Question 333: Ray Optics

Question

The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is 20 cm. The focal length of lenses are

Accepted Answer

Magnifying power, m = $${{{f_0}} \over {{f_e}}}$$ = 9 ..........(1)
where fo and fe are the focal lengths of the objective and eyepiece respectively
Also,f
o
+ f
e
= 20 cm ...(2)
On solving (1) and (2), we get
f
o
= 18 cm, f
e
= 2 cm

Answer

10 cm, 10 cm

Answer

15 cm, 5 cm

Answer

18 cm, 2 cm

Answer

11 cm, 9 cm

Ray Optics

The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is 20 cm. The focal length of lenses are

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