A motor car is fitted with a convex driving mirror of focal length `25 cm`. Another motor car is `10 m` away from the driving mirror of the first car. Calculate the position of second car seen in the driving mirror. If the second car is overtaking the first car with a relative speed of `20m//s`, how fast will the image of the second car be moving ?
Text Solution
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Here, `f = + 25 cm = (1)/(4)m, u = -10 m` From mirror formula `(1)/(v) + (1)/(u) = (1)/(f)` `(1)/(v) = (1)/(f) - (1)/(u) = (4)/(1) + (1)/(10) = (41)/(10)` `v = (10m)/(41) = 0.244 m` As `v` is positive, a virtual and erect image of second car will be formed on the other side of convex mirror of 1st car. Now, `(du)/(dt) = 20 m//s, f = (1)/(4)m, u = - 10 m`. The speed of the image of second car `(du)/(dt) = (-f^(2))/(u - f)^(2) xx (du)/(dt) = -((1//4)/(-10 - 1//4))^(2) xx 20` `= - 1.9 xx 10^(-2)m//s` Negative sign shows that as second car overtakes the first car, its image moves away from the convex driving mirror.
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