Galileo devised a simple terrestrial telescope that produces an upright image. I

Question

Galileo devised a simple terrestrial telescope that produces an upright image. It consists of a converging objective lens and a diverging eyepiece at opposite ends of the telescope tube. For distant objects, the tube length is the objective focal length less the absolute value of the eyepiece focal length.
(a) Does the user of the telescope see a real or virtual image?

Virtual Image



(b) Where is the final image? (For values of infinity, enter 999 as your answer.)
qe = infinity (999)

(c) If a telescope is to be constructed with a tube 9.5 cm long and a magnification of 2.40, what are the focal lengths of the objective and eyepiece?
feyepiece = cm
fobjective = cm




I'm having trouble with part c of this question. Any help would be great! Thank you.

Explanation / Answer

Concept: In Terrestial telescope objectibve is converging lens and the eye piece is diverging lens Here the object forms a real image and inverted image of a very far object This image actas as virtual object to the eye piece thus, pe = - fe Apply lens maker's formual for the eye piece 1/ qe = 1/pe - 1/fe where qe is th eimage distance ,              pe is the object distnce            = - 1/fe + 1/fe    qe = (Infinite) (a) Since parallel rays emerging form the eye piece            eye will observes the virtual image (b) since , qe = .thus, final image is at infinity (c) Lenght of the telescope L = 9.5 cm         Mangification: M = 2.4             but , Mangification: M = fo / fe              L = fe + fo =9.5 cm                 since,2.4 = fo /fe                         fo = 2.4 fe        For eye piece gives a virtual image        it shoul be considered as -ve sing                thus,   2.4 fe - fe = 9.5 cm                        fe = 6.78 cm               so,    fo = 2.4(6.78)                            = 16.28cm    Thus, focal lenght of the eye piece = fe = - 6.78 cm             foacl lenght of the objective lens = fo = 16.28 cm In Terrestial telescope objectibve is converging lens and the eye piece is diverging lens Here the object forms a real image and inverted image of a very far object This image actas as virtual object to the eye piece thus, pe = - fe Apply lens maker's formual for the eye piece 1/ qe = 1/pe - 1/fe where qe is th eimage distance ,              pe is the object distnce            = - 1/fe + 1/fe    qe = (Infinite) (a) Since parallel rays emerging form the eye piece            eye will observes the virtual image (b) since , qe = .thus, final image is at infinity (c) Lenght of the telescope L = 9.5 cm         Mangification: M = 2.4             but , Mangification: M = fo / fe              L = fe + fo =9.5 cm                 since,2.4 = fo /fe                         fo = 2.4 fe        For eye piece gives a virtual image        it shoul be considered as -ve sing                thus,   2.4 fe - fe = 9.5 cm                         fo = 2.4 fe        For eye piece gives a virtual image        it shoul be considered as -ve sing                thus,   2.4 fe - fe = 9.5 cm                        fe = 6.78 cm               so,    fo = 2.4(6.78)                            = 16.28cm    Thus, focal lenght of the eye piece = fe = - 6.78 cm             foacl lenght of the objective lens = fo = 16.28 cm

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