Unit 06: Optics • Chapter 09 (Ray Optics)
Optical Instruments: Microscope & Telescope
Visual angle physics, simple magnifier, compound microscope, astronomical refracting telescope, Cassegrain reflector, and resolving criteria.
Socho ek chidiya (bird) ko: jab woh aasmaan mein 1 km door udti hai toh ek chote til (dot) jaisi dikhti hai, par wahi chidiya jab aapke haath par aakar baithti hai toh badi aur detailed dikhti hai. Kya chidiya ka physical size badal gaya? Bilkul nahi! Fark sirf Visual Angle ($\theta$) ka hai jo usne aapki retina par banaya.
Optical instruments linear height nahi, balki retina par banne wale visual angle ko bada karte hain (Angular Magnification $m = \beta/\alpha$).
• Microscope ek nanhe object ko hamare paas (near point $D = 25\text{ cm}$) laakar use two stages mein mega-magnify karta hai (pehle Objective bada real image banata hai, fir Eyepiece use magnifying glass ki tarah aur bada karta hai).
• Telescope infinity par baithe vishalkay objects (Moon/Stars) se aane wali parallel rays ko collect karke focal plane par ek compact real image banata hai, jise eyepiece aankh ke paas laakar bada visual angle provide karta hai!
1. Compound Microscope
Two converging lenses: Objective ($f_o$ small, aperture small) + Eyepiece ($f_e$ moderate):
Where $L$ is the optical tube length (distance between second focal point of objective and first focal point of eyepiece).
2. Astronomical Telescope (Refracting)
Objective ($f_o$ large, aperture large) + Eyepiece ($f_e$ small):
Negative sign indicates an inverted final image relative to the distant celestial object. Normal adjustment provides relaxed, strain-free eye viewing.
Ray Diagram and Magnifying Power of a Compound Microscope
Fig 9.5A: Compound microscope ray diagram for final image formation at the Near Point ($D = 25\text{ cm}$).
1. The total magnifying power $m$ of a compound microscope is the product of the linear magnification of the objective ($m_o$) and the angular magnification of the eyepiece ($m_e$):
2. For the objective lens, the linear magnification is $m_o = \frac{h'}{h} = \frac{v_o}{u_o}$. Since the object is placed just beyond $f_o$ ($u_o \approx -f_o$) and the real image is formed very close to the eyepiece ($v_o \approx L$, the optical tube length):
3. The eyepiece functions identically to a simple magnifier. Its angular magnification is:
• Case (a) When the final image is formed at the Near Point ($v = -D$):
• Case (b) When the final image is formed at Infinity (Normal Adjustment):
Design Criterion: To achieve massive magnifying power, both $f_o$ and $f_e$ must be exceptionally small ($f_o, f_e \ll D$).
Ray Diagram and Magnifying Power of an Astronomical Telescope (Refracting Type)
Fig 9.5B: Astronomical telescope ray diagram in Normal Adjustment ($v = \infty$); tube length $L = f_o + f_e$.
1. The magnifying power $m$ of a telescope is defined as the ratio of the angle subtended by the final image at the eye ($\beta$) to the angle subtended by the distant object at the unaided eye ($\alpha$):
2. For small paraxial angles $\alpha \approx \tan\alpha$ and $\beta \approx \tan\beta$:
3. Therefore, the general magnifying power is:
• Normal Adjustment (Relaxed Eye, Final Image at Infinity): Here, $u_e = f_e$:
• Near Point Adjustment (Final Image at $D = 25\text{ cm}$): By lens formula for eyepiece ($\frac{1}{-D} - \frac{1}{-u_e} = \frac{1}{f_e} \implies \frac{1}{u_e} = \frac{1}{f_e}\left(1 + \frac{f_e}{D}\right)$):
Design Criterion: For large magnification, objective focal length must be large ($f_o \gg f_e$) and eyepiece focal length must be small.
Reflecting Telescope (Cassegrain Type) & Advantages over Refractors
In modern observational astronomy, refracting telescopes suffer from severe fundamental limitations. The Cassegrain reflecting telescope replaces the heavy glass objective with a large parabolic primary concave mirror.
1. Zero Chromatic Aberration
Mirrors reflect all wavelengths according to the laws of reflection without dispersion ($r = i$). Hence, reflecting telescopes are completely free from chromatic aberration (color fringing).
