Question 278: Ray Optics

Question

The driver sitting inside a parked car is watching vehicles approaching from behind with the help of his side view mirror, which is a convex mirror with radius of curvature $\mathrm{R}=2 \mathrm{~m}$. Another car approaches him from behind with a uniform speed of $90 \mathrm{~km} / \mathrm{hr}$. When the car is at a distance of 24 m from him, the magnitude of the acceleration of the image of the car in the side view mirror is ' $a$ '. The value of $100 a$ is __________ $\mathrm{m} / \mathrm{s}^2$.

Accepted Answer

Given, R = 2m, So, f = 1m (as $f=\frac{R}{2}$)
$${V_0} = 90\,km/hr = 90 	imes {{1000} \over {3600}} = 25\,m/s$$
$v_0$ is uniform, so $${a_0} = 0$$
$$u =  - 24\,m$$
Using mirror formula,
$${1 \over v} + {1 \over u} = {1 \over f}$$
$$ \Rightarrow {1 \over v} + {1 \over { - 24}} = {1 \over 1} \Rightarrow {1 \over v} = 1 + {1 \over {24}} = {{25} \over {24}}$$
$$ \Rightarrow v = {{24} \over {25}}m$$
$$m = {{ - v} \over u}$$
by differentiating mirror formula,
$$ - {1 \over {{v^2}}}{{dv} \over {dt}} - {1 \over {{u^2}}}{{du} \over {dt}} = 0$$
$$ \Rightarrow  - {1 \over {{v^2}}}{v_I} - {1 \over {{u^2}}}{v_0} = 0$$ ...... (1)
$$ \Rightarrow {v_I} = {{ - {v^2}} \over {{u^2}}}{v_0} =  - {{{{24}^2}} \over {{{25}^2}}}$$
$$ \Rightarrow {v_I} = {{ - 1} \over {25}}$$ m/s
by differentiating (1) w.r.t. time,
$${{ - 1} \over {{v^2}}}{a_i} + {2 \over {{v^3}}}v_I^2 - {1 \over {{u^2}}}{a_0} + {2 \over {{u^3}}}v_0^2 = 0$$
$${a_I} = {2 \over v}v_I^2 + {{2{v^2}} \over {{u^3}}}v_0^2$$
$$ = {2 \over {24}}{1 \over {{{25}^2}}} + {2 \over { - {{24}^3}}}{{{{24}^2}} \over {{{25}^2}}}{25^2}$$
$$ = {1 \over {12}}\left[ {{1 \over {25}} - 1} ight] = {1 \over {12}} 	imes {{ - 24} \over {25}} =  - {2 \over {25}}$$
$$ \Rightarrow 100\,{a_I} = {{ - 2} \over {25}} 	imes 100 = 8\,m/{s^2}$$

Ray Optics

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