Q.An angular magnification (magnifying power) of $30\text{X}$ is desired using an objective of focal length $1.25\ \text{cm}$ and an eyepiece of focal length $5\ \text{cm}$. How will you set up the compound microscope?

[!TLDR]

From $f_o+f_e=27$ and $f_o/f_e=8$, the eyepiece focal length is 3 cm — option (B).

Concept

For an astronomical telescope in normal adjustment, the objective and eyepiece are separated by $L=f_o+f_e$, and the magnifying power is $m=f_o/f_e$ (CBSE/NCERT Class 12 Ray Optics).

Solution

Given $L=f_o+f_e=27$ cm and $m=\dfrac{f_o}{f_e}=8$, so $f_o=8f_e$. Substituting:

$8f_e+f_e=27\ \Rightarrow\ 9f_e=27\ \Rightarrow\ f_e=3\ \text{cm}.$ …

Step 1 — Eyepiece (final image at $D=25$ cm).

The final virtual image is on the same side as the object, so $v_e=-25\text{ cm}$, $f_e=6.25\text{ cm}$.

So the intermediate image sits $5\text{ cm}$ in front of the eyepiece.

Step 2 — Locate the intermediate image relative to the objective.

The lenses are $20\text{ cm}$ apart, so the objective's real image is at

Step 3 — Objective (find the object distance). …

The key idea is that in normal adjustment, the telescope is set so that the final image is at infinity. This means the eyepiece is positioned so that the intermediate image formed by the objective lies exactly at the focal point of the eyepiece. Consequently, the distance between the objective and the eyepiece (the tube length) is simply the sum of their focal lengths:

The magnifying power $M$ of an astronomical telescope in normal adjustment is defined as the ratio of the angle subtended by the image to the angle subtended by the object, and it simplifies to

We are given $L = 36\ \text{cm}$ and $M = 8$. So we have two equations in two unknowns — a straightforward system.

1. Set up the equations From the magnifying power:

From the tube length:

2. Substitute and solve Replace $f_o$ in the second equation:

Then

3. Check against the options The pair $(f_o = 32\ \text{cm}, f_e = 4\ \text{cm})$ matches option (C). …

The telescope is used on a near object: the object sits $300\,\text{cm}$ from the objective, and the final image is formed at the least distance of distinct vision, $D=25\,\text{cm}$.

Objective ($f_o=60\,\text{cm}$, $u_1=-300\,\text{cm}$):

Eyepiece ($f_e=5\,\text{cm}$, final image $v_2=-25\,\text{cm}$ at the near point): …

Concept & Intuition

An astronomical telescope in normal adjustment means the final image is formed at infinity — the eyepiece is set so that the light rays emerging from it are parallel. This happens when the image formed by the objective lies exactly at the first focal point of the eyepiece. Consequently, the distance between the two lenses (the tube length) is simply the sum of their focal lengths:

The magnifying power (angular magnification) in this setting is given by the ratio of the objective’s focal length to the eyepiece’s focal length:

We are given $L = 32\ \text{cm}$ and $M = 7$. That gives us two equations in two unknowns — straightforward algebra.

Step-by-step solution

1. Write the two conditions From normal adjustment:

From magnifying power:

2. Express one variable in terms of the other From (2):

3. Substitute into the length equation $$$$

Concept: simple microscope (magnifying glass). When the final image is formed at the least distance of distinct vision D (image at the near point), the magnifying power is

M = 1 + D/f.

Given D = 25 cm, f = 10 cm:

M = 1 + 25/10 = 1 + 2.5 = 3.5. …

1. Set up the geometry

The lens sits between the object and the screen, and it forms a real image on the screen. Working with numerical (unsigned) distances:

• object distance $= u$
• image distance $= v$
• they must add up to the object–screen separation:

2. Use the magnification

For a thin lens the numerical magnification of a real image is

3. Solve for $u$ and $v$

Substituting (2) into (1):

4. Apply the lens formula

In magnitudes, the thin-lens relation for a real object and real image is

Now substitute, noting the denominator $u+v$ is just $x$:

5. Sanity checks …

Concept: astronomical telescope in normal adjustment (parallel rays / relaxed eye).

Magnifying power m = f_o/f_e = 9

Tube length L = f_o + f_e = 20 cm

Substituting f_o = 9 f_e: …

Step 1 — Read the geometry

From the figure: light converges towards $P$, and the concave lens is inserted at the dotted line through $O$, with $OP = 12\ \text{cm}$ measured along the axis, on the far (outgoing) side of the lens.

Step 2 — Identify the virtual object

If the lens were absent, the rays would meet at $P$. Because the rays are already converging when they strike the lens, they never actually diverge from a real object point. The point $P$ — where they would have met — acts as a virtual object.

In the Cartesian sign convention (light travelling left → right, distances measured from the optical centre, rightward positive):

The positive $u$ is the whole trick of the question — a real object would give a negative $u$.

Step 3 — Apply the lens formula

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ …