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

  • 10
What are you looking for?