Question

A particle is oscillating on the x-axis with an amplitude 2 cm about the point $x_0=10cm$, with a frequency $\omega$. A concave mirror of focal length 5 cm is placed at the origin, Find the incorrect statement.

• A. The image execute periodic motion.
• B. the image executes non periodic motion.
• C. The turning points of the image are asymmethic w.r.t. the image of the point at x=10 cm.
• D. The distance between the turning points of the oscillation of the image is $\frac{100}{21}$

Answer 1

Answer: (B)

the image executes non periodic motion.

Given, object distance $\left(u\right)=-10cm$

focal length $\left(f\right)=-5cm$

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$(mirror formuli

where, $v=$ image distance

$\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

$\Rightarrow \frac{1}{v}=-\frac{1}{5}+\frac{1}{10}=\frac{1}{10}$

$\Rightarrow v=-10$

And

$v=10cm,u_1=-8cm,u_2=-12cm$

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u_2}=-\frac{1}{5}+\frac{1}{12}=-\frac{7}{60}$

$\Rightarrow v_2=-\frac{60}{7}cm$

$\frac{1}{v_1}=\frac{1}{f}-\frac{1}{u_1}=\frac{1}{5}+\frac{1}{8}=\frac{-3}{40}$

$\Rightarrow v_1=-\frac{40}{3}cm$

Difference between $\begin{array}{cc}{V}_1\&V_2& =K_2-V_1\\ & =\frac{-60}{7}+\frac{4Q}{3}\end{array}$

$\begin{array}{c}\Rightarrow V_2-V_1=\frac{100}{21}\\ \text{so, the image executes periodic}\\ \text{motionand the turning points of}\\ \text{the image of the point at}x=10cm.\\ \text{so, option (B) is incorrect.}\end{array}$