Why a lens magnifies
What the eye actually perceives as "size" is the angle an object subtends at the eye, not its true linear dimension. A distant object looks small because it subtends a small angle; bring it closer and the angle grows. The closest distance at which a normal eye can see an object clearly and without strain is the near point, taken as $D \approx 25\,\text{cm}$ (also called the least distance of distinct vision).
An object cannot be brought nearer than $D$ for clear viewing with the unaided eye. A converging lens removes this limit: it lets the object sit much closer than $25\,\text{cm}$ while still forming an image the relaxed eye can focus. The lens therefore increases the angle subtended, and that angular increase is what we call magnifying power.
The maximum angle a height $h$ can subtend at the unaided eye is $\theta_o \approx h/D$, reached when the object sits at the near point.
The simple microscope
A simple microscope, or simple magnifier, is a single converging lens of small focal length used to view small nearby objects. The lens is held near the object — one focal length away or less — with the eye positioned close to the lens on the other side. The object is placed within the focal length, so the image formed is virtual, erect and magnified, on the same side as the object.
Two image positions are standard. If the object is placed exactly at the focus ($u = -f$), the image forms at infinity, which the relaxed eye views most comfortably. If the object is moved slightly inside the focus, the image is virtual and closer than infinity; pushed to the most useful limit, the image sits at the near point $D$, giving the largest magnification but with some strain on the eye. NIOS calls the infinity case normal adjustment.
The object lies inside the focus; emerging rays diverge, and their backward extensions (dashed) locate an enlarged, erect, virtual image on the object side.
Magnifying power of a simple microscope
Image at the near point. The linear magnification is $m = v/u = v\left(\dfrac{1}{f}-\dfrac{1}{v}\right)\!=\!1-\dfrac{v}{f}$ from the lens equation. By the sign convention, $v$ is negative and equal in magnitude to $D$, so the magnification becomes
$$m = 1 + \frac{D}{f}$$
With $D \approx 25\,\text{cm}$, a magnification of six requires $f = 5\,\text{cm}$. This $m$ is also the ratio of the angle subtended by the image to that subtended by the object when placed at $D$ for comfortable viewing.
Image at infinity (normal adjustment). Here we compare angles directly. Without the lens, the object at the near point subtends $\theta_o \approx h/D$. With the object at the focus ($u = -f$), the image is at infinity and the angle subtended is $\theta_i \approx h/f$. The angular magnification is therefore
$$m = \frac{\theta_i}{\theta_o} = \frac{D}{f}$$
This is exactly one less than the near-point value. The viewing is more relaxed, and since $D/f$ is usually a few units, the difference of one is small. NCERT adopts the infinity convention for all further discussion of microscopes and telescopes.
$1 + D/f$ versus $D/f$ — read the image position
Both formulas are correct for a simple microscope; only the image position decides which to use. Near point (maximum magnification, slight strain) gives $m = 1 + D/f$; infinity / normal adjustment / relaxed eye gives $m = D/f$. Examiners pick the wording deliberately. If a question says "image at the near point" or "least distance of distinct vision", the answer must carry the extra $+1$.
Near point → $m = 1 + \dfrac{D}{f}$ | Infinity (relaxed) → $m = \dfrac{D}{f}$.
Calculate the magnifying power of a simple microscope having a focal length of $2.5\,\text{cm}$.
For the image at the near point, $m = 1 + \dfrac{D}{f} = 1 + \dfrac{25}{2.5} = 1 + 10 = 11$.
A simple microscope is limited: for realistic focal lengths the maximum magnification is about nine or less, and a convex lens magnifies up to roughly twenty times only with very short focal lengths. For larger magnifications, two lenses are combined.
The microscope formulas rest on the thin-lens equation and focal length. Revise the lens maker's formula if $1/v - 1/u = 1/f$ feels rusty.
The compound microscope
A compound microscope uses two converging lenses placed coaxially at the two ends of a tube. The lens nearer the object is the objective — short aperture, short focal length $f_o$. The lens nearer the eye is the eyepiece (eye lens), of focal length $f_e$. The object is placed just beyond the focus of the objective (between $F$ and $2F$), so the objective forms a real, inverted, magnified image inside the tube.
