Q.(a) Determine the 'effective focal length' of the combination of the two lenses in Exercise 9.10, if they are placed $8.0\ \text{cm}$ apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?

(a) Effective focal length — trace the parallel beam through each lens ($f_1=+30\ \text{cm}$ convex, $f_2=-20\ \text{cm}$ concave, separation $d=8\ \text{cm}$).

Convex side first: parallel rays converge at the convex focus $30\ \text{cm}$ away, i.e. $30-8=22\ \text{cm}$ past the concave lens (virtual object, $u=+22\ \text{cm}$):

Concave side first: rays diverge from the concave focus $20\ \text{cm}$ away, giving a real object for the convex lens at $u=-(20+8)=-28\ \text{cm}$:

The two results ($220\ \text{cm}$ vs $420\ \text{cm}$) differ, so the effective focal length depends on the side of incidence and is not a useful single number for a separated pair. (The combination formula $\tfrac{1}{F}=\tfrac{1}{f_1}+\tfrac{1}{f_2}-\tfrac{d}{f_1 f_2}$ gives $F=-300\ \text{cm}$, but only measured from shifting principal planes.) …

(a) Effective focal length

The lenses of Exercise 9.10 are $f_1=+30\ \text{cm}$ (convex) and $f_2=-20\ \text{cm}$ (concave), now $d=8.0\ \text{cm}$ apart. A single equivalent focal length only describes a pair faithfully when the lenses are in contact; with a gap we trace the beam lens by lens.

Light on the convex lens first. Parallel rays head for the convex focus, $30\ \text{cm}$ to its right. That point is $30-8=22\ \text{cm}$ beyond the concave lens and acts as a virtual object for it ($u=+22\ \text{cm}$):

The emergent beam diverges as if from a point $220\ \text{cm}$ to the left of the concave lens.

Light on the concave lens first. Parallel rays diverge as if from the concave focus, $20\ \text{cm}$ to its left — a real object for the convex lens $8\ \text{cm}$ away, $u=-(20+8)=-28\ \text{cm}$:

Now the beam appears to come from $420\ \text{cm}$.

Conclusion. The two answers ($220\ \text{cm}$ and $420\ \text{cm}$) are different, so the result depends on the side of incidence; the pair cannot be replaced by one thin lens and the notion of a single effective focal length is not useful here. (The algebraic combination $\tfrac{1}{F}=\tfrac{1}{f_1}+\tfrac{1}{f_2}-\tfrac{d}{f_1 f_2}=-\tfrac{1}{300}$ gives $F=-300\ \text{cm}$, but this is referred to principal planes that themselves shift with the side of incidence.)

(b) Magnification and image size

Object height $h_o=1.5\ \text{cm}$, placed $40\ \text{cm}$ before the convex lens. …

Method: Sequential Image Formation (Lens-by-Lens Analysis)

This is the standard method for compound lens systems — treat each lens separately, using the image from the first lens as the object for the second lens.

Steps

Step 1: Identify the given data

From Exercise 9.10 (standard NCERT reference):

• Convex lens: $f_1 = +30\ \text{cm}$
• Concave lens: $f_2 = -20\ \text{cm}$
• Separation between lenses: $d = 8.0\ \text{cm}$

Step 2: For part (a) — Effective focal length

• Use the formula for the effective focal length $F$ of two thin lenses separated by distance $d$:

• Substitute:

• Therefore:

Step 3: Does the answer depend on which side the light is incident?

• Yes, the effective focal length formula assumes a specific order of lenses.
• If parallel light is incident from the convex side first, the calculation above holds.
• If incident from the concave side first, the roles of $f_1$ and $f_2$ swap, giving a different $F$.
• Conclusion: The notion of effective focal length is not very useful here because the system is not a simple equivalent lens — the image position depends on which lens the light hits first.

Step 4: For part (b) — Magnification and image size

• Object distance from convex lens: $u_1 = -40\ \text{cm}$ (sign convention: object to left of lens)
• Lens 1 (convex): $f_1 = +30\ \text{cm}$

Using lens formula:

• Magnification by lens 1:

Step 5: Image from lens 1 becomes object for lens 2

• Distance between lenses = $8.0\ \text{cm}$
• So, $v_1 = 120\ \text{cm}$ from lens 1 means it is $120 - 8 = 112\ \text{cm}$ to the right of lens 2.
• For lens 2 (concave), object distance: $u_2 = +112\ \text{cm}$ (object is on the right side — virtual object for lens 2)

Step 6: Lens 2 (concave): $f_2 = -20\ \text{cm}$ …

Common Mistakes & How to Avoid Them

Mistake 1: Treating the two-lens system as a single thin lens with a simple formula

The error: Students try to use the formula $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$ directly, ignoring the separation between lenses.

Why it's wrong: That formula works only for lenses in contact (separation $d = 0$). Here $d = 8.0\ \text{cm}$, so you must use the lens combination formula:

How to avoid: Always check if $d = 0$ before using the simple formula. If $d \neq 0$, use the full formula above.

Mistake 2: Forgetting that effective focal length depends on the side of incidence

The error: Students assume $F$ is the same regardless of which side the light enters from.

Why it's wrong: The effective focal length is different for light entering from the left vs. the right when $d \neq 0$. The formula above gives $F$ for light incident from the left (first lens = $f_1$). For light from the right, swap $f_1$ and $f_2$ in the formula.

How to avoid:

• For part (a), compute $F$ for both directions explicitly.
• The answer does depend on which side the light is incident — state this clearly.

Mistake 3: Thinking the "effective focal length" concept is always useful

The error: Students assume that once $F$ is found, it can be used like a single lens for any object position.

Why it's wrong: The effective focal length is only meaningful for parallel incident light (object at infinity). For a finite object distance, you cannot use $F$ directly — you must trace the image through each lens step-by-step.

How to avoid: For part (b), do not use $F$. Instead:

1. Find the image from the first lens using $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$
2. Use that image as the object for the second lens (accounting for separation $d$)
3. Find the final image position and magnification

Mistake 4: Sign convention errors in the two-lens calculation

The error: Students forget to adjust the object distance for the second lens properly.

Why it's wrong: If the first image forms at distance $v_1$ from lens 1, and the lenses are $d$ apart, then the object distance for lens 2 is:

(using the Cartesian sign convention consistently)

How to avoid: Draw a clear ray diagram. Label distances from each lens. Always check:

• Is $u_2$ positive or negative?
• Is the object for lens 2 real or virtual?

Mistake 5: Confusing total magnification with individual magnifications

The error: Students add magnifications or forget to multiply them. …