Physics, asked by BrainlyHelper, 1 year ago

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.

Answers

Answered by prmkulk1978
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 Harshikesh16726
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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