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.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 22 on this concept.

CBSE 2026Set 55/1/11 markMCQ
$10\ \text{cm}$
›Reveal solutionSolution

Cutting a symmetric concave lens through its middle (perpendicular to the principal axis) leaves each piece with only one curved surface, so each plano-concave piece has half the power — and therefore double the focal length: 20 cm, option (A).

Concept and Intuition

The lens maker's formula relates a thin lens's focal length to its two radii of curvature and the refractive index of the material:
$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
A symmetric biconcave lens has two curved surfaces, and each contributes equally to the total power. When the lens is cut through its middle by a plane perpendicular to the principal axis, each piece keeps one original curved surface and gains a flat face. A flat surface has an infinite radius of curvature, so it contributes nothing to the power — each piece is left with only half the original bending power.

Step-by-Step Solution

1. Apply the formula to the original biconcave lens.

Use the New Cartesian sign convention with light travelling left to right. For a biconcave lens of equal radii of magnitude $R$ : the first surface's centre of curvature lies on the incident (left) side, so $R_1 = -R$ ; the second surface's centre of curvature lies on the outgoing (right) side, so $R_2 = +R$ . Then
$\frac{1}{f} = (\mu-1)\left(\frac{1}{-R} - \frac{1}{+R}\right) = -\frac{2(\mu-1)}{R}$
The negative sign confirms a diverging lens. With $f = -10\ \text{cm}$ :
$\frac{\mu-1}{R} = \frac{1}{20}\ \text{cm}^{-1}$
2. Apply it to one plano-concave piece.

Each piece keeps one curved surface ( $R_1 = -R$ ) and gains a flat cut face ( $R_2 = \infty$ ):
$\frac{1}{f'} = (\mu-1)\left(\frac{1}{-R} - \frac{1}{\infty}\right) = -\frac{\mu-1}{R} = -\frac{1}{20}\ \text{cm}^{-1}$
3. Read off the result.
$f' = -20\ \text{cm}$
Each piece is still diverging, with a focal length of magnitude $20\ \text{cm}$ — double that of the original lens.

Tip

Shortcut: the two identical curved surfaces of a symmetric lens contribute equal power. Removing one (replacing it with a flat face) halves the power $P = 1/f$ , and halving the power doubles the focal length: $|f'| = 2 \times 10 = 20\ \text{cm}$ . No radii needed.

Watch out

A common mistake is to think that cutting the lens "in half" halves the focal length, giving $5\ \text{cm}$ . It is the power that halves, not the focal length — so the focal length doubles. (Note the cut here is perpendicular to the principal axis; cutting along the axis instead leaves both curved surfaces intact and does not change the focal length at all.)

✓Final answer
$The focal length of each plano-concave lens is 20 cm, which corresponds to option (A) .$
CBSE 2026Set 55/3/11 markMCQ
$\mu$
›Reveal solutionSolution
We calculate the focal lengths of the plano-convex and equi-concave lenses separately using the lens maker's formula, applying the correct sign conventions for radii of curvature. Then, we combine these focal lengths to find the equivalent focal length of the system. The focal length of the combination is $\boxed{-\dfrac{R}{\mu-1}}$ .

Figure — plano-convex and equi-concave lens combination

Concept and Intuition

To find the focal length of a combination of thin lenses kept in contact, we first need to determine the focal length of each individual lens. The fundamental tool for this is the Lens Maker's Formula.

Lens Maker's Formula

The focal length $f$ of a thin lens made of a material with refractive index $\mu$ (relative to the surrounding medium, usually air, for which $\mu_{air} = 1$ ) is given by:
$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Sign Convention for Radii of Curvature

We will use the following convention for $R_1$ and $R_2$ in the lens maker's formula, assuming light travels from left to right:

$R_1$ (First Surface):

If the first surface is convex (bulges towards the right), $R_1$ is positive ( $+R$ ).

If the first surface is concave (bulges towards the left), $R_1$ is negative ( $-R$ ).

If the first surface is flat (plano), $R_1$ is infinite ( $\infty$ ).

$R_2$ (Second Surface):

If the second surface is convex (bulges towards the left), $R_2$ is negative ( $-R$ ).

If the second surface is concave (bulges towards the right), $R_2$ is positive ( $+R$ ).

If the second surface is flat (plano), $R_2$ is infinite ( $\infty$ ).

Watch out

The sign convention for $R_1$ and $R_2$ is a common source of error. Always be consistent with the convention you choose. The one outlined above ensures that converging lenses have positive focal lengths and diverging lenses have negative focal lengths when $\mu > 1$ .

