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.You need to login to perform this action.
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