Question: The magnifying power of an astronomical telescope in normal adjustment is 29 and the objective and the eyepiece are separated by a distance of 150 em. Find the focal lengths of the two lenses.
The correct answer is –
M = (f_o)/(f_e) x (25 cm)/(D)
where f_o is the focal length of the objective lens, f_e is the focal length of the eyepiece, and D is the distance between the objective lens and the eyepiece.
Given that the magnifying power of the telescope is 29 and the distance between the objective lens and the eyepiece is 150 cm (note that the unit should be cm, not em), we can write:
29 = (f_o)/(f_e) x (25 cm)/(150 cm)
Simplifying, we get:
29 = (f_o)/(f_e) x (1/6)
Multiplying both sides by 6f_e, we get:
174f_e = f_o
We also know that the distance between the objective lens and the eyepiece is 150 cm. Therefore:
D = f_o + f_e
Substituting f_o = 174f_e into this equation, we get:
D = 174f_e + f_e = 175f_e
Now, we have two equations with two unknowns:
29 = (f_o)/(f_e) x (25 cm)/(150 cm) D = 175f_e
Substituting f_o = 174f_e into the first equation, we get:
29 = (174f_e)/(f_e) x (25 cm)/(150 cm)
Simplifying, we get:
f_e = (25 cm)/(174/29) = 3.7 cm
Substituting this value of f_e into the second equation, we get:
D = 175f_e = 175 x 3.7 cm = 647.5 cm
Finally, we can use the equation f_o = 174f_e to find the focal length of the objective lens:
f_o = 174f_e = 174 x 3.7 cm = 644.4 cm
Therefore, the focal length of the objective lens is 644.4 cm and the focal length of the eyepiece is 3.7 cm.
