{R_1}
For the second surface, the image formed by the first surface acts as a virtual object, with refraction from n₂ to n₁: \frac{n_1}{v} - \frac{n_2}{$v_1 = \frac{n_1 - n_2}{R_2}$
Adding both equations eliminates the intermediate image distance v₁ and yields the general thin lens formula: \frac{1}{v} - \frac{1}{u} = (\frac{n_2}{n_1} - 1)(\frac{1}{R_1} - \frac{1}{R_2})

Result: \frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})

Equivalent Focal Length of Two Thin Lenses Separated by Distance d

Conditions: Two thin lenses of focal lengths f₁ and f₂ separated by a distance d in air

Trace a paraxial ray parallel to the principal axis at height h₁ incident on the first lens.
Determine the deviation and the height h₂ at which the ray strikes the second lens: $h_2 = h_1 (1 - \frac{d}{f_1})$
Calculate the total deviation produced by the system to find the reciprocal of the equivalent focal length F.

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
Lens Maker's Formula	$$\frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$	f = focal length, μ_rel = n_lens / n_medium, R₁, R₂ = radii of curvature Metres (m)	Thin lens, paraxial rays
Mirror Focal Length	$$f = \frac{R}{2}$$	f = focal length, R = radius of curvature Metres (m)	Spherical mirror, paraxial approximation
Thin Lens Formula	$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$	f = focal length, v = image distance, u = object distance Metres (m)	Thin lens in uniform medium
Combination of Lenses in Contact	$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} + \dots$$	F = equivalent focal length, f₁, f₂ = individual focal lengths Metres (m)	Thin lenses in tight contact
Lenses Separated by Distance d	$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$$	F = equivalent focal length, d = separation distance Metres (m)	Coaxial thin lenses separated by distance d
Displacement Method Focal Length	$$f = \frac{L^2 - d^2}{4L}$$	L = distance between object and screen, d = distance between two lens positions Metres (m)	L > 4f
Lens in Liquid Medium	$$\frac{f_m}{f_a} = \frac{(n_g - n_a)n_m}{(n_g - n_m)n_a}$$	f_m, f_a = focal lengths in medium and air, n_g, n_m, $n_a = refractive indices$ Dimensionless ratio	Same lens immersed in a liquid
Power of a Lens	$$P = \frac{1}{f}$$	P = optical power, f = focal length in metres Dioptres (D)	Focal length expressed in metres

Graphs and experiments

u-v Graph for a Convex Lens

Draw: A hyperbola plot of image distance v versus object distance u, or a plot of 1/v versus 1/u yielding a straight line with intercepts equal to 1/f.

Interpret: The straight-line plot of 1/v versus 1/u has a slope of +1 and intercepts on both axes equal to 1/f.

Exam use: Tested conceptually for slope interpretation and graphical determination of focal length.

Displacement Method Experiment

Draw:

Interpret: Allows precise calculation of focal length f without locating the optical center directly.

Exam use: Numerical problems based on the formula $f = (L^{2} - d^{2})/4L$.

Variation of Focal Length with Wavelength (Chromatic Aberration)

Draw: Plot of focal length f versus wavelength λ showing decrease of focal length toward the violet end of the spectrum.

Interpret: Refractive index depends on wavelength (μ ∝ 1/λ² via Cauchy's relation), causing shorter wavelengths to focus closer to the lens.

Exam use: Conceptual questions on chromatic aberration and dispersive power.

Solved numericals

TGT core

A double convex lens has radii of curvature of 20 cm each. If the refractive index of the glass is 1.5, calculate its focal length in air.

Identify given values: $R_1 = +20 cm$, $R_2 = -20 cm$, $\mu = 1.5$.
Substitute values into the Lens Maker's Formula: \frac{1}{f} = (1.5 - 1) \left(\frac{1}{20} - \frac{1}{-20}\right).
Compute: \frac{1}{f} = 0.5 \times \left(\frac{2}{20}\right) = 0.5 \times \frac{1}{10} = \frac{1}{20}.

Answer: f = 20 cm

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: $f_1 = +20 cm = 0.2 m$, $f_2 = -30 cm = -0.3 m$.
Apply the combination formula: \frac{1}{F} = \frac{1}{20} + \frac{1}{-30} = \frac{3 - 2}{60} = \frac{1}{60} cm^{-1}.
Calculate power: $P = \frac{1}{F (in m)} = \frac{1}{0.6} D = +1.67 D$.

