Question

A point object lies inside a transparent solid sphere of radius 20 cm and of refractive index n = 2. When
the object is viewed from air through the nearest surface it is seen at a distance 5 cm from the surface.
Find the apparent distance of object when it is seen through the farthest curved surface.
20cm

Ans is 80 plz ecplain correctly ​

Answer 1

Explanation:

−μ

1

+

v

μ

2

=

R

μ

2

−μ

1

...1

Here object is placed in the medium of refrective index μ

1

and a spherical surface of radius of curvature'R' seperate it from medium of refrective index μ

2

since,

μ

2

=2,μ

1

=1

u=?,v=5cmR=20cm

from eq 1 we get

20

1−2

=

5

1

−

u

2

20

−1

−

5

1

=−

u

2

u=8cm

Now in second case,

μ=40−8=32cmv=?

Thus,

20

1−2

=

v

1

−

32

2

=

v

1

=

20

−1

+

32

2

=

v

1

=

1

=

80

v=80cm