question_answer

The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eyepiece is found to be 20 cm. The focal lengths of lenses are:

A)  18 cm, 2 cm   

B)  11 cm, 9 cm

C)  10 cm, 10 cm  

D)  15 cm, 5 cm

Correct Answer: A

Solution :

 Key Idea: Length of tube in normal adjustment is equal to sum of focal length of eyepiece and objective. To see with relaxed eye final image should be formed at infinity. Under normal adjustment \[{{\mu }_{e}}={{f}_{e}}\] \[M=\frac{{{f}_{o}}}{{{f}_{e}}}\] \[L={{f}_{o}}+{{f}_{e}}\] Given, \[M=\frac{{{f}_{o}}}{{{f}_{e}}}=9\] \[\Rightarrow \] \[{{f}_{o}}=9\,{{f}_{e}}\] ... (i) and \[L=20={{f}_{o}}+{{f}_{e}}\] \[\Rightarrow \] \[9\,{{f}_{e}}+{{f}_{e}}=20\] \[\Rightarrow \] \[{{f}_{e}}=2\,cm\] ... (2) From Eq. (i), we get \[{{f}_{o}}=9\,{{f}_{e}}\] \[{{f}_{o}}=9\times 2=18\,cm\] Note: In order to increase the magnifying power of a telescope the focal length of objective lens \[({{f}_{o}})\] should be large and the focal length of eyepiece \[({{f}_{e}})\] should be small.