Question

The magnifying power of a telescope in its normal adjustment is 20. If the length of the telescope is 105 cm in this adjustment, the focal length of the objective lens is

• A. 100cm
• B. 5cm
• C. 80cm
• D. 95cm

Answer 1

Answer: (A)

For normal adjustment, magnification is given by $m=\frac{f_o}{f_e}$ and the length of the telescope tube is $f_0+f_e$ . Let the focal length of objective and eye lenses are $f_o$ and $f_e$ respectively. Therefore, magnification $m=\frac{f_o}{f_e},$ $20=\frac{f_o}{f_e}\Rightarrow f_o=20f_e$ And $f_o+f_e=105,$ On solving, $f_o=100$ cm and $f_e=5cm$ .