In short

For a thin convex lens of focal length f, an object at signed distance u from the optical centre forms an image at signed distance v satisfying \tfrac{1}{v} - \tfrac{1}{u} = \tfrac{1}{f}.

Three standard methods to find f (in the CBSE/ISC lab):

  1. Distant-object method — point the lens at a distant tree or window; catch the sharp image on a screen. The screen-to-lens distance is f directly. Quick but accurate only to about 5%.
  2. u–v method — set up on an optical bench. For each object distance u, slide the screen to find a sharp image at v. Tabulate (u, v) for 5–8 positions, then use f = uv/(u-v) (or, better, plot).
  3. 1/v-vs-1/u plot — the u–v data plotted as a straight line with slope +1 and intercepts 1/f on both axes. Read f from either intercept.
\boxed{\;\frac{1}{v} \;=\; \frac{1}{u} + \frac{1}{f}\;}\qquad\text{(Cartesian convention; plot of } 1/v \text{ vs } 1/u).

For a concave mirror of focal length f, the same u–v method works with \tfrac{1}{v} + \tfrac{1}{u} = \tfrac{1}{f}. Plot of 1/v vs 1/u is a straight line with slope -1 and both intercepts at 1/f.

For two thin lenses in contact with focal lengths f_1 and f_2, the equivalent focal length F satisfies \tfrac{1}{F} = \tfrac{1}{f_1} + \tfrac{1}{f_2}. Measure F directly; knowing f_1, you can extract f_2 even if f_2 is of a concave lens (otherwise hard to measure alone, since a concave lens alone forms no real image).

The master technique in every one of these experiments is removal of parallax — slide the screen (or a pin) until its image coincides with the object image no matter how you move your eye. That is the signature of a sharp image, and no lens-bench student should settle for less.

The optical bench in any Class 12 physics lab is a wooden rail about 1.5 metres long, with a millimetre scale down one side and four sliding uprights on it — each upright carries a pin, a lens holder, a mirror holder, or a small white screen. There is a slot for a bright source (usually a bulb behind a wire-mesh object, or an illuminated cross-wire). You can spend half an hour fussing with this bench to measure the focal length of a lens to three significant figures, and the result will be more accurate than any ₹2000 digital measurement device on the market. That is because the bench exploits two deeply clever ideas: parallax as a null criterion for "sharp image," and linear plotting to extract the focal length from many measurements at once, smearing out random error.

This article walks through the three CBSE lab methods for the focal length of a convex lens, the adaptation for a concave mirror, the lens-combination trick for finding the focal length of a concave lens (which has no real image of its own, so cannot be measured directly), and the error sources that separate a good reading from a publishable one. The physics itself — the lens formula 1/v - 1/u = 1/f and the mirror formula 1/v + 1/u = 1/f — is derived in Thin Lenses and Spherical Mirrors. Here, the focus is on how to measure f.

Parallax — the master technique

Before any method, learn the one trick that separates a lab-book answer from a junk answer: removal of parallax.

Point at a distant object — a window across the room. Close one eye and hold up a finger close to your face. Move your head left and right. The window moves relative to the finger. This apparent shift is parallax, and it happens because the finger and the window are at different distances — your eye sees them against different parts of the background.

Now bring the finger to the exact same distance as the window (impossible physically, but bear with me). Move your head. The two no longer shift relative to each other — they move together. Zero relative shift means zero parallax; zero parallax means same distance.

On the optical bench, you exploit this. The "image" of a pin formed by a convex lens is, at the right moment, at a definite distance v from the lens. If you slide a second pin (an image pin) along the bench and observe it alongside the first pin's image, the image pin is at the right position when moving your eye side to side produces no relative shift between the image-of-pin-1 and the real pin-2. That is the condition for v to be the true image distance.

Parallax demonstrationTwo sketches side by side. Left sketch: an eye views two objects A and B at different distances; moving the eye head shows B shifting relative to A. Right sketch: A and B at the same distance; moving the eye shows no shift. Different distances — parallaxeyeABA and B shift relative to each other Same distance — no parallaxeyeABA and B move together
Parallax. Left: objects at different distances appear to shift relative to each other as the eye moves. Right: same distance, no relative shift. This is the null criterion used throughout the optical bench — you slide a pin (or a screen) until the parallax vanishes, and then you read the position on the bench scale.