Answer: $F = +60 cm$, $P = +1.67 D$

PGT extension

A convex lens of refractive index 1.5 has a focal length of 20 cm in air. What will be its focal length when immersed in a liquid of refractive index 1.33?

Use the lens focal length ratio formula: \frac{f_m}{$f_a = \frac{(n_g - 1)}{\left(\frac{n_g}{n_m} - 1\right)}$.
Substitute values: $n_g = 1.5$, $n_m = 1.33 = \frac{4}{3}$, $f_a = 20 cm$.
Compute denominator: \frac{1.5}{4/3} - $1 = \frac{4.5}{4} - 1 = \frac{0.5}{4} = \frac{1}{8}$.
Compute ratio: \frac{f_m}{20} = \frac{1.5 - 1}{1/8} = \frac{0.5}{1/8} = 4.
Solve for f_m: $f_m = 20 \times 4 = 80 cm$.

Answer: $f_m = 80 cm$

Assumptions and limitations

Paraxial ray approximation is assumed, which limits validity to small angles of incidence where sin(θ) ≈ θ.
Lenses are considered thin, meaning thickness is neglected compared to radii of curvature and object/image distances.

Common misconceptions

Common error: The focal length of a lens changes when it is inverted.

Correct: The focal length remains exactly the same upon inversion.

Focal length depends only on the radii of curvature and refractive indices, which do not change when the lens is flipped.

Common error: A convex lens always converges light regardless of the surrounding medium.

Correct: A convex lens can behave as a diverging lens if immersed in a medium with a higher refractive index than the lens material.

The relative refractive index term (μ_rel - 1) becomes negative, reversing the optical action.

Quick revision

Lens Maker's Formula: \frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})
Lens Maker's Formula: \frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)
Mirror Focal Length: f = \frac{R}{2}

Must remember

The Lens Maker's Formula relates the focal length of a thin lens to its relative refractive index and the radii of curvature of its surfaces: $\frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
The focal length of a spherical mirror is strictly half of its radius of curvature under paraxial conditions: $f = \frac{R}{2}$, and it is independent of the surrounding medium.
When two thin lenses of focal lengths $f_1$ and $f_2$ are placed in close contact, the equivalent focal length is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.
When two thin lenses are separated by a distance $d$, their equivalent focal length is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$.
The displacement method (U-V method) uses the formula $f = \frac{L^2 - d^2}{4L}$ to find the focal length of a convex lens for an object-screen distance $L > 4f$.
A convex lens can behave as a diverging lens if it is immersed in a medium whose refractive index is higher than that of the lens material.

Glossary

Lens Maker's FormulaAn equation relating focal length to the relative refractive index and radii of curvature of the lens surfaces.

Relative Refractive Index ($\mu_{rel}$)The ratio of the refractive index of the lens material to that of the surrounding medium ($n_{lens} / n_{medium}$).

Optical Power ($P$)The reciprocal of the focal length expressed in metres, measured in dioptres (D).

Displacement MethodAn experimental technique using two distinct lens positions between a fixed object and screen to determine focal length.

Spherical Mirror vs Thin Lens

Parameter	Spherical Mirror	Thin Lens
Medium Dependency	Independent of surrounding medium	Dependent on surrounding medium via $\mu_{rel}$
Focal Length Formula	$f = \frac{R}{2}$	$\frac{1}{f} = (\mu_{rel} - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
Refracting Surfaces	Single reflecting surface	Two refracting surfaces

Practice Quiz

Choose an option to reveal the explanation.

1. What is the relation between the focal length ($f$) and radius of curvature ($R$) of a spherical mirror under paraxial conditions?

Explanation: Under paraxial conditions, the focal length of a spherical mirror is strictly half of its radius of curvature, given by $f = \frac{R}{2}$.

2. Does the focal length of a spherical mirror change when it is immersed in water?

Explanation: Reflection does not depend on the speed of light in the surrounding medium. Therefore, the focal length of a spherical mirror remains unaffected when immersed in water or any other medium.

3. Which of the following expressions represents the Lens Maker's Formula for a thin lens?

Explanation: The Lens Maker's Formula is given by $\frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$, where $R_1$ and $R_2$ follow standard sign conventions.

4. Two thin lenses of focal lengths $f_1$ and $f_2$ are placed in tight contact. What is the equivalent focal length ($F$)?

Explanation: When two lenses are in contact, the reciprocal of the equivalent focal length is the sum of individual reciprocals: $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$, which simplifies to $F = \frac{f_1 f_2}{f_1 + f_2}$.

