At what distance eye 'E' will observe the fourth image (after four refractions from plane surfaces) of object 'O' from itself?
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Solution
The fourth image will be formed when the incident ray emerging from object 'O' suffers the refraction at 4 interfaces.
Here, 2 different mediums are present : μ1=1.5 and μ2=2
⇒ shift produced in the direction of incident ray from 'O' will be,
Stotal=S1+S2
⇒Stotal=t1(1−1μ1)+t2(1−1μ2)
or Stotal=10(1−13/2)+10(1−12)
⇒Stotal=103+5=253cm
Initially the distance between object (O) & eye (E) is,
d=10×5=50cm
⇒ Fourth image will form at a distance d' from 'E',
∴d′=d−Stotal=50−253=1253cm
Hence, option (c) is the correct answer.
Why this question?
Caution: Here observer is in air and two different media are also separated by 10cm of air, thus shift will not be produced by 10cm layer of air.