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JEE Mains 04-Sep-2020 Shift 1 Solved Paper
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© examsnet.com
Question : 21
Total: 75
In a compound microscope, the magnified virtual image is formed at a distance of
25
c
m
from the eye-piece. The focal length of its objective lens is
1
c
m
.
If the magnification is 100 and the tube length of the microscope is
20
c
m
,
then the focal length of the eye-piecelens (in
c
m
) is _______.
[4 Sep 2020 Shift 1]
Your Answer:
Validate
Solution:
for first lens
=
1
v
1
−
1
−
x
=
1
1
⇒
v
1
=
x
x
−
1
also magnification
|
m
1
|
=
|
v
1
u
1
|
=
1
x
−
1
for
2
nd
lens this is acting as object so
u
2
=
−
(
20
−
v
1
)
=
−
(
20
−
x
x
−
1
)
and
v
2
=
−
25
c
m
angular magnification
|
m
A
|
=
|
D
u
2
|
=
25
|
u
2
|
Total magnification
m
=
m
1
m
A
=
100
(
1
x
−
1
)
(
25
20
−
x
x
−
1
)
=
100
25
20
(
x
−
1
)
−
x
=
100
⇒
1
=
80
(
x
−
1
)
−
4
x
⇒
76
x
=
81
⇒
x
=
81
76
⇒
u
2
=
−
(
20
−
81
76
81
76
−
1
)
=
−
19
5
now by lens formula
1
−
25
−
1
−
19
5
=
1
f
e
⇒
f
e
=
25
×
19
106
≈
4.48
c
m
© examsnet.com
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