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KVPY SA Exam 2015 Question Paper
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© examsnet.com
Question : 69
Total: 80
A point object is placed 20 cm left of a convex lens of focal length f = 5 cm (see the figure). The lens is made to oscillate with small amplitude A along the horizontal axis. The image of the object will also oscillate along the axis with
amplitude
A
9
, out of phase with the oscillations of the lens.
amplitude
A
3
, out of phase with the oscillations of the lens.
amplitude
A
3
, in phase with the oscillations of the lens.
amplitude
A
9
, in phase with the oscillations of the lens
Validate
Solution:
1
f
=
1
v
−
1
u
1
v
=
1
f
+
1
u
v
=
fu
f
+
u
m
=
v
u
=
f
f
+
u
As lens is oscillating with small amplitude A.
∴ Image will oscillate with m
2
A
When lens move left then O will come near to lens thus I will go away. Thus image is oscillating out of phase with respect to lens.
m
=
5
5
−
20
⇒
5
−
15
=
−
1
3
Amplitude of image =
(
1
3
)
2
A
=
A
9
© examsnet.com
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