Lenses in Contact: Power and Focal Length

1. The Concept: One System, One Effective Lens

When two or more thin lenses are placed in direct contact, they behave as a single lens with a new, combined focal length and power. To understand how this works, we'll follow the path of a light ray through the system. The logic is identical to what we used in the Lens Maker's formula derivation: the image formed by the first lens becomes the object for the second.

To visualize this, let's watch a short video that illustrates the concept using ray diagrams.

As the video demonstrates, if we have two convex lenses, $L_1$ and $L_2$, in contact:

1. Lens $L_1$ takes the object $O$ and forms an image $I_1$.
2. This image $I_1$ then acts as a virtual object for the second lens, $L_2$.
3. Lens $L_2$ takes this virtual object $I_1$ and forms the final image $I$.

The combination of lenses effectively takes the object $O$ and produces the final image $I$, acting like a single lens with an equivalent focal length, $F_{eq}$.

2. Deriving the Formulas for Combined Lenses

Now, let's translate this visual concept into mathematical formulas. You'll find this derivation is a straightforward application of the thin lens formula you already know.

Equivalent Focal Length

We'll apply the thin lens formula, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$, to each lens sequentially.

For the first lens, $L_1$, with focal length $f_1$:

• Object distance = $u$
• Image distance = $v_1$ (the position of the intermediate image $I_1$)
  $\frac{1}{f_1}=\frac{1}{v_1}-\frac{1}{u}\cdots (1)$

For the second lens, $L_2$, with focal length $f_2$:

• Object distance = $v_1$ (since $I_1$ is the object for $L_2$)
• Image distance = $v$ (the position of the final image $I$)
  $\frac{1}{f_2}=\frac{1}{v}-\frac{1}{v_1}\cdots (2)$

Now, we simply add equations (1) and (2):
$\left(\frac{1}{f_1}\right)+\left(\frac{1}{f_2}\right)=\left(\frac{1}{v_1}-\frac{1}{u}\right)+\left(\frac{1}{v}-\frac{1}{v_1}\right)$
The intermediate image term $\frac{1}{v_1}$ cancels out:
$\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{v}-\frac{1}{u}$

We know that for the equivalent lens with focal length $F_{eq}$, it takes an object at $u$ and forms a final image at $v$. So, for the combination, $\frac{1}{F_{eq}}=\frac{1}{v}-\frac{1}{u}$.

Therefore, we arrive at the formula for the equivalent focal length:
$\frac{1}{F_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}$
This can be generalized for any number of thin lenses in contact:
$\frac{1}{F_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}+\cdots$

Equivalent Power

The concept of optical power simplifies this relationship even further. Recall that the power of a lens in dioptres (D) is the reciprocal of its focal length in meters: $P=1/f$.

From our focal length formula:
$\frac{1}{F_{eq}}={\sum }_i\frac{1}{f_i}$
It follows directly that the equivalent power is simply the algebraic sum of the individual powers:
$P_{eq}=P_1+P_2+P_3+\cdots$

This additive property makes power a very convenient quantity to work with when dealing with lens combinations.

To reinforce these formulas and their derivation, please review the following resource. It also highlights crucial points for problem-solving.

3. Application and NEET-Style Problems

This is where we apply the theory to the kind of vigorous mathematical problems you'll face in the NEET exam. The key is strict adherence to the sign convention.

• Convex Lens (Converging): Focal length $f$ is positive. Power $P$ is positive.
• Concave Lens (Diverging): Focal length $f$ is negative. Power $P$ is negative.

Example 1: Convex and Concave Combination

Let's solve a typical problem involving a mix of lenses.
Problem: A convex lens of focal length 20 cm is placed in contact with a concave lens of focal length 40 cm. Find the focal length and power of the combination.

Solution:

1. Assign signs to focal lengths:

   • Convex lens: $f_1=+20$ cm = +0.2 m
   • Concave lens: $f_2=-40$ cm = -0.4 m

2. Calculate equivalent focal length ($F_{eq}$):
   $\frac{1}{F_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{+20}+\frac{1}{-40}=\frac{2-1}{40}=\frac{1}{40}$
   $F_{eq}=+40\text{ cm}$
   Since the equivalent focal length is positive, the combination acts as a converging lens.

3. Calculate equivalent power ($P_{eq}$):
   We can do this in two ways:

   • Using $F_{eq}$: $P_{eq}=\frac{1}{F_{eq}(\text{in m})}=\frac{1}{+0.4}=+2.5\text{ D}$
   • By summing individual powers:
     • $P_1=\frac{1}{f_1(\text{in m})}=\frac{1}{0.2}=+5\text{ D}$
     • $P_2=\frac{1}{f_2(\text{in m})}=\frac{1}{-0.4}=-2.5\text{ D}$
     • $P_{eq}=P_1+P_2=(+5)+(-2.5)=+2.5\text{ D}$
       Both methods give the same result. The combination has a power of +2.5 Dioptres.

