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#astronomical optics

2 public questions tagged with this topic.

Why does the image in a refracting telescope appear inverted without additional optics?

**Simple microscope** magnification M = 1 + D/f when image at D=25 cm near point, M = D/f when image at infinity (relaxed eye), f focal length (cm), D least distance of distinct vision 25 cm. For f=4 cm, image at 25 cm, M=1+25/4=7.25, angular magnification ratio of angle subtended by image to that by object at D. In a refracting telescope, the objective lens forms a real, inverted image of the distant object at its focal plane. The eyepiece magnifies this inverted image without reinverting it, so the final image remains inverted unless an additional lens or prism system is used. Substituting v

Ref: NCERT > Physics Book > Ray Optics > Optical Instruments - Simple and Compound Microscope

A telescope has an objective of focal length \( 160 \, \text{cm} \) and an eyepiece of focal length \( 8 \, \text{cm} \)

**Convex lens image formation**: object beyond 2F (u>2f) real inverted diminished between F and 2F, at 2F same size at 2F, between F and 2F magnified beyond 2F, at F image at infinity, within F virtual erect magnified same side. For f=15 cm, u=30 cm=2f, m = v/u =30/30=1? Actually v=30 cm, m=-1, same size inverted. Magnifying power: m = (f_o/f_e) . f_o = 160 cm , f_e = 8 cm . m = (160/8) = 20 . Substituting values gives 20, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror),

Ref: NCERT > Physics Book > Ray Optics > Thin Lenses - Lens Formula, Magnification and Power