An object approaches a convergent lens from the left of the lens with a uniform speed 5 m/s and stops at the f

NCERT Exemplar · Q3

Q. An object approaches a convergent lens from the left of the lens with a uniform speed 5 m/s and stops at the focus. The image

(a) moves away from the lens with an uniform speed 5 m/s.

(b) moves away from the lens with an uniform accleration.

(d) moves towards the lens with a non-uniform acceleration.

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Concept understanding — Lens Maker's Formula

The Intuition: Why a Lens Bends Light

A lens works because light slows down when it enters glass. When a wavefront hits a curved surface at an angle, different parts of it slow down at different moments, and the wavefront bends. The stronger the curvature, the more it bends.

A lens has two surfaces. Each surface bends light by an amount that depends on its radius of curvature $R$ and the refractive index $n$ of the glass. The net bending — the focal length $f$ — is the combined effect of both surfaces.

If you had a single spherical surface separating air from glass, its contribution to bending power is $\frac{n-1}{R}$ . A lens has two such surfaces: light goes from air into glass at the first surface, then from glass back into air at the second. Because the two surfaces face opposite directions relative to the travelling light, their radii typically carry opposite signs.

Note

This uses the New Cartesian Sign Convention (the one used in NCERT and CBSE): all distances are measured from the optical centre, and the direction the incident light travels in is taken as positive. So $R$ is positive if the centre of curvature lies on the side the light is travelling towards (the outgoing side), and negative if it lies on the side the light is travelling from (the incident side).

The Precise Statement

For a thin lens (thickness negligible compared to the radii), the Lens Maker's Formula is:

\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

where:

$f$ is the focal length of the lens (positive for converging, negative for diverging)
$n$ is the refractive index of the lens material relative to the surrounding medium (usually air)
$R_1$ is the radius of curvature of the first surface (the one light reaches first)
$R_2$ is the radius of curvature of the second surface

\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

How to Apply It: A Worked Example

Take a biconvex lens made of glass ( $n = 1.5$ ) with both surfaces having the same radius of curvature magnitude, $20$ cm.

Light travels left to right. The first surface bulges toward the incoming light, so its centre of curvature lies to the right of the surface — on the side the light is travelling towards. By the rule above, $R_1 = +20$ cm.

The second surface also bulges outward (away from the lens), so its centre of curvature lies to the left of that surface — on the side the light is travelling from. So $R_2 = -20$ cm.

Plug in:

\frac{1}{f} = (1.5 - 1)\left(\frac{1}{20} - \frac{1}{-20}\right) = 0.5 \times \left(\frac{1}{20} + \frac{1}{20}\right) = 0.5 \times \frac{2}{20} = \frac{1}{20}

So $f = +20$ cm. Positive means converging — correct for a biconvex lens.

Watch out

The most common mistake is getting the sign of $R_2$ wrong. For a biconvex lens, $R_1$ is positive and $R_2$ is negative. For a biconcave lens, it's the reverse: $R_1$ negative, $R_2$ positive. Always sketch the lens and mark where each surface's centre of curvature actually sits.

Why the Formula Works (Brief Derivation)

The derivation applies the refraction-at-a-single-spherical-surface formula twice — once at each surface — and combines the results. For a thin lens, the image formed by the first surface acts as the object for the second surface.

Each surface contributes a "power" $P = \dfrac{n_2 - n_1}{R}$ , where $n_1$ and $n_2$ are the refractive indices on either side of that surface. At the first surface, $n_1 = 1$ (air), $n_2 = n$ (glass). At the second, $n_1 = n$ , $n_2 = 1$ . Adding the two contributions (with the sign convention above) gives the Lens Maker's Formula.

When Does It Break?

The formula assumes:

The lens is thin — thickness is negligible compared to the radii.
Light rays are paraxial — close to the axis, making small angles with it.
The surrounding medium is air on both sides (if not, $n$ is replaced by the ratio $n_{\text{lens}}/n_{\text{medium}}$ ).

In exam problems these assumptions almost always hold. The formula is the direct route from a lens's geometry to its focal length — and, by extension, to how strongly it converges or diverges light.

The Lens Maker's Formula is one of the most important derivations in the NCERT Class 12 Physics Ray Optics chapter, and 'lens maker's formula derivation class 12 physics' or 'lens maker's formula important questions' are frequently searched terms during board and JEE Main preparation. Getting the sign convention for R1 and R2 right, as shown here, is also the single most common source of error in JEE Main and NEET optics numericals.

Because the lens equation is nonlinear, an object approaching a convergent lens at constant speed does not produce an image moving at constant speed - as the object nears the focus, the image's speed grows without bound and its acceleration keeps changing. This matches option (c): the image moves away from the lens with a non-uniform acceleration.

Setting up with the lens formula

Using the Cartesian sign convention (light travels left to right, distances measured from the lens): the object is real and to the left, so its distance $u$ is negative; the image distance is $v$ ; the focal length of a convergent lens is $f>0$ . The thin-lens formula is

\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \quad\Longrightarrow\quad v = \frac{uf}{u+f}.