Example 2: NEET Previous Year Questions

Let's tackle some actual questions from past NEET papers. These often test conceptual understanding or combine multiple concepts.

NEET 2023 Question:
Two thin lenses are of the same focal lengths (f), but one is convex and the other one is concave. When they are placed in contact with each other, the equivalent focal length of the combination will be...
(a) Zero
(b) f/4
(c) f/2
(d) Infinite

Solution:

• Convex lens: $f_1=+f$
• Concave lens: $f_2=-f$
• Apply the formula:
  $\frac{1}{F_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{f}+\frac{1}{-f}=\frac{1}{f}-\frac{1}{f}=0$
• If $\frac{1}{F_{eq}}=0$, then $F_{eq}=\infty$.
• The correct answer is (d) Infinite. The combination acts like a plane glass slab, with zero power. You can find this question and its solution in the provided resource LINK under Section 1, Q2.

NEET 2017 Question (Advanced Application):
This problem cleverly combines the Lens Maker's formula from our last lesson with today's topic.

Two identical glass (${\mu }_g=3/2$) equi-convex lenses of focal length f each are kept in contact. The space between the two lenses is filled with water (${\mu }_w=4/3$). The focal length of the combination is...
(a) f/3
(b) f
(c) 4f/3
(d) 3f/4

Solution Strategy:

1. Realize the system is now three lenses in contact:

   • Lens 1: The first equi-convex glass lens ($f_1$).
   • Lens 2: A plano-concave water lens formed in the middle ($f_{water}$).
   • Lens 3: The second equi-convex glass lens ($f_3$).

   The equivalent focal length will be $\frac{1}{F_{eq}}=\frac{1}{f_1}+\frac{1}{f_{water}}+\frac{1}{f_3}$.

2. Analyze the lenses:

   • For the glass lenses: $f_1=f_3=f$. Let the radii of curvature be $R$ and $-R$.
     Using the Lens Maker's formula in air: $\frac{1}{f}=({\mu }_g-1)(\frac{1}{R}-\frac{1}{-R})=(\frac{3}{2}-1)(\frac{2}{R})=(\frac{1}{2})(\frac{2}{R})=\frac{1}{R}$. So, $f=R$.

   • For the water lens: It is a plano-concave lens with radii $-R$ and $+R$.
     Using the Lens Maker's formula for the water lens (in air, for now):
     $\frac{1}{f_{water}}=({\mu }_w-1)(\frac{1}{-R}-\frac{1}{+R})=(\frac{4}{3}-1)(\frac{-2}{R})=(\frac{1}{3})(\frac{-2}{R})=\frac{-2}{3R}$.
     Since $R=f$, we have $\frac{1}{f_{water}}=\frac{-2}{3f}$.

3. Combine the focal lengths:
   $\frac{1}{F_{eq}}=\frac{1}{f_1}+\frac{1}{f_{water}}+\frac{1}{f_3}=\frac{1}{f}+\left(\frac{-2}{3f}\right)+\frac{1}{f}$
   $\frac{1}{F_{eq}}=\frac{2}{f}-\frac{2}{3f}=\frac{6-2}{3f}=\frac{4}{3f}$
   $F_{eq}=\frac{3f}{4}$

• The correct answer is (d) 3f/4. This problem is a perfect example of how NEET combines concepts. You can find this question in resource LINK under Section 3, Q5.

Conclusion

Great job working through these concepts and problems. You have now learned how to analyze systems of thin lenses in contact, a crucial skill for understanding more complex optical devices.

Key Takeaways:

• A combination of thin lenses in contact acts as a single equivalent lens.
• The equivalent reciprocal focal length is the sum of the individual reciprocal focal lengths: $\frac{1}{F_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}+\cdots$.
• It's often simpler to work with power, as equivalent power is the direct algebraic sum of individual powers: $P_{eq}=P_1+P_2+\cdots$.
• Strict adherence to the sign convention (+ for convex/converging, - for concave/diverging) is non-negotiable for getting correct answers.
• Complex systems can be broken down into simpler components, like in the water-filled lens problem.

Preview of the Next Lesson:
We have mastered the case where lenses are in direct contact. But what if they are separated by a distance, as they are in a telescope or microscope? In our next lesson, we will analyze image formation for systems of lenses separated by a distance. This adds another layer of complexity but will bring us much closer to understanding real-world optical instruments.