Combined Focal Length of Two Thin Lenses in Contact

When two thin lenses are placed in contact, they behave like a single lens having an equivalent focal length. This equivalent focal length depends on the focal lengths of the individual lenses and helps simplify image-formation calculations in lens systems.

The focal length of a single lens that produces the same optical effect as the given combination of lenses kept in contact.

\[\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}\]

For more than two thin lenses in contact:

Given

• Two thin lenses are placed in contact.
• Their focal lengths are f₁ and f₂.
• The final combination acts like a single lens of focal length f.

Derivation idea

• The first lens forms an image of the object.
• That image acts as the object for the second lens.
• Combining the two lens equations gives the equivalent focal length formula.

Result obtained

Using the lens formula successively for the two lenses, the final relation becomes:

Interpretation

• If both lenses are converging, the combination becomes more powerful.
• If one lens is converging and the other is diverging, the net effect depends on their magnitudes.

Use the standard Cartesian sign convention in lens problems:

• Distances are measured from the optical centre.
• Distances measured in the direction of incident light are positive.
• Distances measured opposite to the direction of incident light are negative.
• Convex lens focal length is positive; concave lens focal length is negative in the usual convention used in ray optics.

Step 1: Image formed by the first lens

Given:

• Object distance, u₁ = −30 cm
• Focal length, f₁ = +10 cm

Using the lens formula,

• \[ \frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1} \]
• \[ \frac{1}{v_1} - \frac{1}{-30} = \frac{1}{10} \]
• \[ v_1 = 15\ \text{cm} \]

The image is formed 15 cm to the right of the first lens.

Step 2: The image acts as an object for the second lens

The second lens is 5 cm away from the first lens.

So, object distance for the second lens:

• u₂ = 15 − 5 = 10 cm

Since the object is on the right side of the second lens, it is a virtual object.

Given:

• u₂ = +10 cm
• f₂ = −10 cm

Using the lens formula,

\[ \begin{aligned} \frac{1}{v_2} - \frac{1}{10} &= \frac{1}{-10},\\ \frac{1}{v_2} &= 0,\\ v_2 &= \infty. \end{aligned} \]

The image is formed at infinity.

Step 3: The image acts as an object for the third lens

An object at infinity forms its image at the focus of the third lens.

Given:

• u₃ = ∞
• f₃ = +30 cm

Using the lens formula,

\[ \begin{aligned} \frac{1}{v_3} - \frac{1}{\infty} &= \frac{1}{30},\\ v_3 &= 30\ \text{cm}. \end{aligned} \]

The final image is formed 30 cm to the right of the third lens.

Wearing two optical lenses together is similar to combining their optical effects into one net effect. This is why lens combinations are studied through equivalent focal length and power addition.

Application areas

• Spectacle lens combinations.
• Optical instruments using multiple lenses.
• Simplified treatment of compound lens systems at the school level.