Combination of Thin Lenses in Contact

When two thin lenses with focal lengths $f_1$ and $f_2$ are placed coaxially in contact, the focal length $F$ of the combination is given by:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$
Step-by-step Solution

Identify the properties of the plano-convex lens (Lens 1).

Refractive index: $\mu$

First surface: Flat. According to our sign convention, $R_1 = \infty$ .

Second surface: Convex. It bulges towards the left (as seen from the second surface, or its center of curvature is to the left). According to our sign convention, $R_2 = -R$ .

Calculate the focal length of the plano-convex lens ( $f_1$ ).

Using the lens maker's formula:

$\frac{1}{f_1} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Substitute the values for $R_1$ and $R_2$:
$\frac{1}{f_1} = (\mu - 1) \left( \frac{1}{\infty} - \frac{1}{-R} \right)$ $\frac{1}{f_1} = (\mu - 1) \left( 0 + \frac{1}{R} \right)$ $\frac{1}{f_1} = \frac{\mu - 1}{R}$
Therefore, the focal length of the plano-convex lens is:
$f_1 = \frac{R}{\mu - 1}$
This is a positive focal length, as expected for a converging lens.

3. Identify the properties of the equi-concave lens (Lens 2).

* Refractive index: $\mu$

* First surface: Concave. It bulges towards the left. According to our sign convention, $R_1 = -R$ .

* Second surface: Concave. It bulges towards the right. According to our sign convention, $R_2 = +R$ .

Calculate the focal length of the equi-concave lens ( $f_2$ ). Using the lens maker's formula:

$\frac{1}{f_2} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Substitute the values for $R_1$ and $R_2$:
$\frac{1}{f_2} = (\mu - 1) \left( \frac{1}{-R} - \frac{1}{+R} \right)$ $\frac{1}{f_2} = (\mu - 1) \left( -\frac{1}{R} - \frac{1}{R} \right)$ $\frac{1}{f_2} = (\mu - 1) \left( -\frac{2}{R} \right)$ $\frac{1}{f_2} = -\frac{2(\mu - 1)}{R}$
Therefore, the focal length of the equi-concave lens is:
$f_2 = -\frac{R}{2(\mu - 1)}$
This is a negative focal length, as expected for a diverging lens.

5. Calculate the focal length of the combination ( $F$ ).

Using the formula for lenses in contact:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$
Substitute the calculated values for $f_1$ and $f_2$:
$\frac{1}{F} = \frac{\mu - 1}{R} + \left( -\frac{2(\mu - 1)}{R} \right)$ $\frac{1}{F} = \frac{\mu - 1}{R} - \frac{2(\mu - 1)}{R}$
Combine the terms:
$\frac{1}{F} = \frac{(\mu - 1) - 2(\mu - 1)}{R}$ $\frac{1}{F} = \frac{-\left(\mu - 1\right)}{R}$
Therefore, the focal length of the combination is:
$F = -\frac{R}{\mu - 1}$
✓Final answer
$\boxed{-\dfrac{R}{\mu-1}}$
CBSE 2026Set DS1 mark
$Q. In which the power of a lens will be large — in air or water?$
›Reveal solutionSolution
Power is larger in air, because the glass–water relative refractive index is smaller than the glass–air one.

Concept. By the lens-maker's formula the power of a lens depends on the refractive index of the lens relative to its surroundings:

$P = \frac1f = \left({}_{m}n_g - 1\right)\left(\frac1{R_1}-\frac1{R_2}\right),$

where ${}_{m}n_g = n_g/n_m$ is the index of glass with respect to the medium.

In air ( $n_m \approx 1$ ): ${}_{a}n_g \approx 1.5$ , so $({}_{a}n_g-1) \approx 0.5$ .

In water ( $n_m \approx 1.33$ ): ${}_{w}n_g = 1.5/1.33 \approx 1.13$ , so $({}_{w}n_g-1) \approx 0.13$ .

Since the factor $({}_{m}n_g-1)$ is much smaller in water, the focal length increases and the power decreases in water. Hence the power is larger in air.

✓Final answer
$The lens has larger power in air (in water its focal length increases and power falls).$
CBSE 2025Set D1 markMCQ
$Q.A convex lens is dipped in a liquid, whose refractive index is equal to the refractive index of the material of the lens. Then its focal length will (A) become zero (B) become infinity (C) reduce (D) increase$
›Reveal solutionSolution

Lensmaker's formula has a factor (n_lens/n_medium − 1); if the two indices are equal this factor is zero, so 1/f = 0 and f → ∞.