The modern alternative is to look for a sharp image on a white screen. A parallax method uses pins; a screen method uses a screen. The two are equivalent in the final precision (about 1 mm on a well-set bench), but parallax is more sensitive to small errors and tends to give a crisper null — at the cost of requiring more practice to master.

Method 1 — focal length of a convex lens by the distant-object method

The idea

Parallel rays from a distant object converge at the second focus F_2 after passing through a convex lens. So if you catch the image of a distant object on a screen, the screen sits exactly at the focal plane — the screen-to-lens distance is the focal length.

The procedure

Step 1. Stand at an open window. Hold the convex lens between your eye and a distant tree (at least 20 m away — a window across the compound, a telephone pole, a tall tree line).

Step 2. Hold a small white screen (a piece of card) a few cm behind the lens.

Step 3. Adjust the screen-to-lens distance until the image of the tree on the screen is sharp. The image will be inverted and diminished.

Step 4. Measure the screen-to-lens distance with a metre rule. This is f.

The approximation you are making

The lens formula for an object at distance \lvert u\rvert gives image distance

v \;=\; \frac{uf}{u+f}.

For \lvert u\rvert = 20 m = 2000 cm and f = 20 cm, v = (-2000)(20)/(-2000 + 20) = -40000/-1980 = 20.20 cm — a 1% deviation from f = 20 cm. For \lvert u\rvert = 10 m, the deviation is about 2%. The method is therefore accurate only to a few percent, which is why you follow it up with the u–v method.

Why this gets you started anyway

The distant-object method gives you a quick estimate of f — perhaps f \approx 18\ \text{cm} — within thirty seconds. This tells you, for the u–v method, what range of object distances to use: you need \lvert u\rvert between f and 3f or so to get a real image on the bench, so for f = 18 cm you should place the object somewhere between 20 cm and 60 cm from the lens. Without the rough f, you would hunt blindly.

Method 2 — the u–v method on an optical bench

The setup

Optical bench setup for u-v method on a convex lensA long horizontal bench with a scale running along the top. From left to right: an illuminated object pin at position xO, a lens at position xL, and a screen at position xI. The distances u = xL - xO (object to lens) and v = xI - xL (lens to image) are labelled. A bulb behind the object pin illuminates it. 020406080100120 cm bulb object pin at x_O = 20 lens at x_L = 45 screen at x_I = 102 |u| = x_L − x_Ov = x_I − x_L
Optical bench for the $u$-$v$ method. A bulb illuminates the object pin. The convex lens sits a distance $\lvert u\rvert$ to its right. The screen is moved until a sharp inverted image appears at distance $v$ to the right of the lens. All positions are read off the bench scale.

Place a bulb and an object pin at the left end of the bench (around the 20 cm mark, say). Place the lens upright at around 45 cm. Move a small white screen from the right until a sharp, inverted, real image appears on the screen. Record the three bench positions x_O (object pin), x_L (lens), x_I (screen).

Then \lvert u\rvert = x_L - x_O (always positive here; the signed u is -\lvert u\rvert because the object is to the left of the lens in the Cartesian convention) and v = x_I - x_L (positive, image is to the right).

The procedure — five to eight measurements

Step 1. Keep the object pin fixed. Move the lens to roughly \lvert u\rvert = 25 cm from the object (slightly larger than the rough f you got from Method 1). Move the screen until a sharp image forms. Record x_L, x_I.

Step 2. Move the lens to \lvert u\rvert = 30 cm, find the new screen position. Record.

Step 3. Repeat for \lvert u\rvert = 35, 40, 50, 60, 80 cm. Five to eight readings in total.

Step 4. For each row, compute \lvert u\rvert and v, and then f = \lvert u\rvert v / (\lvert u\rvert + v) (derived below). Tabulate.

Step 5. Average the values of f — or better, plot 1/v vs 1/u (Method 3 below).

Deriving the formula f = uv/(u - v)

Start from the lens formula, sign-correctly:

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

For a real object and real image, u = -\lvert u\rvert and v = +\lvert v\rvert (Cartesian signs). Plug in:

\frac{1}{v} - \frac{1}{-\lvert u\rvert} \;=\; \frac{1}{v} + \frac{1}{\lvert u\rvert} \;=\; \frac{1}{f}.

