A bio-convex lense (refractive index 3/2) has radii of curvature 20 cm each. It is fitted to a hole in a large box filled with water (refractive index = 4/3). A point object is placed outside the box at a distance of 40 cm from lens on its axis. Distance of image formed in water is:
(a) + 160 cm
(b) + 60 cm
(c) - 100 cm
(d) + 120 cm
Correct option: a) + 160 cm
Here,
u = -40 cm
R1 = 20 cm (for 1st surface)
R2 = -20 cm (for 2nd surface)
n1 = 1
n2 = 1.5
n3 = 4/3
Applying Gauss optical formula on the first surface, we get a virtual object, this object acts as the object for the second surface. Let v’ = virtual object distance.
n2/v’ – n1/u = (n2 – n1)/R1
1.5/v’- 1/(-40) = (1.5 – 1)/20
=> 1/v’ = 0 ---1.
For second surface, ray moving from denser (glass) to rarer (water)
n3/v – n2/v’ = (n3 – n2)/R2
(4/3)/v – (3/2)×0 = (4/3 – 3/2)/(-20) using 1.
=> v = 160 cm