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The focal lengths of the objective and eyepiece of a compound microscope are 8mm and 2.5 cm respectively . A man with normal near point (25 cm)can focus distinctly the microscope . What is the separation between the lenses and magnification power of the instrument ? |
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Answer» SOLUTION :Forthe OBJECTIVE, `f_o =0.8 CM andu_(o) =-0.9 cm ` From `1/v-1/u=1/f`, we get `v_o =(0.8xx(-0.9))/(-0.9+0.8)=7.2 cm ` For the EYEPIECE ,`f_e =2.5 cm and v_e=-25 cm ` `therefore u_e=(v_ef_e)/(f_e-v_e)=(-25xx2.5)/(2.5+25)=-(25)/(11)=-2.27 cm ` `therefore` The distance between the two lenses `=v_(o)+u_e=7.2+2.27=9.47 cm ` `therefore` Magnification power , `m=(v_o)/(u_o)xx(v_e)/(u_e)=(7.2)/(0.9)xx(25/25)/(11)=88` |
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