JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Geometrical Optics

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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 of the side view mirror is 'a'. The value of 100a is ______

m / s^{2}

Answer: 8

Solution

The convex mirror has radius

R = 2

m, so its focal length is

f = 1

m (positive, by the sign convention for a convex mirror). The approaching car is at

u = - 24

m and closes in at

V = 90 km/hr = 25 m/s

, so

\frac{d u}{d t} = + 25 m/s

(u increasing toward zero as the gap shrinks).

From the mirror formula

v = \frac{u f}{u - f}

, differentiating with respect to

u

\frac{d v}{d u} = \frac{f ( u - f ) - u f}{( u - f ) ^{2}} = - \frac{f ^{2}}{( u - f ) ^{2}} .

Since the object moves at constant speed (

d u / d t = V

is constant), the image's acceleration comes from differentiating a second time:

\frac{d ^{2} v}{d t ^{2}} = V^{2} \frac{d ^{2} v}{d u ^{2}} = V^{2} \cdot \frac{2 f ^{2}}{( u - f ) ^{3}} .

Substituting

f = 1

u = - 24

(so

u - f = - 25

) and

V = 25

a_{I} = \frac{( 25 ) ^{2} \times 2 \times 1 ^{2}}{( - 25 ) ^{3}} = \frac{1250}{- 15625} = - 0.08 m/s^{2} .

The magnitude of the image acceleration is

0.08 m/s^{2}

, so

100 a = 100 \times 0.08 = 8.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Integer answer

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JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Geometrical Optics

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