A particle executes a simple harmonic motion of amplitude 1.0 cm along the principal axis of a convex lens of focal length 12 cm. The mean position of oscillation is at 20 cm from the lens. Find the amplitude of oscillation of the image of the particle.
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Answered by
4
Given :
f= 12cm
a)when the particle[ object ] is at 19cm , let the imge be formed at v1
1/f=1/v-1/u
1/f1+1/u1=1/v1
1/12-1/19
=19-12/228
1/v1=7/228
v1=228/7
v1=32.57cm
b) when particle is at 21cm then image is formed at V2
1/f=1/v-1/u
1/12-1/21= 1/v2
21-12/252=1/v2
v2=252/9
v2=28cm
Amplitude of oscillation of the image particle = v1-v2/2= 32.57- 28/2=4.57/2=2.285cm
Answered by
1
Answer:
When the object is at 19cm from the lens, let the image will be at, v
1
.
⇒
v
1
1
−
μ
1
1
=
f
1
⇒
v
1
1
−
−19
1
=
12
1
⇒v
1
=32.57cm
Again, when the object is at 21cm from the lens, let the image will be at v
2
⇒
v
2
1
−
u
2
1
=
f
1
⇒
v
2
1
+
21
1
=
12
1
⇒v
2
=28cm
∴ Amplitude of vibration of the image is A=
2
A
′
B
′
=
2
v
1
−v
2
⇒A=
2
32.57−28
=2.285cm
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