Why: substituting the signed values converts the lens-formula subtraction into an addition in magnitudes. This is the working form — everyone in the CBSE lab uses 1/\lvert u\rvert and 1/v with a plus sign.

Solve for f:

\frac{1}{f} \;=\; \frac{\lvert u\rvert + v}{\lvert u\rvert v} \quad\Longrightarrow\quad f \;=\; \frac{\lvert u\rvert v}{\lvert u\rvert + v}.

Why: combine the two fractions over the common denominator \lvert u\rvert v, then reciprocate. The formula is symmetric in \lvert u\rvert and v — which reflects the physical fact that the object and image are interchangeable (swap them, and the same lens gives the same pair).

For convenience throughout the lab, you drop the absolute-value bars and use the magnitudes u and v with an understanding that both are positive quantities here:

\boxed{\;f \;=\; \frac{uv}{u+v}\;}\qquad\text{(magnitudes, for real object → real image on a convex lens).}

Method 3 — the 1/v-vs-1/u linear plot

Why a plot beats averaging

A single row of the u–v table gives one estimate of f. Averaging seven rows gives a more reliable answer, but it does not tell you whether the data are consistent. If one of the rows has \lvert u\rvert read off wrong (the student accidentally subtracted the wrong two numbers), simply averaging hides the outlier.

A plot exposes the outlier. When the lens formula 1/v + 1/u = 1/f (magnitudes) is recast as

\frac{1}{v} \;=\; -\,\frac{1}{u} + \frac{1}{f},

the graph of 1/v (vertical axis) versus 1/u (horizontal axis) is a straight line with slope -1 and both intercepts equal to 1/f. Plot your seven data points; any single row that lies off the line is suspicious — probably a mis-read pin or a lens that was misaligned in that measurement.

Reading f from the plot

Two independent readings of f — from the two intercepts — and a check on the slope. Any two agree to ~1% and the slope is within ~5% of -1 → publish. If the intercepts differ, the lens has a non-negligible aberration and f is not uniquely defined.

Interactive: slide the object distance, watch the point move on the plot

Interactive: 1/v vs 1/u plot for a convex lens of f = 20 cmLinear plot of 1/v on the vertical axis versus 1/u on the horizontal axis, both in inverse centimetres. A straight line descending from (0, 0.05) to (0.05, 0) intersects both axes at 1/f = 0.05, corresponding to f = 20 cm. A draggable red point slides along the axis to show the corresponding 1/v value live. 1/u (cm⁻¹)1/v (cm⁻¹) 0.010.020.030.040.05 0.010.020.030.040.05 1/f = 0.05 (y-intercept)1/f = 0.05 (x-intercept) drag along 1/u axis
The $1/v$-vs-$1/u$ plot for a convex lens with $f = 20$ cm. Seven sample data points (dark dots) lie on the straight line $1/v = 1/f - 1/u$. The intercepts on both axes are $1/f = 0.05\ \text{cm}^{-1}$, giving $f = 20$ cm. The readout shows, for any draggable $1/u$, the corresponding $u$, $v$, and $f = uv/(u+v)$ — which always returns 20 cm because every point is on the line.

Why slope = -1 is the check that the lens is thin

The lens formula 1/v - 1/u = 1/f becomes, in lab magnitudes, 1/v + 1/u = 1/f — a straight line with slope -1 on the (1/u, 1/v) plane. If the lens is not thin — if the two spherical surfaces are separated by a non-negligible distance — the formula picks up a second-order correction and the plot bends slightly. A slope of -0.92 or -1.08 is a signature of a thick lens; a slope of -1.00 \pm 0.02 says the thin-lens approximation is good to about 2%.

Method 4 — focal length of a concave mirror

The idea carries over with two small changes.