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

Explanation: The optical power $P = \frac{1}{f}$ is measured in dioptres (D) when the focal length $f$ is expressed in metres.

6. What happens to a biconvex glass lens (refractive index $1.5$) when it is immersed in a liquid of refractive index $1.65$?

Explanation: When immersed in a medium with a higher refractive index than the lens, $\mu_{rel} = \frac{n_{lens}}{n_{medium}} < 1$, making $(\mu_{rel} - 1)$ negative. Thus, a convex lens behaves as a diverging lens.

7. Two thin coaxial lenses of focal lengths $f_1$ and $f_2$ are separated by a distance $d$. What is the equivalent focal length $F$?

Explanation: For two coaxial thin lenses separated by a distance $d$, the equivalent focal length formula is $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$.

8. In the displacement method to find the focal length of a convex lens, what is the minimum distance required between the object and the screen?

Explanation: The displacement method requires the distance between the fixed object and screen to be strictly greater than $4f$ ($L > 4f$) to obtain two distinct real image positions.

9. In the displacement method, if $L$ is the distance between the object and screen, and $d$ is the distance between the two lens positions, the focal length $f$ is given by:

Explanation: The standard formula derived from lens equations for the displacement method is $f = \frac{L^2 - d^2}{4L}$.

10. If a lens has a focal length of $+0.5\text{ m}$, what is its optical power?

Explanation: Power $P = \frac{1}{f} = \frac{1}{0.5\text{ m}} = +2.0\text{ D}$.

11. What is the sign of the focal length of a concave mirror under standard Cartesian sign conventions?

Explanation: A concave mirror has its focus in front of the reflecting surface, meaning its focal length is measured against the direction of incident light, resulting in a negative sign.

12. If a thin lens is cut along its principal axis into two identical halves, how does the focal length of each half compare to the original lens?

Explanation: Cutting a lens along its principal axis does not change the radii of curvature of the surfaces or the refractive index, so the focal length of each half remains identical to the original lens.

13. What happens to the focal length of a glass convex lens if the wavelength of incident light is increased from violet to red?

Explanation: Refractive index decreases with an increase in wavelength (Cauchy's relation: $\mu = A + \frac{B}{\lambda^2}$). Since $\frac{1}{f} \propto (n - 1)$, a smaller refractive index for red light results in a larger focal length compared to violet light.

14. What is the equivalent power of two lenses with powers $+4\text{ D}$ and $-2\text{ D}$ placed in contact?

Explanation: The total power of lenses in contact is the algebraic sum of their individual powers: $P = P_1 + P_2 = +4\text{ D} + (-2\text{ D}) = +2\text{ D}$.

15. Which of the following assumptions is fundamental in deriving the standard Lens Maker's Formula?

Explanation: The Lens Maker's Formula is derived assuming paraxial rays (small angles) and a thin lens where thickness is neglected compared to radii of curvature.

PYQ Patterns

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

1. A biconvex lens has radii of curvature of magnitude $20\text{ cm}$ each and is made of glass of refractive index $1.5$. What is its focal length in air?

Explanation: Using the Lens Maker's Formula $\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$, with $R_1 = +20\text{ cm}$ and $R_2 = -20\text{ cm}$: $\frac{1}{f} = (1.5 - 1)\left(\frac{1}{20} - \frac{-1}{20}\right) = 0.5 \times \frac{2}{20} = \frac{1}{20}$, giving $f = 20\text{ cm}$.

2. Two thin lenses of focal lengths $+10\text{ cm}$ and $-15\text{ cm}$ are placed in contact. The focal length of the combination is:

Explanation: Using $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{10} + \frac{1}{-15} = \frac{3 - 2}{30} = \frac{1}{30}$, so $F = +30\text{ cm}$.

3. A convex lens of focal length $f$ is cut into two identical halves perpendicular to its principal axis. What is the focal length of each resulting part?

Explanation: When cut perpendicular to the principal axis, one of the flat surfaces has an infinite radius of curvature ($R_2 = \infty$). Applying the Lens Maker's Formula shows that the focal length of each half becomes $2f$.

4. In a displacement method experiment, the distance between the object and screen is $100\text{ cm}$. For two positions of the lens, sharp images are formed on the screen separated by $20\text{ cm}$. What is the focal length of the lens?

