Q.(a) Draw a labelled ray diagram of compound microscope, when final image forms at the least distance of distinct vision.

The question gives you a telescope with an objective of focal length $f_o = 144$ cm and an eyepiece of focal length $f_e = 6.0$ cm. You need the magnifying power and the length of the telescope tube.

The key idea is that a telescope is used to view distant objects. Light from a faraway object arrives as nearly parallel rays. The objective lens forms a real, inverted image at its focal plane. The eyepiece then acts as a magnifier to view that image.

For relaxed viewing (normal adjustment), the eyepiece is adjusted so that the final image is at infinity. This means the image formed by the objective must lie exactly at the focal point of the eyepiece. So the distance between the two lenses — the tube length — is simply the sum of their focal lengths.

The magnifying power (or angular magnification) is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended by the object when viewed directly. For a telescope in normal adjustment, this ratio simplifies beautifully to the ratio of the focal lengths.

Let's apply this directly.

1. Magnifying power:

2. Tube length:

So the magnifying power is 24 and the tube length is 150 cm. …

Why focal lengths matter for magnification

A compound microscope uses two converging lenses in tandem. The objective forms a real, magnified image of the tiny object, and the eyepiece acts as a magnifying glass to view that intermediate image. The total magnification is the product of the magnifications produced by each lens, so to maximize it we need to understand how each focal length enters the formula.

The objective's magnification depends on how far the real image forms relative to the object distance. For a small object placed just beyond the focal point of the objective, the image distance $v_o$ is much larger than the object distance $u_o$, giving magnification $m_o = \frac{v_o}{u_o}$. In the standard microscope setup, the object is very close to the focus, so $u_o \approx f_o$, and the image forms near the other end of the tube at distance $L$ (the tube length). This gives:

The eyepiece magnifies the intermediate image like a simple magnifying glass. When the final image is at the near point $D$ (typically 25 cm), the angular magnification is:

For large $D$ compared to $f_e$, this simplifies to $m_e \approx \frac{D}{f_e}$.

Step-by-step analysis

1. Total magnification is inversely proportional to both focal lengths. From the formula above, $M \propto \frac{1}{f_o f_e}$. To maximize $M$, we need both $f_o$ and $f_e$ to be as small as possible. This immediately rules out options (C) and (D), which suggest large focal lengths.

2. Compare the two focal lengths. Now we must decide between options (A) and (B): should $f_e > f_o$, or $f_o > f_e$?

3. Physical constraints of microscope design. In a practical compound microscope, the objective must form a real image at a reasonable distance (the tube length $L$, typically 15–20 cm). If $f_o$ were too large, the object would need to be placed far from the lens, defeating the purpose of examining a small specimen up close. The objective typically has $f_o$ in the range of a few millimetres to about 2 cm. …

A compound microscope uses two converging lenses: an objective lens with a short focal length and a short aperture, and an eyepiece lens with a larger focal length and aperture. The objective forms a real, inverted, and magnified intermediate image. This intermediate image then acts as the object for the eyepiece, which functions like a simple magnifier to produce the final, highly magnified virtual image.

For the most suitable viewing by a relaxed eye, the final image formed by the eyepiece is at infinity. This condition is met when the intermediate image formed by the objective lens falls exactly at the first focal point of the eyepiece.

The total magnification ($M$) of a compound microscope is the product of the linear magnification produced by the objective ($M_o$) and the angular magnification produced by the eyepiece ($M_e$):

$M = M_o \times M_e$

Let's break down the calculation for each part.

1. Identify Given Parameters and Standard Values:

   • Focal length of the objective, $f_o = 1 \text{ cm}$.
   • Focal length of the eyepiece, $f_e = 2 \text{ cm}$.
   • Tube length of the microscope, $L = 10 \text{ cm}$.
   • Least distance of distinct vision (for a relaxed eye, the final image is at infinity, but the eyepiece magnification is still referenced to $D$), $D = 25 \text{ cm}$.

2. Understand the Definition of Tube Length ($L$) for the Formula:

   The term "tube length" can sometimes be ambiguous. In the context of the standard formula for compound microscope magnification, $M = \left(\frac{L}{f_o}\right) \left(\frac{D}{f_e}\right)$, the tube length $L$ is specifically defined as the distance between the second focal point of the objective ($F_o'$) and the first focal point of the eyepiece ($F_e$). This is crucial for the formula to hold true.

   Watch out

   If "tube length" were interpreted as the physical distance between the objective and eyepiece lenses, the calculation would be different and would not lead to any of the given options. For competitive exams, when the formula $M = (L/f_o)(D/f_e)$ is implied by the options, assume $L$ refers to the distance between the internal focal points.

3. Calculate the Magnification of the Objective ($M_o$):

   For a relaxed eye, the intermediate image ($I_1$) formed by the objective must be at the first focal point of the eyepiece ($F_e$).

   Given our definition of $L$ (distance between $F_o'$ and $F_e$), the image $I_1$ is formed at a distance $v_o = f_o + L$ from the objective lens.

   The object for the objective ($O$) is placed just outside its first focal point ($F_o$).

   Using the lens formula for the objective:

   $\frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o}$

   $\frac{1}{u_o} = \frac{1}{v_o} - \frac{1}{f_o} = \frac{1}{f_o + L} - \frac{1}{f_o} = \frac{f_o - (f_o + L)}{f_o(f_o + L)} = \frac{-L}{f_o(f_o + L)}$ …

The sign of magnifying power in optical instruments tells us about the orientation of the final image relative to the object. A negative magnifying power means the image is inverted; a positive one means it is erect. This is a convention used consistently in ray optics.

For a compound microscope, the objective lens forms a real, inverted, and magnified image of the object. This intermediate image then acts as the object for the eyepiece, which works as a simple magnifier. The eyepiece produces a virtual, magnified image that is erect with respect to the intermediate image. But because the intermediate image itself was already inverted, the final image ends up inverted relative to the original object.

Let’s trace the orientation step by step.

1. The objective lens inverts the object.

   The objective has a short focal length and forms a real image. For a real image formed by a convex lens, the image is always inverted with respect to the object. So the intermediate image is inverted.

2. The eyepiece does not re-invert.

   The eyepiece is used as a simple magnifier — the object (the intermediate image) is placed just inside its focal point. A simple magnifier always produces a virtual image that is erect relative to the object placed before it. So the eyepiece takes the inverted intermediate image and produces a final image that is erect with respect to that intermediate image.

3. Net orientation: inverted.

   If the intermediate image is inverted, and the eyepiece keeps that orientation (erect relative to the intermediate image), the final image remains inverted relative to the original object.

   Mathematically, the magnifying power $M$ of a compound microscope is given by:

   $M = -\frac{L}{f_o} \cdot \frac{D}{f_e}$ …