By the lensmaker's formula in a medium,

1/f = (n_lens/n_medium − 1)(1/R₁ − 1/R₂)

If the liquid's refractive index equals the lens material's index, then n_lens/n_medium = 1, so the factor (1 − 1) = 0.

Hence 1/f = 0, which means f = ∞. The lens becomes optically 'invisible' — light passes straight through without bending, exactly as if there were no lens.

✓Final answer
$(B) become infinity.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. When monochromatic red light is used instead of blue light in a convex lens, its focal length$
(a) does not change
(b) increases
(c) decreases
(d) remain same
›Reveal solutionSolution

Red light has a lower refractive index than blue (dispersion), and $f \propto 1/(n-1)$ , so lower $n$ gives a larger $f$ .

By the lens maker's formula, $\dfrac{1}{f} = (n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$ , so $f \propto \dfrac{1}{(n-1)}$ for fixed geometry. Due to dispersion, the refractive index of a material is slightly higher for blue light than for red light ( $n_{blue} > n_{red}$ , since blue light bends more). Using red light instead of blue therefore means a smaller $(n-1)$ , giving a larger focal length — the focal length increases when red light is used instead of blue.

✓Final answer
$(b) increases.$
CBSE 2024Set FS1 mark
$Q. Find the ratio of focal length of lens in air and that of lens when it is immersed in liquid.$
›Reveal solutionSolution

$\dfrac{f_{air}}{f_{liquid}}=\dfrac{n_g-n_l}{n_l\,(n_g-1)}$ , where $n_g$ , $n_l$ are the refractive indices of glass and the liquid.

Concept. The lens maker's formula uses the index of the lens relative to its surroundings:

$\frac{1}{f}=\left(\,{}_{med}n_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right).$

In air ( $n_{air}=1$ ): $\dfrac{1}{f_{air}}=(n_g-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right).$

In liquid: the glass index relative to the liquid is $n_g/n_l$ , so

$\frac{1}{f_{liquid}}=\left(\frac{n_g}{n_l}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right).$

Divide.

$\frac{f_{air}}{f_{liquid}}=\frac{(n_g/n_l)-1}{n_g-1}=\frac{n_g-n_l}{n_l\,(n_g-1)}.$

Since $n_l>1$ , the liquid lens has the larger focal length (weaker lens).

✓Final answer
$\dfrac{f_{air}}{f_{liquid}}=\dfrac{n_g-n_l}{n_l\,(n_g-1)}$
CBSE 2024Set ANNUAL1 mark
$Q. The power of a lens is greater in water or air?$
›Reveal solutionSolution
A glass lens is more powerful in air than in water, because water's refractive index is closer to that of glass.

Power of a lens, $P = 1/f$ , and from the lens maker's formula:

$\frac{1}{f} = \left(\frac{{}_a\mu_g}{{}_a\mu_m} - 1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

where ${}_a\mu_m$ is the refractive index of the surrounding medium (relative to air).

In air, ${}_a\mu_m = 1$ , so the relative refractive index of glass w.r.t. the medium is large ( $\approx 1.5$ ), giving a small $f$ and large $P$ .

In water, ${}_a\mu_m \approx 1.33$ , close to glass's $1.5$ , so the relative refractive index is small; $f$ increases and $P$ decreases (the lens stays converging, since glass is still denser than water, but it becomes much weaker).

✓Final answer
$Power is greater in air. In water the effective refractive index of glass relative to the surrounding medium drops (water's index is close to glass's), which increases the focal length and reduces the power.$
CBSE 2024Set ANNUAL1 markMCQ
$Q.The refractive index of the material of a double equiconvex lens is 2.5. If R be its radius of curvature, then its focal length is$
(a) 0
(b) R/3
(c) 2R
(d) 3R.
›Reveal solutionSolution

Applying the lens maker's formula to an equiconvex lens (equal radii of curvature, opposite sign) with refractive index 2.5 gives $f = R/3$ .

The lens maker's formula is

$\dfrac{1}{f} = (\mu - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$

For a double equiconvex lens, both surfaces bulge outward with the same radius of curvature magnitude $R$ . Using the standard sign convention (distances measured from the optical centre, in the direction of incident light being positive): the first surface is convex towards the incoming light, so $R_1 = +R$ ; the second surface's centre of curvature lies on the incoming side, so $R_2 = -R$ .

