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TS EAMCET 5-Aug-2021 Shift 2 Question Paper
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© examsnet.com
Question : 104
Total: 160
A point source is located at a distance of 20 cm from the front surface of a symmetrical glass biconvex lens with equal radii of curvature 5 cm. The distance at which image formed from the rear surface of this lens is
[Given refractive index of the glass is 1.5]
20
3
cm
10
3
cm
5
cm
10
cm
Validate
Solution:
Given,
Object distance from biconvex lens,
u
=
−
20
c
m
Radii of curvature,
R
=
5
c
m
Refractive index,
µ
=
1.5
As we know that,
1
f
=
(
µ
−
1
)
(
1
R
1
−
1
R
2
)
where,
f
is focal length and
R
1
,
R
2
are radii of curvatures of both surfaces.
i.e.
R
1
=
R
2
=
R
∴
1
f
=
(
1.5
−
1
)
(
1
R
+
1
R
)
=
1
2
⋅
2
R
=
1
R
∴
f
=
R
=
5
c
m
By using lens formula,
1
f
=
1
v
−
1
u
where,
v
is image distance.
∴
1
v
=
1
f
+
1
u
=
1
5
−
1
20
=
3
20
⇒
v
=
20
∕
3
c
m
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