draw a lens in differentfive rarer images
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Answer:
Consider a convex lens (or concave lens) of absolute refractive index μ
2
to be placed in a rarer medium of absolute refractive index μ
1
.
Considering the refraction of a point object on the surface XP
1
Y, the image is formed at I
1
who is at a distance of V
1
.
CI
1
=P
1
I
1
=V
1
(as the lens is thin)
CC
1
=P
1
C
1
=R
1
CO=P
1
O=u
It follows from the refraction due to convex spherical surface XP
1
Y
−u
μ
1
+
v
1
μ
2
=
R
1
μ
2
−μ
1
..........(i)
The refracted ray from A suffers a second refraction on the surface XP
2
Y and emerges along BI. Therefore I is the final real image of O.
Here the object distance is
u=CI
1
≈P
2
I
1
=V
P
1
P
2
is very small
Let CI≈P
2
I=V
(Final image distance)
Let R
2
be radius of curvature of second surface of the lens.
It follows from refraction due to concave spherical surface from denser to rarer medium that
v
1
−μ
2
+
v
μ
1
=
R
2
μ
1
−μ
2
=
−R
2
μ
2
−μ
1
............(ii)
Adding (i)and(ii)
u
−μ
1
+
v
μ
2
=(μ
2
−μ
1
)(
R
1
1
−
R
2
1
)
or
μ
1
(
v
1
−
u
1
)=(μ
2
−μ
2
)(
R
1
1
−
R
2
1
)
But
v
1
−
u
1
=
f
1
and
μ
1
μ
2
=μ
Thus we get=
f
1
=(μ−1)(
R
1
1
−
R
2
1
)
solution