Explanation: Using $f = \frac{L^2 - d^2}{4L}$, where $L = 100\text{ cm}$ and $d = 20\text{ cm}$: $f = \frac{100^2 - 20^2}{4 \times 100} = \frac{10000 - 400}{400} = \frac{9600}{400} = 24\text{ cm}$? Wait, let's recalculate: $9600 / 400 = 24\text{ cm}$. Let's verify options. Option C is $24\text{ cm}$. Correct index is 2.

5. A lens of refractive index $1.5$ has a focal length of $20\text{ cm}$ in air. When it is immersed in a liquid of refractive index $1.33$, its focal length becomes:

Explanation: The relative refractive index $\mu_{rel} = \frac{n_g}{n_m}$ decreases when immersed in water ($1.33$) compared to air ($1.0$). Since $\frac{1}{f} \propto (\mu_{rel} - 1)$, a smaller relative refractive index decreases the optical convergence, thereby increasing the focal length.

6. Two thin lenses of focal lengths $f_1$ and $f_2$ are separated by a distance $d$. If the combination behaves as a system with infinite focal length, what is the relation between $d$, $f_1$, and $f_2$?

Explanation: For equivalent focal length $F = \infty$, $\frac{1}{F} = 0$. From $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} = 0$, we get $\frac{d}{f_1 f_2} = \frac{f_1 + f_2}{f_1 f_2}$, which yields $d = f_1 + f_2$.

Timed Topic Test

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

1. What is the focal length of a plane glass slab of refractive index $1.5$?

Explanation: Both surfaces of a plane glass slab are flat, meaning their radii of curvature are infinite ($R_1 = \infty, R_2 = \infty$). Substituting into the Lens Maker's Formula gives $\frac{1}{f} = 0$, so $f = \infty$.

2. A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens material. Its focal length will:

Explanation: When $n_{lens} = n_{medium}$, $\mu_{rel} = 1$, making $(\mu_{rel} - 1) = 0$. Thus $\frac{1}{f} = 0$, so the focal length becomes infinite and the lens acts as a plane sheet of glass.

3. The power of a thin convex lens is $+5\text{ D}$. When it is placed in contact with a thin concave lens, the combination has a power of $+2\text{ D}$. What is the focal length of the concave lens?

Explanation: Total power $P = P_1 + P_2 \implies +2 = +5 + P_2 \implies P_2 = -3\text{ D}$. The focal length of the concave lens is $f_2 = \frac{1}{P_2} = \frac{1}{-3}\text{ m} = -0.333\text{ m} = -33.3\text{ cm}$.

4. Which of the following optical systems is independent of chromatic aberration by design?

Explanation: An achromatic doublet combines lenses of suitable materials and focal lengths to eliminate chromatic aberration for two specific wavelengths.

5. If the radius of curvature of each surface of a symmetric double convex lens is $R$ and its refractive index is $1.5$, its focal length is:

Explanation: Using $\frac{1}{f} = (1.5 - 1)\left(\frac{1}{R} - \frac{1}{-R}\right) = 0.5 \times \frac{2}{R} = \frac{1}{R}$, which gives $f = R$.

6. What is the equivalent focal length of two lenses of focal length $f$ placed in contact coaxially?

Explanation: $\frac{1}{F} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}$, which gives $F = \frac{f}{2}$.

7. Under what condition does a convex lens form a virtual image of the same size as the object?

Explanation: A convex lens forms a virtual image when the object is placed within the focal length, but that virtual image is always magnified, never the same size as the object (unlike a plane mirror).

8. In the displacement method, if the object-to-screen distance $L$ is less than $4f$, what is the nature of the roots for the lens positions?

Explanation: The quadratic equation yielding the lens positions in the displacement method has a discriminant proportional to $(L^2 - 4fL)$. If $L < 4f$, the discriminant is negative, resulting in imaginary roots (meaning no real lens positions exist).

9. How does the power of a lens vary when green light is replaced by red light?

Explanation: Red light has a smaller refractive index in glass than green light ($n_r < n_g$). Since power $P = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$, a smaller refractive index leads to a smaller optical power.

10. A convex lens of glass ($n = 1.5$) has a focal length of $10\text{ cm}$ in air. What is its focal length when immersed in carbon disulfide ($n = 1.63$)?

Explanation: Since $n_{lens} < n_{medium}$ ($1.5 < 1.63$), the relative refractive index $\mu_{rel} < 1$, making $(\mu_{rel} - 1)$ negative. Thus, the focal length becomes negative, indicating a diverging action.