Compound Lens Systems: NEET Problem Solving

Hello! Welcome back to our study of Ray Optics.

In our last lesson, we established the fundamental step-by-step method for analyzing lens systems where the lenses are separated by a distance. We saw that the key is to treat the system sequentially: find the image from the first lens, and then use that image as the object for the second lens.

Today, we will put that method into vigorous practice. Your goal for this lesson is to apply your knowledge to solve a variety of NEET-style problems involving compound lens systems. We will tackle different combinations of lenses and scenarios that are commonly featured in your exam. This session is all about building your problem-solving speed and accuracy, just as you requested.

Let's begin by reviewing the core principles:

Image from Lens 1 is Object for Lens 2: The position of the object for the second lens ( $u_{2}$ ) is found using the image position from the first lens ( $v_{1}$ ) and the distance between the lenses ( $d$ ). The formula is $u_{2} = d - v_{1}$ . Remember to be careful with signs!
Total Magnification: The overall magnification of the system is the product of the individual magnifications: $M_{t o t a l} = m_{1} \times m_{2}$ .

1. Broadening the Scope: Combining Converging and Diverging Lenses

So far, we've mostly seen systems with two converging lenses. A very common exam scenario involves a combination of a converging (convex) and a diverging (concave) lens. The method remains the same, but you must be careful with the negative focal length of the diverging lens.

This video provides an excellent, detailed walkthrough of such a problem. It solves for the final image characteristics and also includes a ray diagram to help visualize the process.

Multiple Two Lens System with Diverging and Converging Lens

Let's watch this worked example from The Organic Chemistry Tutor to see how to handle a system with one converging and one diverging lens.

Watch the video from the beginning until 11:43, and then from 12:38 to the end. The ray diagram section in between is a good visual, but we will focus on the calculation. Pay close attention to: (03:08) The calculation for the first lens. (07:20) How the object distance for the second (diverging) lens is calculated. (08:00) The use of the negative focal length for the diverging lens. (09:30) The calculation of the final magnification and image properties.

Summary of the video's solution:

System Setup: A convex lens ( $f_{1} = + 3$ cm) and a concave lens ( $f_{2} = - 4$ cm) are separated by $d = 18$ cm. An object is placed at $u_{1} = 6$ cm from the convex lens.
Lens 1 (Convex): Using the lens formula, the first image is formed at $v_{1} = + 6$ cm. This is a real, inverted image ( $m_{1} = - 1$ ).
Lens 2 (Concave): The image at $v_{1} = 6$ cm becomes the object for the second lens. The object distance is $u_{2} = d - v_{1} = 18 - 6 = 12$ cm.
Final Image: Using the lens formula for the second lens with $f_{2} = - 4$ cm and $u_{2} = + 12$ cm, the final image is formed at $v_{2} = - 3$ cm.
- The negative sign for $v_{2}$ means the final image is virtual and formed on the same side as the object for that lens (i.e., to the left of the concave lens).
Final Magnification: The magnification by the second lens is $m_{2} = - v_{2} / u_{2} = - (- 3) /12 = + 0.25$ . The total magnification is $M = m_{1} \times m_{2} = (- 1) \times (+ 0.25) = - 0.25$ . The final image is inverted (relative to the original object) and reduced.

2. Reinforcing a Tricky Case: The Virtual Object

In our last lesson, we introduced the concept of a virtual object, which occurs when the first lens forms an image behind the second lens ( $v_{1} > d$ ). This forces us to use a negative object distance ( $u_{2}$ ) for the second lens. This is a crucial concept to master for complex NEET problems.

Let's analyze another example to solidify your understanding.

Physics - Optics: Lenses (1 of 5) Lens Combinations - Two Converging Lenses

This video by Michel van Biezen tackles a problem where the intermediate image forms behind the second lens, creating a virtual object. This is a perfect reinforcement of this important scenario.

Please watch the entire video. Focus on the moment at (04:33) where the concept of the virtual object is introduced and its distance (S_2) is calculated. Notice how the sign is handled.