Given $\mu = 2.5$ :

$\dfrac{1}{f} = (2.5-1)\left(\dfrac{1}{R} - \dfrac{1}{-R}\right) = 1.5 \times \dfrac{2}{R} = \dfrac{3}{R}$

$f = \dfrac{R}{3}$

✓Final answer
$The focal length is R/3. Choice (b).$
CBSE 2024Set ANNUAL1 markMCQ
$Q. The focal length of a glass (μ=1.5) lens in air is 20cm. If it is dipped in water (μ = 4/3), its focal length in water will be -$
(a) 80 cm
(b) 40 cm
(c) 60 cm
(d) 20 cm
›Reveal solutionSolution

A lens's focal length depends on its refractive index relative to the surrounding medium; a converging lens weakens (longer $f$ ) when moved from air into a denser medium like water.

Lensmaker's equation: $\dfrac{1}{f} = \left(\dfrac{\mu_{lens}}{\mu_{medium}} - 1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$

In air: $\mu_{rel} = 1.5/1 = 1.5$

$\dfrac{1}{20} = (1.5-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right) \Rightarrow \left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right) = \dfrac{1}{10}$

In water: $\mu_{rel} = 1.5/(4/3) = 1.125$

$\dfrac{1}{f_w} = (1.125-1)\times\dfrac{1}{10} = 0.125\times0.1 = \dfrac{1}{80}$

$f_w = 80\text{ cm}$

✓Final answer
$= 80\text{ cm}$
CBSE 2023Set BS1 mark
$n = \dfrac{3}{2}$
›Reveal solutionSolution

With glass and surrounding liquid of the same index, $f\to\infty$ ; the lens stops acting as a lens.

The lens-maker's formula in a medium uses the relative refractive index $n_{rel}=\dfrac{n_{glass}}{n_{liquid}}$ :

$\frac{1}{f}=(n_{rel}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right).$

Here $n_{glass}=n_{liquid}=\dfrac{3}{2}$ , so $n_{rel}=1$ and $(n_{rel}-1)=0$ . Therefore $\dfrac{1}{f}=0\Rightarrow f=\infty$ . There is no refraction at the surfaces (no change of speed on entering the glass), so the convex lens becomes optically inactive, transmitting light like a plane parallel glass slab.

✓Final answer
$f=\infty$
CBSE 2023Set F1 markMCQ
$Q.The radius of curvature of each surface of a biconvex lens is 20 cm and the refractive index of the material of the lens is 1.5. The focal length of the lens is (A) 20 m (B) 1/20 m (C) 20 cm (D) 1/20 cm$
›Reveal solutionSolution

Lens maker's formula gives f = 20 cm for a biconvex lens with R = 20 cm and n = 1.5.

For a thin lens, the lens maker's formula is

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

For a biconvex lens with each surface radius 20 cm, using sign convention $R_1 = +20$ cm and $R_2 = -20$ cm:

$\dfrac{1}{f} = (1.5-1)\left(\dfrac{1}{20} - \dfrac{1}{-20}\right) = 0.5\left(\dfrac{1}{20} + \dfrac{1}{20}\right) = 0.5 \times \dfrac{2}{20} = \dfrac{1}{20}\ \text{cm}^{-1}$

$f = 20\ \text{cm}$

✓Final answer
$(C) 20 cm.$
CBSE 2023Set ANNUAL1 markMCQ
$Q.A convex lens is immersed in a liquid of refractive index greater than that of glass. It will behave as a -$
(a) convergent lens
(b) divergent lens
(c) plane glass
(d) homogeneous liquid
›Reveal solutionSolution

When the surrounding medium is optically denser than the lens material, a convex lens's effect reverses.

By the lens maker's formula, $\dfrac{1}{f} = \left(\dfrac{n_{lens}}{n_{medium}}-1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$ . For a convex lens in air, $n_{lens}>n_{medium}$ , so $\left(\dfrac{n_{lens}}{n_{medium}}-1\right)>0$ and $f$ is positive (converging). If the lens is immersed in a liquid whose refractive index is greater than that of the glass ( $n_{medium}>n_{lens}$ ), then $\left(\dfrac{n_{lens}}{n_{medium}}-1\right)$ becomes negative, flipping the sign of $f$ — the convex-shaped lens now behaves as a diverging (divergent) lens, even though its physical shape is unchanged.

✓Final answer
$(b) It behaves as a divergent lens.$

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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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