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In an astronomical telescope, the focal lengths of the objective and the eye piece are 100cm and 5cm respectively . If the telescope is focussed on a scale 2m from the objective, the final image is formed at 25cm from the eye. Calculate (i) the magnification and (ii) the distance between the objective and the eyepiece. |
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Answer» Solution :`f_0=100cm ,f_e=5cm` To find the image distance due to objective `mu_0=-2m=-200cm, v_0=?` For a lens `1/f=1/v-1/u, 1/+100=1/v_0-1/(-200)` `1/v_0=1/100-1/200=1/200, v_0=200cm` MAGNIFICATION of objective `m_0=v_0/v=200/(-200)=-1` To find the object distance for the eyepiece `v_e=-25cm, u_e=?` For a lens `1/f=1/v-1/u,1/(+5)=1/(-25)-1/u` `1/u_e=-1/25 -1/5=- 6/25, u_e=- 25/6cm` Magnification of the eyepiece `m_e=v_e/u_e= (-25 TIMES 6)/(-25)=6` (i) Magnification of the telescope `=m_0 times m_e=-1 times 6=-6` (II) Distance between the objective and the eyepiece= `v_0+|u_e|=200+25/6 =204.2cm` |
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