This first image acts as the object for the eyepiece, which is positioned so that the first image lies at or just within its focus. The eyepiece then behaves like a simple microscope, producing a final virtual, enlarged image at infinity (or at the near point for slightly higher magnification). Because the objective already inverts the image and the eyepiece does not re-invert it, the final image is inverted with respect to the original object.
The objective forms a real inverted image (height $h'$) near the eyepiece focus $F_e$; the eyepiece, acting as a magnifier, sends rays out parallel for a final image at infinity. $L$ is the tube length.
Magnifying power of a compound microscope
The total magnification is the product of the objective and eyepiece magnifications, $m = m_o \times m_e$. The objective gives a linear magnification
$$m_o = \frac{h'}{h} = \frac{L}{f_o}$$
where $h'$ is the size of the intermediate image, $f_o$ the objective focal length, and $L$ the tube length — the distance between the second focal point of the objective and the first focal point of the eyepiece. The eyepiece acts as a simple microscope on this image, contributing $m_e = 1 + D/f_e$ when the final image is at the near point, or $m_e = D/f_e$ when it is at infinity.
Combining for the final image at infinity gives the standard NEET formula:
$$m = m_o\, m_e = \frac{L}{f_o}\cdot\frac{D}{f_e}$$
High magnification means short $f_o$ and short $f_e$
In $m = \dfrac{L}{f_o}\cdot\dfrac{D}{f_e}$, both focal lengths sit in the denominator, so a large magnification needs small $f_o$ and small $f_e$ together with a large tube length $L$. This is the opposite of the telescope, where the objective focal length is made large. A common slip is to assume the eyepiece formula always carries the $+1$: it does only for the near-point case ($m_e = 1 + D/f_e$); for the infinity case it is $m_e = D/f_e$.
Microscope: small $f_o$, small $f_e$, large $L$. Telescope: large $f_o$. Do not confuse the two.
| Feature | Simple microscope | Compound microscope |
|---|---|---|
| Lenses | One converging lens | Objective + eyepiece |
| Objective role | — | Real, inverted, magnified intermediate image |
| Final image | Virtual, erect, magnified | Virtual, inverted, magnified |
| m (near point) | m = 1 + D/f | m = (L/f_o)(1 + D/f_e) |
| m (at infinity) | m = D/f | m = (L/f_o)(D/f_e) |
| Typical magnification | Up to about 9 (limited) | Hundreds (compounded) |
Worked examples
A compound microscope has an objective of $f_o = 1.0\,\text{cm}$, an eyepiece of $f_e = 2.0\,\text{cm}$, and a tube length $L = 20\,\text{cm}$. Find the magnification for the final image at infinity ($D = 25\,\text{cm}$).
$m = \dfrac{L}{f_o}\cdot\dfrac{D}{f_e} = \dfrac{20}{1.0}\times\dfrac{25}{2.0} = 20 \times 12.5 = 250$.
What focal length convex lens is needed for a simple microscope of magnification six when the image is at the near point ($D = 25\,\text{cm}$)?
$m = 1 + \dfrac{D}{f}\Rightarrow 6 = 1 + \dfrac{25}{f}\Rightarrow \dfrac{25}{f} = 5 \Rightarrow f = 5\,\text{cm}$, matching the NCERT statement.
Formula summary
| Quantity | Formula | Condition |
|---|---|---|
| Simple microscope | m = 1 + D/f | Image at near point (D ≈ 25 cm) |
| Simple microscope | m = D/f | Image at infinity (relaxed eye) |
| Objective magnification | m_o = L/f_o | L = tube length |
| Eyepiece (near point) | m_e = 1 + D/f_e | Final image at near point |
| Eyepiece (infinity) | m_e = D/f_e | Final image at infinity |
| Compound (infinity) | m = (L/f_o)(D/f_e) | Product m_o × m_e |
Microscope essentials
- The eye perceives angular size; the near point $D \approx 25\,\text{cm}$ is the closest comfortable viewing distance.
- Simple microscope: one converging lens, virtual erect magnified image; $m = 1 + D/f$ (near point) or $m = D/f$ (infinity).
- The infinity value is exactly one less than the near-point value; infinity viewing is more relaxed.
- Compound microscope: objective (small $f_o$) makes a real inverted intermediate image; eyepiece magnifies it. Final image is virtual and inverted.
- Total magnification is the product: $m = (L/f_o)(D/f_e)$ at infinity. Small $f_o$, small $f_e$, large $L$ give high magnification.