Geometrical OpticsHard
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
Options
A.11 cm, 9 cm
B.10 cm, 10 cm
C.15 cm, 5 cm
D.18 cm, 2 cm
Solution
f0 + fe = 20 ....(i)
....(ii)
Solving F0 = 10 cm
Fe = 2 cm
....(ii)Solving F0 = 10 cm
Fe = 2 cm
Create a free account to view solution
View Solution FreeMore Geometrical Optics Questions
A ray of light is incident normally on one face of 30o - 60o - 90o prism of refraction index 5/3 immersed in meter water...The focal length of a concave mirror is 12 cm. Where should an object of length 4 cm be placed, so that a real image of ...When light is refracted from air into glass...An object is lying at a distance of 90 cm from a concave mirror of focal length 30 cm. The position and nature of image ...A ray of light propagates from glass (refractive index = 3/2) to water ( refractive index = 4/3). The value of the criti...