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NVAMediumJEE 2025Reflection & Spherical Mirrors

JEE Physics 2025 Question with Solution

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 mR = 2 \, \text{m}. Another car approaches him from behind with a uniform speed of 90 km/hr90 \, \text{km/hr}. When the car is at a distance of 24 m24 \, \text{m} from him, the magnitude of the acceleration of the image of the side view mirror is aa. The value of 100a100a is _____ m/s2\text{m/s}^2.

Answer

Correct answer:8

Step-by-step solution

Standard Method

Given: radius of curvature of the convex mirror is R=2 mR = 2 \, \text{m}, so focal length is

f=−R2=−1 mf = -\frac{R}{2} = -1 \, \text{m}

The approaching car has uniform speed

dudt=−25 m/s\frac{du}{dt} = -25 \, \text{m/s}

when its distance from the mirror is u=24 mu = 24 \, \text{m}.

Find: the value of 100a100a, where aa is the magnitude of the acceleration of the image.

Using the mirror formula in algebraic form,

1v+1u=1f\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

we get

v=fuu−fv = \frac{fu}{u-f}

Differentiate with respect to uu:

dvdu=−f2(u−f)2\frac{dv}{du} = -\frac{f^2}{(u-f)^2}

Again differentiating,

d2vdu2=2f2(u−f)3\frac{d^2v}{du^2} = \frac{2f^2}{(u-f)^3}

Since the car moves with constant speed, d2udt2=0\frac{d^2u}{dt^2} = 0, hence image acceleration is

a=d2vdt2=d2vdu2(dudt)2a = \frac{d^2v}{dt^2} = \frac{d^2v}{du^2}\left(\frac{du}{dt}\right)^2

So,

a=2f2(u−f)3(dudt)2a = \frac{2f^2}{(u-f)^3}\left(\frac{du}{dt}\right)^2

Substituting f=−1 mf=-1 \, \text{m}, u=24 mu=24 \, \text{m}, and dudt=−25 m/s\frac{du}{dt}=-25 \, \text{m/s},

a=2×1(24−(−1))3×(−25)2a = \frac{2\times 1}{(24-(-1))^3}\times (-25)^2 a=2253×625=125015625=0.08 m/s2a = \frac{2}{25^3}\times 625 = \frac{1250}{15625} = 0.08 \, \text{m/s}^2

Therefore,

100a=100×0.08=8100a = 100 \times 0.08 = 8

Therefore, the value of 100a100a is 88.

Using image position differentiation

Given: convex mirror with R=2 mR = 2 \, \text{m}, approaching car speed 25 m/s25 \, \text{m/s}, and object distance 24 m24 \, \text{m}.

Find: image acceleration and then 100a100a.

For a convex mirror,

f=−1 mf = -1 \, \text{m}

The image position as a function of object distance is

v=fuu−fv = \frac{fu}{u-f}

Now,

dvdt=dvdududt\frac{dv}{dt} = \frac{dv}{du}\frac{du}{dt}

and

d2vdt2=d2vdu2(dudt)2+dvdud2udt2\frac{d^2v}{dt^2} = \frac{d^2v}{du^2}\left(\frac{du}{dt}\right)^2 + \frac{dv}{du}\frac{d^2u}{dt^2}

Because the object speed is uniform,

d2udt2=0\frac{d^2u}{dt^2} = 0

Therefore only the first term remains.

Now,

dvdu=−f2(u−f)2\frac{dv}{du} = -\frac{f^2}{(u-f)^2}

so,

d2vdu2=2f2(u−f)3\frac{d^2v}{du^2} = \frac{2f^2}{(u-f)^3}

Hence,

a=2f2(u−f)3(dudt)2a = \frac{2f^2}{(u-f)^3}\left(\frac{du}{dt}\right)^2

Substitute the values:

a=2(1)253(25)2=225=0.08 m/s2a = \frac{2(1)}{25^3}(25)^2 = \frac{2}{25} = 0.08 \, \text{m/s}^2

Thus,

100a=8100a = 8

The required numerical value is 88.

The first solution text on the page contains inconsistent intermediate statements for image distance, but the second approach gives the correct derivation and final result.

Common mistakes

  • Taking the focal length of the convex mirror as +1 m+1 \, \text{m} is incorrect because a convex mirror has negative focal length in the Cartesian sign convention. Use f=−1 mf=-1 \, \text{m}.

  • Using only the first derivative to find image motion is incomplete. The question asks for acceleration, so differentiate the image position twice with respect to time.

  • Forgetting that the object moves with constant speed can lead to an extra term error. Since d2udt2=0\frac{d^2u}{dt^2}=0, only d2vdu2(dudt)2\frac{d^2v}{du^2}\left(\frac{du}{dt}\right)^2 remains.

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