The mirror formula, signs, and the working equation

The spherical-mirror formula in Cartesian convention is

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

For a concave mirror and a real object forming a real image, both u and v are negative (object and image on the same side as the incident light, to the left of the pole in the CBSE convention) and f is negative. All three quantities being negative means that in magnitudes the formula becomes exactly the same:

\frac{1}{\lvert v\rvert} + \frac{1}{\lvert u\rvert} \;=\; \frac{1}{\lvert f\rvert} \quad\Longrightarrow\quad \lvert f\rvert \;=\; \frac{\lvert u\rvert \lvert v\rvert}{\lvert u\rvert + \lvert v\rvert}.

Procedure — mirror on an optical bench

Mount the concave mirror on one upright. Place an illuminated object pin on another upright, a slight distance away from the mirror. The reflected image is on the same side as the object, so you cannot use a screen behind the mirror — you use a second pin (the "image pin") that you position by parallax removal.

Optical bench for concave mirror u-v methodHorizontal bench with a concave mirror on the right at position xM, an object pin on the left at position xO, and an image pin between them at position xI. Distances u and v are both measured from the mirror, both on the same side. 020406080100 cm object pin (x_O) image pin (x_I) mirror (x_M)
Concave-mirror optical bench. The object pin sits at $x_O$, the mirror pole at $x_M$, and the image pin is slid until parallax with the image of the object vanishes — its position is $x_I$. Then $\lvert u\rvert = x_M - x_O$ and $\lvert v\rvert = x_M - x_I$.

Step 1. Get a rough f by the distant-object method: point the mirror at a distant window, catch the sharp image on a screen held in front of it. Measure the screen-to-mirror distance. This is \lvert f\rvert to about 5% — typically 10–15 cm for a school concave mirror.

Step 2. Place the object pin at \lvert u\rvert \approx 2\lvert f\rvert (this is the position for \lvert v\rvert = \lvert u\rvert, giving unit magnification — easy to find as a reference).

Step 3. Look into the mirror over the top of the object pin. You see an inverted image of the object pin somewhere between the mirror and the object. Slide the image pin (a second pin, placed in front of the mirror) until, as you move your eye side to side, the image pin and the image of the object pin do not shift relative to each other. No parallax — the image pin is at the image distance.

Step 4. Record x_O, x_I, x_M. Compute \lvert u\rvert = x_M - x_O and \lvert v\rvert = x_M - x_I.

Step 5. Repeat for \lvert u\rvert from 1.5\lvert f\rvert to 4\lvert f\rvert in 5–7 steps.

Step 6. Plot 1/\lvert v\rvert vs 1/\lvert u\rvert. For a concave mirror, the relation 1/\lvert v\rvert + 1/\lvert u\rvert = 1/\lvert f\rvert gives a straight line with slope -1 and both intercepts at 1/\lvert f\rvert. Read \lvert f\rvert from the intercepts.

Why parallax is indispensable here

You cannot catch the image on a screen behind the concave mirror — the image is in front of the mirror, on the same side as the object. A screen placed in front of the mirror would block the incoming light. A second pin, thin enough to see past, is the only practical way to locate the image. This is why the concave-mirror experiment uses two pins from the start, and it is also why parallax removal — rather than screen sharpness — is the definitive null criterion in Indian physics labs.

Method 5 — equivalent focal length of a lens combination

The idea

Two thin lenses of focal lengths f_1 and f_2 in contact (separation much less than either focal length) form a combination whose equivalent focal length F obeys

\frac{1}{F} \;=\; \frac{1}{f_1} + \frac{1}{f_2}.

Derived in Lensmaker's Equation and Combinations. If one of the lenses is a concave lens (negative f_2), you cannot measure f_2 directly — a concave lens alone forms only a virtual image, which no screen can catch. But if you combine it with a convex lens of known f_1 such that F > 0 (the combination is still converging), you can measure F on the bench by the u–v method. Rearranging:

\frac{1}{f_2} \;=\; \frac{1}{F} - \frac{1}{f_1} \quad\Longrightarrow\quad f_2 \;=\; \frac{F f_1}{f_1 - F}.

Why: the combination formula solved for f_2. If F < f_1 (the combination is less converging than the convex lens alone, which happens when you add a weak concave lens to a strong convex lens), then f_1 - F > 0 and f_2 > 0 — so the unknown is actually a weak converging lens. If F > f_1, then f_1 - F < 0 and f_2 < 0 — the unknown is diverging, with the magnitude given by the formula.

