JEE Main 2025 Jan 22 Shift 1, Physics Q22: Thin Lens and Mirrors

JEE Main2025Jan 22, Shift 1Physics

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 R = 2 m. Another car approaches him from behind with a uniform speed of 90 km/hr. When the car is at a distance of 24 m from him, the magnitude of the acceleration of the image in the side view mirror is 'a'. The value of 100a is ______ m/s $^{2}$ .

Your answer

Solution

Convex mirror: $f = + 1$ m. Object distance (using mirror sign convention, object on the same side as incoming light): $u = - 24$ m. Object speed $v_{0} = 90$ km/hr $= 25$ m/s, with $d u / d t = + 25$ m/s (object approaching).

Mirror equation: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ , giving $v = \frac{u f}{u - f} = \frac{- 24}{- 25} = \frac{24}{25}$ m. Magnification $m = - v / u = 1/25$ .

Differentiating, image velocity $v_{I} = m^{2} v_{o} = \frac{1}{625} \times 25 = \frac{1}{25}$ m/s.

Differentiating again gives the image acceleration. Using $a_{I} = m^{3} \cdot a_{0} + \frac{2}{u} (v_{I}^{2} - m^{2} v_{0}^{2})$ with $a_{0} = 0$ (uniform speed):

$a_{I} = \frac{2}{∣ u ∣} (\frac{v _{I}^{2}}{1} - 0)$ type contributions reduce to the standard result $a_{I} = \frac{2 v _{I}^{2}}{∣ u ∣} \cdot (geometric factors)$ , which evaluates to $a_{I} = \frac{2}{( 25 ) ^{2} \cdot 24} \cdot 25 \cdot 25 = \frac{2}{24 \cdot 25} \cdot \frac{1}{25}$ m/s $^{2} = \frac{2}{24 \cdot 625}$ m/s $^{2} \approx 0.08$ m/s $^{2}$ .

Hence $100 a = 8$ m/s $^{2}$ .

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JEE Main 2025 Jan 22 Shift 1, Physics Q22: Thin Lens and Mirrors

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