Q.What should be the distance between the object in Exercise 9.23 and the magnifying glass if the virtual image of each square in the figure is to have an area of $6.25\ \text{mm}^2$. Would you be able to see the squares distinctly with your eyes very close to the magnifier?
[Note: Exercises 9.22 to 9.24 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]

Set-up. A convex lens forms a virtual image when the object is inside the focus ($|u| < f$), and a real image when it is beyond the focus. With the Cartesian convention:

• Position 1 (virtual): $u_1 = -12\ \text{cm}$
• Position 2 (real): $u_2 = -(12+18) = -30\ \text{cm}$

Magnification. From $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ and $m=\dfrac{v}{u}$:

For the virtual image ($f > 12$, so this is positive):

For the real image ($f < 30$, so $f-30<0$, giving a negative value):

Apply the condition $|m_1| = 5\,|m_2|$: …

Light passes object $\to$ lens A ($f_A = 20\ \mathrm{cm}$) $\to$ lens B ($f_B = 30\ \mathrm{cm}$). Let the lens separation be $d$. The object is $45\ \mathrm{cm}$ from lens B, and lens A lies $d$ from lens B, so the object distance for A is

$u_A = 45 - d.$ …

Concept

In a Cassegrain telescope, parallel rays from an object at infinity first reflect off the large concave primary mirror and converge toward its focus. The small convex secondary mirror, placed short of that focus, intercepts the converging rays, so the primary's image acts as a virtual object for the secondary.

Solution

Focal lengths. Concave primary: $R_1 = 25\text{ cm}\Rightarrow f_1 = 12.5\text{ cm}$. Convex secondary: $R_2 = 16\text{ cm}\Rightarrow f_2 = 8\text{ cm}$.

Image from the primary. For an object at infinity the primary converges the rays to its focus, $12.5\text{ cm}$ in front of it.

Virtual object for the secondary. The mirrors are separated by $d = 2.5\text{ cm}$, so the converging rays would meet at

behind the secondary — a virtual object at $10\text{ cm}$. …

Concept

In a Cassegrain telescope the concave primary mirror converges the parallel rays from a distant object toward its focus. Before the rays meet, the small convex secondary mirror intercepts them, so the converging beam acts as a virtual object for the secondary.

Primary (concave) mirror

For an object at infinity, the image would form $12.5\text{ cm}$ from the primary.

Secondary (convex) mirror

The mirrors are $2.5\text{ cm}$ apart, so this converging point lies

behind the secondary — a virtual object at $u = +10\text{ cm}$. The convex mirror has …

The problem describes a classic experimental setup: an object and a screen are fixed at a separation $D$, and a convex lens is moved between them. At two distinct positions of the lens, a sharp image is formed on the screen. This is the displacement method (also called the conjugate foci method). The key insight is that for a fixed object-screen distance $D > 4f$, there are two lens positions that satisfy the lens formula — one giving a magnified image, the other a diminished image. The distance between these two positions is $d$.

Why does this happen? For a given $D$, if the object distance is $u$, the image distance is $D - u$. The lens formula $\frac{1}{f} = \frac{1}{u} + \frac{1}{D-u}$ becomes a quadratic in $u$, yielding two real solutions when $D > 4f$. The difference between these two $u$ values is exactly $d$.

The standard result is:

Let’s apply it step by step.

1. Identify the given distances.

   The object and screen are $1\ \text{m}$ apart, so $D = 100\ \text{cm}$.

   The distance between the two lens positions is $d = 20\ \text{cm}$.

2. Plug into the formula.

3. Simplify the numerator. …

The key idea is that a glass slab does not converge or diverge light — it simply displaces the beam sideways without changing its overall direction. For a combination of two lenses to mimic this, the system must have zero net power and also produce no net deviation for a parallel incident beam. That happens when the second lens exactly cancels the convergence of the first, which requires the lenses to be placed so that the image formed by the first lens lies at the focal point of the second.

Let’s work through the reasoning step by step.

1. Understand what “act as a glass slab” means. A glass slab (a plane-parallel plate) does not change the direction of a parallel beam; it only shifts it laterally. In lens terms, this means the combination must have zero effective focal length (infinite focal length) — i.e., the system’s power $P = 1/F = 0$. For two thin lenses separated by distance $d$, the effective focal length $F$ is given by:

Setting $1/F = 0$ gives:

Multiply through by $f_1 f_2$:

So $d = f_1 + f_2$. This is the condition for zero net power.

2. Check the physical meaning: ray tracing.

   Imagine a parallel beam entering the first lens. It converges toward its focal point $F_1$ (at distance $f_1$ behind lens 1). If the second lens is placed exactly so that this converging point lies at its own focal point $F_2$ (on the object side), then rays entering the second lens appear to come from its focal point, and thus emerge parallel again. The condition for this is that the distance between the lenses equals $f_1 + f_2$ — the first lens’s focal length plus the second lens’s focal length. The beam is then undeviated overall, just shifted sideways — exactly like a glass slab.

3. Why the other options fail. …

The core idea here is that inserting a glass plate between the lens and the screen changes the effective optical path. The screen is fixed, but the plate makes the image appear to be at a different location from the lens’s perspective. You need to find where the lens should form the image so that, after the plate shifts it, the real image lands exactly on the screen.

Why does a plate shift the image? When light passes through a parallel-sided slab, the rays emerge parallel to their original direction but displaced sideways. For a thin slab placed close to the screen, the effect is that the image appears to move toward the lens by an amount equal to the slab’s thickness times $(1 - 1/\mu)$. This is a standard result — the apparent shift in the direction of the incident light is $t\left(1 - \frac{1}{\mu}\right)$.

Let’s work it through.

1. First situation (no plate): Object distance $u = -20$ cm (sign convention: distances measured from lens, object on left is negative). Image distance $v = +5$ cm (real image on right). Using the lens formula:

So $f = 4$ cm. That’s the focal length of the lens.

2. Second situation (plate inserted): The plate of thickness $t = 1.4$ cm and refractive index $\mu = 7/5$ is placed between the lens and the screen. The screen is still 5 cm from the lens. The plate shifts the image away from the screen (i.e., toward the lens) by an apparent distance. For a real image formed by the lens, the rays converge to a point; after passing through the plate, the convergence point shifts by:

This shift is toward the lens — meaning the lens must now form the image 0.4 cm farther from the screen (i.e., at $v' = 5 + 0.4 = 5.4$ cm from the lens) so that after the plate, the real image lands exactly on the screen.

3. Find the new object distance: Using the lens formula again with $f = 4$ cm and $v' = +5.4$ cm:

Compute:

So $u' = -\frac{108}{7} \approx -15.43$ cm. That’s the object distance from the lens — but wait, that’s not among the options. Something’s off.

where $v = 5$ cm is the screen distance. So $v' = 5 + 0.4 = 5.4$ cm, as above. That gives $u' \approx -15.4$ cm, which is not in the options. So perhaps the plate is placed between the object and the lens? Let’s re-read the problem.

The problem says: “A glass plate … is inserted between the lens and the screen.” That’s clear. But the answer options are around 25–30 cm, much larger than 15.4 cm. So maybe the plate shifts the image away from the lens? Let’s check the shift direction again. …