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WBJEE 2019 Physics Solved Paper
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© examsnet.com
Question : 28
Total: 40
A point object is placed on the axis of a thin convex lens of focal length 0.05 m at a distance of 0.2 m from the lens and its image is formed on the axis. If the object is now made to oscillate along the axis with a small amplitude of A cm, then what is the amplitude of oscillation of the image ?
[ you may assume,
1
1
+
x
≈ 1 - x, where x << 1 ]
4
A
9
×
10
−
2
m
5
A
9
×
10
−
2
m
A
3
×
10
−
2
m
A
9
×
10
−
2
m
Validate
Solution:
According to the question, we can draw the following diagram
Given,
u
=
−
0.2
m
and
f
=
0.05
m
As we know that,
1
f
=
1
v
−
1
u
.....(i)
1
v
=
1
f
+
1
u
1
v
=
1
0.05
−
1
0.2
1
v
=
100
5
−
10
2
1
v
=
20
−
5
⇒
v
=
1
15
m
Now, differentiating eq. (i), we get
−
d
v
v
2
=
−
d
u
u
2
∴
d
v
=
d
u
v
2
u
2
A
max
=
A
×
(
1
15
)
2
×
(
1
(
−
0.2
)
2
)
A
max
=
A
×
1
225
×
25
A
max
=
A
9
Here A is in cm.
Hence,
A
max
=
A
9
×
10
−
2
m
© examsnet.com
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