The object starts far away ( $u\to-\infty$ ) and moves toward the lens at a constant speed of $5\ \text{m/s}$ , stopping right at the focus ( $u\to -f$ ).

Relating image velocity to object velocity

Differentiate the lens equation with respect to time:

-\frac{1}{v^2}\frac{dv}{dt} + \frac{1}{u^2}\frac{du}{dt} = 0 \quad\Longrightarrow\quad \frac{dv}{dt} = \left(\frac{v}{u}\right)^2\frac{du}{dt}.

Since $v/u = f/(u+f)$ (directly from the lens formula above),

\frac{dv}{dt} = \left(\frac{f}{u+f}\right)^2\frac{du}{dt}.

The object moves toward the lens at constant speed $5\ \text{m/s}$ : since $u$ is negative and its magnitude is shrinking, $u$ is increasing, so $\dfrac{du}{dt}=+5\ \text{m/s}$ (constant). So

\frac{dv}{dt} = \frac{5f^2}{(u+f)^2}.

What happens as the object nears the focus

As $u\to -f$ (approaching from $u<-f$ ), the denominator $(u+f)\to0^-$ , so $(u+f)^2\to0^+$ and

\frac{dv}{dt} \longrightarrow +\infty.

The image velocity $dv/dt$ is positive, meaning $v$ increases - the (real) image moves further to the right, i.e. away from the lens, and its speed grows without bound as the object approaches the focus.

Is the acceleration uniform or not?

Differentiate once more:

\frac{d^2v}{dt^2} = \frac{d}{dt}\left[\frac{5f^2}{(u+f)^2}\right] = -\frac{10f^2}{(u+f)^3}\cdot\frac{du}{dt} = -\frac{50f^2}{(u+f)^3}.

Since $u<-f$ throughout the approach, $(u+f)<0$ , so $(u+f)^3<0$ , making $\dfrac{d^2v}{dt^2}>0$ - and, crucially, this second derivative itself keeps changing (it depends on $(u+f)^3$ , which is shrinking toward zero), so the image's acceleration is not constant - it is a genuinely non-uniform acceleration, growing ever larger as the object nears the focus.

Checking the options

(a) "moves away with uniform speed $5\ \text{m/s}$ " - false, the image speed diverges, it isn't constant or even equal to the object's speed.
(b) "moves away with uniform acceleration" - false, we just showed $d^2v/dt^2$ is not constant.
(c) "moves away with a non-uniform acceleration" - true, exactly as derived.
(d) "moves towards the lens" - false, the (real) image moves away from the lens (to larger $v$ ), not towards it.

✓Final answer

Option (c) is correct: the image moves away from the lens with a non-uniform (ever-increasing) acceleration, diverging in speed as the object approaches the focus.

Method: Differentiating the Lens/Mirror Equation to Find Image Velocity and Acceleration

This method solves problems where an object is IN MOTION and you must find how the image moves — its velocity, or whether that velocity/acceleration is uniform — without tracking the image position frame by frame.

Steps

Step 1: Write the governing equation with $u$ and $v$ as functions of time

For a lens (Cartesian sign convention, real object so $u<0$ ):

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

Treat both $u(t)$ and $v(t)$ as time-dependent quantities linked by this one equation at every instant.

Step 2: Differentiate implicitly with respect to time

-\frac{1}{v^2}\frac{dv}{dt} + \frac{1}{u^2}\frac{du}{dt} = 0 \;\Rightarrow\; \frac{dv}{dt} = \left(\frac{v}{u}\right)^2\frac{du}{dt}

This directly relates the image's velocity to the object's velocity through the instantaneous ratio $v/u$ — never assume the image moves at the same speed as the object.

Step 3: Express $v/u$ purely in terms of known quantities

From the lens equation itself, $v/u = f/(u+f)$ , so substitute to get $dv/dt$ as a function of $u$ , $f$ , and the (given, often constant) $du/dt$ alone.

Step 4: Differentiate a second time to check whether the motion is uniform

Differentiate the Step-3 expression again with respect to time. If the result still depends on $u$ (rather than collapsing to a constant), the image's acceleration is genuinely non-uniform — it changes as $u$ changes, even though the object's own velocity was constant.

Step 5 (Applying to this problem): Examine the limiting behaviour

Check what happens as $u$ approaches any special value in the problem (e.g. the focus, $u\to-f$ ). If a factor like $(u+f)$ appears in the denominator, the image's speed and/or acceleration can diverge there — this qualitative limiting check often answers a "uniform vs non-uniform" or "finite vs unbounded" question without needing a specific numeric answer.

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Q.An object approaches a convergent lens from the left of the lens with a uniform speed 5 m/s and stops at the focus. The image

Setting up with the lens formula

Relating image velocity to object velocity

What happens as the object nears the focus

Is the acceleration uniform or not?

Checking the options

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