Key steps from the video:

Lens 1 Analysis: For the first lens ( $f_{1} = 30$ cm) with an object at $S_{1} = 50$ cm, the image is formed at $S_{1}^{'} = + 75$ cm.
The Virtual Object: The lenses are separated by $d = 10$ cm. Since the first image forms at 75 cm, but the second lens is only 10 cm away, the rays hit the second lens while still converging. The point where they would have met becomes a virtual object.
Object Distance for Lens 2: The distance of this virtual object from the second lens is $75 - 10 = 65$ cm. Because it's on the right side of the lens (where images normally form), it's a virtual object, and its distance is negative: $S_{2} = - 65$ cm.
- This perfectly matches our formula: $u_{2} = d - v_{1} = 10 - 75 = - 65$ cm. The formula automatically handles the sign for you!
Final Image Calculation: The rest of the calculation proceeds as usual, using $S_{2} = - 65$ cm in the lens formula for the second lens.

Test your understanding!

Consider the system from the video ( $f_{1} = + 30$ , $f_{2} = + 20$ , $d = 10$ ). If the initial object is moved to $S_{1} = 20$ cm from the first lens, what happens to the intermediate image? Does it become a real or virtual object for the second lens?

Show answer

Find the position of the first image ( $S_{1}^{'}$ ):
$\frac{1}{S _{1}^{'}} = \frac{1}{f _{1}} - \frac{1}{S _{1}} = \frac{1}{30} - \frac{1}{20} = \frac{2 - 3}{60} = - \frac{1}{60}$
So, $S_{1}^{'} = - 60$ cm. This is a virtual image formed 60 cm to the left of the first lens.
Find the object distance for the second lens ( $S_{2}$ ):
This virtual image is located at a distance of $d + ∣ S_{1}^{'} ∣ = 10 + 60 = 70$ cm from the second lens. Since it is to the left, it acts as a real object for the second lens.
Using our formula: $u_{2} = d - v_{1} = 10 - (- 60) = + 70$ cm. The positive sign confirms it's a real object.

3. NEET-Style Problem Gauntlet

Now, let's apply these principles to solve a variety of problems directly from past NEET papers. This is where you hone your exam skills.

Problem 1: The Collimating System (NEET 2021)

Question: A convex lens A with a focal length of 20 cm and a concave lens B with a focal length of 5 cm are kept along the same axis with the distance $d$ between them. If a parallel beam of light falling on A leaves B as a parallel beam, then distance $d$ in cm will be:
(a) 25
(b) 15
(c) 30
(d) 50

Your Thought Process:

A parallel beam incident on lens A will converge at its focal point.
A parallel beam emerging from lens B must have originated from its focal point.
Let's trace the light.

Solution Breakdown:

Lens A (Convex, $f_{A} = + 20$ cm): A parallel beam of light enters lens A. By definition, a convex lens focuses a parallel beam at its second focal point. So, the first image ( $I_{1}$ ) would be formed at a distance $v_{1} = + 20$ cm from lens A.
Lens B (Concave, $f_{B} = - 5$ cm): The final beam emerging from lens B is also parallel. For a diverging lens to produce a parallel beam, the incident rays must be directed towards its focal point on the other side. This means the object for lens B must be a virtual object located at its second focal point. Therefore, $u_{2} = + 5$ cm (it's positive because the object is on the right side of the lens, but it's a virtual object for lens B).
Finding the distance d: The image from lens A ( $I_{1}$ ) must coincide with the focal point of lens B that acts as the object. The distance between the lenses, $d$ , is the distance from lens A to this common point, minus the distance from lens B to this point.
Wait, let's re-think. The image from Lens A is the object for Lens B. So the position of $I_{1}$ and the object position for B must be the same point in space.
- Lens A forms an image at 20 cm from it.
- Lens B needs its object to be at 5 cm from it (to produce a parallel beam).
- Therefore, the distance between the lenses is the distance from A to the focal point, minus the distance from B to that same point.
  Let's visualize:
  [Lens A] ----- $d$ ----- [Lens B]
  The parallel beam hits A and converges towards a point P, which is 20 cm from A.
  This point P must also be the focal point for B, so it is 5 cm from B.
  Therefore, the total distance from A to P is 20 cm. This distance is made up of the distance between the lenses ( $d$ ) and the distance from B to P (5 cm).
  So, $d + 5 cm = 20 cm$ .
  This gives $d = 15$ cm.

Answer: (b) 15. You can verify this in the provided resource LINK, section "2021", Q2.

