Optical instruments — human eye, simple microscope, compound microscope, telescope

1. The Human Eye — why we need instruments at all

WHY $D$ matters: the largest visual angle the unaided eye can get from an object of height $h$ is when the object sits at the near point: $\theta_0 \approx \tan\theta_0 = \frac{h}{D}$ This $\theta_0$ is the reference all "magnifying powers" are compared against.

2. Angular magnification — the master definition

3. Simple Microscope (magnifying glass)

Derivation — image at near point $D$ (maximum magnification)

The lens lets you place the object close (large angle) while throwing a magnified virtual image out to $D$ where the eye can focus.

• Angle with lens: image (or object, paraxially) subtends $\theta = h/u$.
• Naked-eye angle: $\theta_0 = h/D$.

$M = \frac{\theta}{\theta_0} = \frac{h/u}{h/D} = \frac{D}{u}$

Use the lens equation $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$. Put image at near point: $v = -D$ (virtual, same side). Then $\frac{1}{-D} - \frac{1}{u} = \frac{1}{f}\;\Rightarrow\; \frac{1}{u} = -\frac{1}{D}-\frac{1}{f}\;\Rightarrow\; \frac{D}{u} = -\Big(1+\frac{D}{f}\Big).$ Taking magnitudes: $\boxed{M_D = 1 + \frac{D}{f}}$

Why this step? Setting $v=-D$ forces the most strained but sharpest image, giving the biggest $M$.

Image at infinity (relaxed eye)

Place object exactly at focus, $u=-f$, image goes to infinity, eye relaxed: $\boxed{M_\infty = \frac{D}{f}}$ Why smaller? Comfort costs you a factor: $M_D - M_\infty = 1$.

4. Compound Microscope

Derivation

Objective (linear magnification of the real intermediate image): $m_o = \frac{v_o}{u_o} \approx \frac{L}{f_o}$ where $L$ = tube length ≈ distance between the two foci (the intermediate image forms just inside the eyepiece focus). Why $\approx L/f_o$? The object sits just outside $f_o$, so $u_o\approx f_o$ and $v_o \approx L$.

Eyepiece acts as a simple microscope on the intermediate image: $m_e = \frac{D}{f_e}\quad(\text{relaxed eye, image at }\infty)$

Total: $\boxed{M = m_o \, m_e = \frac{L}{f_o}\cdot\frac{D}{f_e}}$ (For image at near point use $m_e = 1 + D/f_e$.)

5. Telescope (astronomical, refracting)

Derivation (normal adjustment, final image at infinity)

Parallel rays from a distant object at angle $\alpha$ form an image in the focal plane of the objective, of height $h$: $\alpha \approx \frac{h}{f_o}$ The eyepiece, with this image at its focus, sends rays out parallel at angle $\beta$: $\beta \approx \frac{h}{f_e}$ Angular magnification: $M = \frac{\beta}{\alpha} = \frac{h/f_e}{h/f_o} = \boxed{\frac{f_o}{f_e}}$ Why long objective, short eyepiece? Big $f_o/f_e$ = big magnification.

Tube length (normal adjustment): $L = f_o + f_e$ (the two foci coincide).

For final image at near point: $M = \dfrac{f_o}{f_e}\Big(1 + \dfrac{f_e}{D}\Big)$.

6. Big-picture comparison (80/20 table)

The 20% that gives 80%: microscopes want short focal lengths and multiply; telescopes want long objective / short eyepiece and use angular magnification.

Flashcards

Connections

• Lens equation and sign conventions
• Linear vs angular magnification
• Resolving power and diffraction limit
• Defects of vision — myopia, hypermetropia
• Reflecting telescope (Cassegrain) vs refracting
• Power of a lens and dioptres

Concept Map

Hinglish (regional understanding)