Question

A particle executes a simple harmonic motion of amplitude $1.0cm$ along the principal axis of convex lens of focal length $12cm$. The mean position of oscillation is at $20cm$ from the lens. Find the amplitude of oscillation of the image of the particle.

Answer 1

When the object is at $19cm$ from the lens, let the image will be at, $v_1$.

$\Rightarrow \frac{1}{v_1}-\frac{1}{{\mu }_1}=\frac{1}{f}$

$\Rightarrow \frac{1}{v_1}-\frac{1}{-19}=\frac{1}{12}$

$\Rightarrow v_1=32.57cm$

Again, when the object is at $21cm$ from the lens, let the image will be at $v_2$

$\Rightarrow \frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f}$

$\Rightarrow \frac{1}{v_2}+\frac{1}{21}=\frac{1}{12}$

$\Rightarrow v_2=28cm$

$\therefore$ Amplitude of vibration of the image is $A=\frac{A^{\prime }{B}^{\prime }}{2}=\frac{v_1-v_2}{2}$

$\Rightarrow A=\frac{32.57-28}{2}=2.285cm$