Optical Instruments — Question 9

Section 34-7 · Physics pages 5688–5689

Question 9 A Galilean telescope uses a virtual object for its negative eyepiece A telescope objective has fo=30 cm and a diverging eyepiece has fe=-5 cm. A distant object supplies parallel input. Set the lens separation so the output is also parallel. d first second fe < 0 Schematic; labels, not drawing lengths, specify the data. (a) Find the separation by treating the objective focus as the eyepiece virtual object. (b) Find the signed angular magnification using M = −fo/fe and compare the length with a positive-eyepiece telescope of the same magnification magnitude.
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Question 9 - Solution A Galilean telescope uses a virtual object for its negative eyepiece d first second fe < 0 (a) Method and calculation qo = fo, pe = d −fo, qe →∞⇒pe = fe. d = fo + fe = 30 −5 = 25 cm . (b) Interpretation and further calculation M = −30/(−5) = +6 upright. dpositive = 30 + 5 = 35 cm ⇒∆d = 10 cm shorter . Physical check The negative eyepiece intercepts rays before they reach the objective focus, so its object distance is negative.

Question 9

A Galilean telescope uses a virtual object for its negative eyepiece

A telescope objective has fo=30 cm and a diverging eyepiece has fe=-5 cm. A distant object supplies parallel input. Set the lens separation so the output is also parallel.

Diagram shown in the original SVG worksheet.

  1. Find the separation by treating the objective focus as the eyepiece virtual object.

  2. Find the signed angular magnification using M=−fo/feM=-f_o/f_e and compare the length with a positive-eyepiece telescope of the same magnification magnitude.

Question 9 - Solution

A Galilean telescope uses a virtual object for its negative eyepiece

Diagram shown in the original SVG worksheet.

(a) Method and calculation

qo=fo,pe=d−fo,qe→∞⇒pe=fe.q_o=f_o,\quad p_e=d-f_o,\quad q_e\to\infty\Rightarrow p_e=f_e.d=fo+fe=30−5=25cm.\boxed{d=f_o+f_e=30-5=25\,\mathrm{cm}}.

(b) Interpretation and further calculation

M=−30/(−5)=+6upright.\boxed{M=-30/(-5)=+6}\quad\text{upright}.dpositive=30+5=35cm⇒Δd=10cmshorter.d_{\mathrm{positive}}=30+5=35\,\mathrm{cm}\Rightarrow\boxed{\Delta d=10\,\mathrm{cm}\ \text{shorter}}.

Physical check

The negative eyepiece intercepts rays before they reach the objective focus, so its object distance is negative.