2. Elimination of Spherical Aberration
By using a parabolic primary mirror instead of a spherical surface, all parallel incident rays converge to a sharp mathematical focus, eliminating spherical aberration.
3. Structural Mechanical Support
A heavy glass lens can only be supported along its outer rim and sags under its own enormous weight. A primary mirror can be supported across its entire back surface.
4. Massive Light Gathering Power
Fabricating large flawless glass lenses beyond 1 metre diameter is practically impossible. Mirrors can be cast in multi-metre diameters, delivering high resolving power.
Trap 1: Microscope vs Telescope Lenses Swap
*Board Question:* Can an astronomical telescope be used as a compound microscope by reversing its lenses?
• NO!
• For a microscope: $f_o$ and $f_e$ are both very small (in mm or few cm) with $f_o < f_e$.
• For a telescope: $f_o$ is very large ($\sim 100\text{ cm}$) and $f_e$ is small ($\sim 2\text{ cm}$).
Swapping them will yield a magnifying power of practically zero or produce extreme aberrations!
Trap 2: Magnification vs Resolving Power of a Telescope
*Exam Trap:* Does increasing the magnification of a telescope increase its ability to distinguish close stars?
• NO! Magnification ($m = f_o/f_e$) only enlarges the angular size.
• The ability to resolve two close celestial objects is Resolving Power ($\text{RP} = \frac{D}{1.22\lambda}$), which depends strictly on the aperture diameter $D$ of the objective, not on magnification!
Trap 3: Normal Adjustment vs Near Point Adjustment
• Near Point ($v = D$): Yields MAXIMUM magnifying power, but the eye muscles are contracted and under accommodation strain.
• Normal Adjustment ($v = \infty$): Magnifying power is slightly lower, but the ciliary eye muscles are completely relaxed, allowing prolonged, fatigue-free viewing.
Trap 4: What Happens if Half of the Telescope Objective is Covered?
*Classic Question:* If the central half of a telescope objective is painted black:
1. The complete image is STILL formed; no part of the image is lost.
2. The brightness (intensity) of the image decreases by 50% because total light collection area is halved.
Problem: A compound microscope consists of an objective lens of focal length $2.0\text{ cm}$ and an eyepiece of focal length $6.25\text{ cm}$ separated by a distance of $15\text{ cm}$. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision ($25\text{ cm}$), and (b) at infinity? What is the magnifying power in each case?
Solution: Given: $f_o = 2.0\text{ cm}, f_e = 6.25\text{ cm}$, total tube separation $d = 15\text{ cm}$.
Case (a) Final image at Near Point ($v_e = -25\text{ cm}$):
Since $v_o + |u_e| = 15\text{ cm} \implies v_o = 15 - 5.0 = 10.0\text{ cm}$. Now for objective:
Magnifying power:
Case (b) Final image at Infinity (Normal Adjustment):
Here $u_e = f_e = 6.25\text{ cm} \implies v_o = 15 - 6.25 = 8.75\text{ cm}$.
Problem: A small telescope has an objective lens of focal length $144\text{ cm}$ and an eyepiece of focal length $6.0\text{ cm}$. (a) What is the magnifying power of the telescope in normal adjustment? (b) What is the separation between the objective and the eyepiece? (c) If this telescope is used to view a $100\text{ m}$ tall tower $3\text{ km}$ away, what is the height of the intermediate image of the tower formed by the objective lens?
Solution: $f_o = 144\text{ cm} = 1.44\text{ m}, f_e = 6.0\text{ cm} = 0.06\text{ m}$.
1. Magnifying power in normal adjustment:
2. Separation (Tube length $L$):
3. Height of tower $h = 100\text{ m}$, distance $u = 3\text{ km} = 3000\text{ m}$.
Angle subtended by tower at objective:
Since the intermediate image is formed at the focal plane ($v \approx f_o$):
Microscope (Normal)
$m = -\frac{L}{f_o}\frac{D}{f_e}$ ($L = v_o + f_e$)
Microscope (Near Pt)
$m = -\frac{L}{f_o}(1 + \frac{D}{f_e})$
Telescope (Normal)
$m = -\frac{f_o}{f_e}$, $L = f_o + f_e$
Telescope Resolving
$\text{RP} = \frac{D}{1.22\lambda} \propto D$