Problem 2: The Liquid Lens (NEET 2019)

Question: Two similar thin equi-convex lenses, of focal length $f$ each, are kept coaxially in contact with each other such that the focal length of the combination is $F_{1}$ . When the space between the two lenses is filled with glycerine (which has the same refractive index $μ = 1.5$ as that of glass) then the equivalent focal length is $F_{2}$ . The ratio $F_{1} : F_{2}$ will be:
(a) 2 : 1
(b) 1 : 2
(c) 2 : 3
(d) 3 : 4

Your Thought Process:

This is a system of lenses in contact. The first case ( $F_{1}$ ) is simple.
The second case ( $F_{2}$ ) involves three lenses: two glass convex lenses and one glycerine lens in the middle. I need to find the focal length of this glycerine lens.

Solution Breakdown:

Case 1 (Focal Length $F_{1}$ ): Two convex lenses of focal length $f$ are in contact.
The equivalent power is $P_{e q} = P_{1} + P_{2} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}$ .
So, the equivalent focal length $F_{1} = \frac{1}{P _{e q}} = \frac{f}{2}$ .
Case 2 (Focal Length $F_{2}$ ): We now have a system of three lenses in contact: Lens1 (convex, glass), Lens2 (concave, glycerine), Lens3 (convex, glass).
- The two glass lenses are equi-convex, so they each have two surfaces with radius of curvature $R$ . Let's assume light comes from the left. For the first lens, the surfaces are $+ R$ and $- R$ .
- The glycerine fills the space between them. This forms a plano-concave lens with radii of curvature $- R$ and $+ R$ .
- Let's find the focal length of the glycerine lens, $f_{g l y}$ , using the Lens Maker's formula:
  $\frac{1}{f _{g l y}} = (μ_{g l y} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
  The first surface has radius $- R$ and the second has radius $+ R$ .
  $\frac{1}{f _{g l y}} = (1.5 - 1) (\frac{1}{- R} - \frac{1}{+ R}) = (0.5) (\frac{- 2}{R}) = - \frac{1}{R}$
- Now let's find the focal length of one of the glass lenses, $f$ :
  $\frac{1}{f} = (μ_{g l a ss} - 1) (\frac{1}{R} - \frac{1}{- R}) = (1.5 - 1) (\frac{2}{R}) = (0.5) (\frac{2}{R}) = \frac{1}{R}$
  So, we find that $f = R$ .
- This means $\frac{1}{f _{g l y}} = - \frac{1}{R} = - \frac{1}{f}$ , so $f_{g l y} = - f$ . The glycerine lens is a diverging lens with the same magnitude of focal length as the convex lenses.
- Now, find the equivalent focal length $F_{2}$ of the three-lens combination:
  $\frac{1}{F _{2}} = \frac{1}{f _{g l a ss}} + \frac{1}{f _{g l y}} + \frac{1}{f _{g l a ss}} = \frac{1}{f} + \frac{1}{- f} + \frac{1}{f} = \frac{1}{f}$
  So, $F_{2} = f$ .
The Ratio $F_{1} : F_{2}$ :
$F_{1} : F_{2} = \frac{f}{2} : f = 1 : 2$

Answer: (b) 1 : 2. This is a classic problem type. You can find similar ones in the resource LINK for years 2016 (Q5) and 2015 (Q3).

Conclusion

Today, we've moved from theory to rigorous application, tackling a range of NEET-level problems on compound lens systems. You've seen how the same fundamental principles can solve problems involving different lens types, virtual objects, and even complex combinations with liquid lenses.

Key Takeaways:

Systematic Approach is Key: Always break down a multi-lens problem into sequential single-lens problems.
Sign Conventions are Non-Negotiable: Master the sign conventions for focal lengths (positive for converging, negative for diverging) and distances (especially for virtual objects).
Recognize Problem Patterns: Problems like the "liquid lens" are common. Recognizing the setup as a combination of three lenses in contact is the crucial first step.
Conceptual Understanding: Questions like the "collimating system" test your core understanding of what focal points represent. Don't just rely on formulas.

Mastering these problems comes down to practice. I encourage you to work through the other problems related to lens combinations in the NEET Previous Year Questions resource.

Preview of the Next Lesson:
We have now fully developed the tools to analyze lens systems. In our next lesson, we will begin our exploration of optical instruments by applying these tools to the simple microscope, analyzing its working principle and calculating its angular magnification.