Astronomical TelescopeJEE Main

Angular magnification f₀/f_e — interactive Physics simulation for IIT-JEE.

Concept

A telescope makes distant objects look angularly bigger. The long-focus objective forms a small real image at its focal plane; the short-focus eyepiece views it. In normal adjustment (relaxed eye) the two focal points coincide: magnification $= f_o/f_e$ and tube length $= f_o + f_e$ . Aperture — not magnification — decides what you can actually see.

Key formula

M = \frac{f_o}{f_e}, \qquad L = f_o + f_e, \qquad \theta_{min} = \frac{1.22\lambda}{D}

Derivation

Distant object subtends angle α; its image (height h) forms at the objective's focus: $\alpha \approx h/f_o$ . The eyepiece shows it at angle $\beta \approx h/f_e$ .

$M = \beta/\alpha = f_o/f_e$ . Resolution is diffraction-limited by the aperture (Rayleigh criterion) — magnifying beyond it gives 'empty magnification'.

Scenarios to explore

Astronomical Telescope — M = f₀/f_e, tube length & resolving power.

Real-world applications

Astronomy — light grasp ∝ D² is why observatories chase giant mirrors.
Terrestrial spotting scopes add an erecting lens (adds 4f to the tube).
Radio telescopes: same 1.22λ/D with λ in metres → km-scale dishes.

JEE exam tips

Normal adjustment: parallel rays in, parallel rays out — final image at infinity.
Image at D instead: M = (f₀/f_e)(1 + f_e/D), slightly bigger.
Brightness ∝ (D/M)² per unit area — high M dims extended objects.

Common mistakes

Microscope/telescope mix-up: telescope wants LONG f₀; microscope wants SHORT.
M = f_e/f₀ (inverted).
Believing more magnification = better — resolution is capped by aperture.

Exam traps to avoid

Astronomical telescope's final image is INVERTED — fine for stars.
Resolving power = 1/θ_min ∝ D — doubling aperture doubles resolving power.