question_answer

The optical length of an astronomical telescope with magnifying power of 10. For normal vision is 44 cm, what is focal length of the objective?

A) 4 cm                                      

B) 40 cm

C) 44 cm                                   

D) 440 cm

Correct Answer: B

Solution :

Key Idea: To see with relaxed eye final image should be formed at infinity. The distance  between the objective and eyepiece is adjusted so that image A'B' formed by  objective is at focus \[{{F}_{e}}'\] of the eyepiece. This adjustment of telescope is called normal adjustment. In this position the length of the telescope is \[{{f}_{o}}+{{f}_{e}}\] (\[{{f}_{o}}\]is foca1 length of objective, \[{{f}_{e}}\]of eyepiece)                 \[\therefore \]   \[44={{f}_{o}}+{{f}_{e}}\]           ?(i) Also, magnifying power \[m=-\frac{{{f}_{o}}}{{{f}_{e}}}\] \[-10=-\frac{{{f}_{o}}}{{{f}_{e}}}\] \[\Rightarrow \]      \[{{f}_{o}}=10{{f}_{e}}\]         ?(ii) Using Eqs. (ii) and (ii), we get \[10{{f}_{e}}+{{f}_{e}}=44\] \[\Rightarrow \]            \[{{f}_{e}}=4\,cm\] Also, focal length of objective is  \[{{f}_{o}}=10{{f}_{e}}\] Therefore, focal length of objective is 40 cm.