I\'m on moble, so these pictures may be hard to see. If you cannot read them ple

Question

I'm on moble, so these pictures may be hard to see. If you cannot read them please let me know. Also please help with these optics problems

fi 20 cm 0 cm 40 cm- 30 cm- An object is located 40 centimeters from the first of two thin converging lenses of focal lengths 20 centimeters and 10 centimeters, respectively, as shown in the figure above. The lenses are separated by 30 centimeters. The final image formed by the two-lens system is located (A) 5.0 cm to the right of the second lens (B) 13.3 cm to the right of the second lens (C) infinitely far to the right of the second lens (D) 13.3 cm to the left of the second lens (E) 100 cm to the left of the second lens

Explanation / Answer

a)   f1 = 20 cm, f2 = 10 cm
   u = -40 cm, d = 30 cm
   Lens formula says that
   1/v - 1/u = 1/f
   For 1st lens
   1/v = 1/20-1/40 = 1/40
   v = 40 cm
   But d between 1st and 2nd lens = 30 cm
   So for 2nd
       u = (40 - 30) cm = 10 cm
   =>   1/v = 1/10 + 1/10 = 1/5
   => v = 5 cm
   Since v is +ve, image is formed 5 cm to the right of 2nd lens

b) Given M = 10
   let object be at a point just before inf, so objective will form inverted image just after the focal point of objective, which is just before the focus of eye piece. So the eye piece will form a virtual image of the invert image leading to a overall inverted image
   => M is -ve
   For telescope M = - fo/fe where fo = focal length of objective lens, fe = focal length of eye piece
   Optical path length d = fo + fe
   Given fo = 1 m
   -10 = -1/fe
   fe = 0.1
   => d = 10.1 m

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