question_answer

An astronomical telescope has an objective of Focal length 100 cm and magnifying power 50. The distance between the two lenses in normal adjustment will be.                                                      [BHU PMT-2002]

A)                  \[R\]                                    

B)                  \[\frac{3r}{4}\]

C)                  \[\frac{256\,R}{81}\]                                     

D)                  \[\frac{81\,R}{256}\]

Correct Answer: B

Solution :

                 Key Idea: in normal adjustment image formed by objective is at focus of eyepiece.                 To see with relaxed eye the final image should be formed at infinity. For this, the distance between the objective and eyepiece is adjusted, so that the image formed by objective is at the focus of eyepiece. This adjustment is called normal adjustment.                                 \[c=v\lambda =\frac{\omega }{2\pi }\lambda \]where \[k=\frac{2\pi }{\lambda }\] is focal length of objective and \[\therefore \] is focal length of eyepieces.                 In this position the length of the telescope is \[c=\frac{\omega }{k}\]                     \[\frac{{{E}_{o}}}{{{E}_{o}}}=c=\frac{\omega }{k}\]                          \[\Rightarrow \]                 And distance between two lenses in normal adjustment=length of telescope \[{{E}_{o}}k={{B}_{o}}\omega \]                 Note : Focal length of objective is large so that a bright image is formed while that of eyepiece is small so that all rays converge at  eyepiece.