Two thin convex lenses of focal lengths f1 and f2 are separated by a horizontal distance d (where d

Q: Why is the object for the second lens considered virtual?

Because the rays from the first lens are converging towards a point $F_1$ that lies *behind* the second lens. They haven't actually met yet when they hit the second lens.

Q: Why is the object distance $u$ positive?

In our sign convention, the direction of incident light is positive. Since the virtual object $F_1$ is to the right of the second lens (the direction light is traveling), the distance is positive.

Q: Why is the object height for the second lens $-\Delta$?

The principal axis of the second lens is at $y = \Delta$. The object $F_1$ is on the x-axis ($y = 0$). Relative to the second lens's axis, the object is $\Delta$ distance *below* it, hence $-\Delta$.

Q: How do we get the final y-coordinate from the image height?

The image height $y_{img}$ is measured relative to the principal axis of the second lens. Since that axis is already at a height $\Delta$ from the origin, the absolute y-coordinate is $\Delta + y_{img}$.

Q: What does the condition $d < f_1$ imply?

It ensures that the focal point of the first lens falls *after* the second lens, making it a virtual object. If $d > f_1$, the rays would converge before the second lens, creating a real object for it.

Question

Two thin convex lenses of focal lengths $f_1$ and $f_2$ are separated by a horizontal distance $d$ (where $d < f_1$, $d < f_2$) and their centres are displaced by a vertical separation $\Delta$ as shown in the figure. Taking the origin of coordinates, $O$, at the centre of the first lens, the $x$ and $y$-coordinates of the focal point of this lens system, for a parallel beam of rays coming from the left, are given by

Options:

(A) $x = \frac{f_1 f_2}{f_1 + f_2}, y = \Delta$

(B) $x = \frac{f_1(f_2 + d)}{f_1 + f_2 - d}, y = \frac{\Delta}{f_1 + f_2}$

(C) $x = \frac{f_1 f_2 + d(f_1 - d)}{f_1 + f_2 - d}, y = \frac{\Delta(f_1 - d)}{f_1 + f_2 - d}$

(D) $x = \frac{f_1 f_2 + d(f_1 - d)}{f_1 + f_2 - d}, y = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a parallel beam of light traveling from the left and striking the first convex lens, $L_1$.

We know from the fundamental principles of optics that a convex lens converges parallel rays to its focal point.

Therefore, the first lens will focus these rays at a point $F_1$, which is located at a distance $f_1$ from the optical center of $L_1$. Since $L_1$ is at the origin $(0,0)$, the coordinates of this intermediate focal point $F_1$ are simply $(f_1, 0)$.

### The Virtual Object

Before these converging rays can actually meet at $F_1$, they are intercepted by the second lens, $L_2$.

This is where the concept of a **virtual object** comes into play. The point $F_1$ acts as a virtual object for $L_2$.

We are given that $L_2$ is placed at a horizontal distance $d$ from $L_1$. Therefore, the distance of this virtual object from $L_2$ is $(f_1 - d)$.

According to our sign convention, since the incident light travels to the right and the virtual object is also to the right of $L_2$, the object distance $u$ is positive.

$$
u = +(f_1 - d)
$$

### Calculating the x-coordinate

Now, we apply the thin lens formula for the second lens to find the final image distance $v$.

$$
\frac{1}{v} - \frac{1}{u} = \frac{1}{f_2}
$$

Substituting our value for $u$, we get:

$$
\frac{1}{v} - \frac{1}{f_1 - d} = \frac{1}{f_2}
$$

Rearranging to solve for $v$:

$$
\frac{1}{v} = \frac{1}{f_2} + \frac{1}{f_1 - d} = \frac{f_1 - d + f_2}{f_2(f_1 - d)}
$$

Inverting this expression gives us the image distance from $L_2$:

$$
v = \frac{f_2(f_1 - d)}{f_1 + f_2 - d}
$$

To find the absolute $x$-coordinate from the origin $O$, we must add the position of $L_2$ (which is $d$) to the image distance $v$.

$$
x = d + v = d + \frac{f_2(f_1 - d)}{f_1 + f_2 - d}
$$

Taking the common denominator and simplifying:

$$
x = \frac{d(f_1 + f_2 - d) + f_2(f_1 - d)}{f_1 + f_2 - d} = \frac{f_1 f_2 + d(f_1 - d)}{f_1 + f_2 - d}
$$

### The Masterstroke: The y-coordinate

This is the part where many students make a mistake. We must carefully consider the reference frames.

The second lens $L_2$ is shifted vertically by a distance $\Delta$. This means its principal axis is the line $y = \Delta$.

Our virtual object $F_1$ lies on the $x$-axis, which is $y = 0$. Therefore, relative to the principal axis of $L_2$, the object is located at a depth of $\Delta$.

This gives us our object height, $y_{obj} = -\Delta$.

To find the final image height, we need the **transverse magnification** $m$ of the second lens.

$$
m = \frac{v}{u} = \frac{\frac{f_2(f_1 - d)}{f_1 + f_2 - d}}{f_1 - d} = \frac{f_2}{f_1 + f_2 - d}
$$

The image height relative to the principal axis of $L_2$ is simply the magnification times the object height.

$$
y_{img} = m \cdot y_{obj} = -m\Delta
$$

Finally, to find the absolute $y$-coordinate from the origin, we must add this relative image height to the height of $L_2$'s principal axis.

$$
y = \Delta + y_{img} = \Delta - m\Delta = \Delta(1 - m)
$$

Substituting our expression for $m$:

$$
y = \Delta\left(1 - \frac{f_2}{f_1 + f_2 - d}\right) = \Delta\left(\frac{f_1 + f_2 - d - f_2}{f_1 + f_2 - d}\right)
$$

$$
y = \frac{\Delta(f_1 - d)}{f_1 + f_2 - d}
$$

And there we have it! The complete coordinates of the final focal point.

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Visualized Solution

$Action of the First Lens$

$Virtual Object for L_{2}$

$Lens Formula for L_{2}$

$Image Distance v$

$Final x -coordinate$

$Transverse Magnification$

$Magnification of L_{2}$

$Final y -coordinate$

The Sigma Insight: Lens

Analyzing the Setup

The Virtual Object

Calculating the x-coordinate

The Masterstroke

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Action of the First Lens

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Virtual Object for L2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lens Formula for L2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Image Distance v

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final x-coordinate

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Transverse Magnification

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnification of L2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final y-coordinate