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# Find the position relative to the lens of the image of the object formed by the system. [ ans is 1]

Asked by Prashant DIGHE 8th July 2019, 5:49 PM
First, let us find the radius of curvature R  from given focal length 30 cm using lens makers formula.

1/f = ( μg - 1) [ (1/R)+(1/R) ]

1/30 = (1.5-1) (2/R)  or R = 30 cm

at spherical surface of air-glass interface,  we have  ......................(1)
where v is the first image distance at right side of lens, u is object distance at left side of lens

substituting the values in eqn.(1),  (1.5/v) -(1/-90) = 0.5/30, we get v = 270 cm

Hence first image is formed at 270 cm from lens at right side.

This is acting as virtual object for spherical surface of glass-water interface

then we have ,   ...................(2)
from eqn.(2), we get v' = 120 cm

Hence second image is formed at 120 cm right side of lens, which is 40 cm behind the plane mirror.

Hence this image is virtual object of mirror and real image (third image) is formed 40 cm in front of mirror.

Now this third image acts as real object to spherical surface of water-glass interface

then we have, ..................(3)
from eqn.(3), we get v'' = -54 cm

Hence fourth image is formed at 54 cm at right side from lens.

Now finally applying same formula for spherical surface of glass-air surface,
.............................(4)
from eqn.(4), we get, vf = -90 cm

Hence position of final image is 90 cm from lens at right side or 10 cm behind the mirror
Answered by Expert 9th July 2019, 10:19 AM
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Tags: refraction

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