Procedure

Step 1. Place the two lenses in contact in a single lens-holder. A thin layer of glycerol (or just a light squeeze) ensures they are optically flush.

Step 2. Run the standard u–v method on this combination. Measure F.

Step 3. Remove the concave lens, measure f_1 of the convex lens alone by the standard method.

Step 4. Solve for f_2.

What this lets you measure

The concave lens that makes up the pair of spectacles worn by a myopic student has a nominal focal length marked in dioptres (-2.75 D, say — focal length f_2 = -1/2.75 \approx -36 cm). Without the combination method, no school lab could verify this value. With a convex lens of known f_1 = 20 cm in contact, the combination has 1/F = 1/20 + 1/(-36) = 0.0500 - 0.0278 = 0.0222\ \text{cm}^{-1}, so F \approx 45 cm — easily measured on the bench. Solve back and f_2 = -36 cm is recovered to within 1%.

Sources of error — what separates a good reading from a great one

Parallax residual

Even after you think you have eliminated parallax, there is always a residual — maybe 1 mm on a 1 metre bench. On a balance where v = 40 cm, this is a 0.25% error in v. Through the formula f = uv/(u+v), it propagates to about a 0.15% error in f. This is the floor — no technique gets below it on an optical bench of this length.

Thick-lens correction

A real convex lens is not infinitely thin. The lens formula 1/v - 1/u = 1/f is measured from the two principal planes of the lens, not from its geometric centre. For a glass lens of 2 cm thickness, the two principal planes are separated by about 1 cm, and each is about 0.5 cm from the geometric centre. If you use the geometric centre to define u and v, every measurement has a systematic error of about 0.5 cm.

The symptom: the 1/v-vs-1/u plot has a slope of -1.00 to high accuracy (the thick-lens correction preserves the slope), but the two intercepts disagree by a few percent — the y-intercept gives one f, the x-intercept gives another. The fix is to measure to each surface of the lens separately and interpolate, or to acknowledge the uncertainty as part of the error bar.

Spherical aberration

A perfect spherical lens does not focus parallel rays to a single point. Rays near the edge of the lens focus slightly closer than rays near the axis — spherical aberration. The consequence in the lab is that the image on the screen has a faint halo; the "sharpest" position of the screen depends on whether you look at the central image or the edge. Spherical aberration typically adds \pm 0.5\% to the uncertainty in f.

Chromatic aberration

The refractive index of glass depends on wavelength, so f is slightly different for red light (f_\text{red}) and blue light (f_\text{blue}). Under white light illumination, the image has coloured fringes; no single screen position is "sharp" for all colours simultaneously. For a crown-glass lens, f_\text{blue} \approx f_\text{red} \times 0.98 — a 2% spread. Using a coloured filter (traditionally, a sodium-vapour lamp in Indian labs gives monochromatic yellow at \lambda = 589 nm) eliminates this and gives a focal length to about 0.3%.

Bench scale errors

A cheap optical bench scale can drift over time — the wood warps, the brass track bends. Calibrate occasionally with a known standard rule. On a good bench, the scale is accurate to \pm 0.5 mm over 1 m — a 0.05% error, usually negligible compared to parallax residual.

Worked examples

Example 1: Convex lens by distant-object method

A Class 12 student in a Jaipur school points a convex lens at a tall neem tree about 50 m away. Holding a piece of cardboard behind the lens, she finds the sharpest image at a distance of 19.8 cm from the lens. An assistant measures again and gets 20.1 cm. Report the focal length with a realistic error estimate.

Step 1. Average the two readings.

f \;\approx\; \frac{19.8 + 20.1}{2} \;=\; 19.95\ \text{cm} \;\approx\; 20.0\ \text{cm}.

Why: two independent observers reduce the random component of the parallax/sharpness judgement. The mean is the best estimate.

Step 2. Estimate the systematic error from the finite object distance.

The true image distance for an object at \lvert u\rvert = 50 m = 5000 cm is

v \;=\; \frac{\lvert u\rvert f}{\lvert u\rvert - f} \;=\; \frac{5000 \times 20}{5000 - 20} \;=\; \frac{100000}{4980} \;=\; 20.08\ \text{cm}.

Why: the derivation of the distant-object approximation: v = f is exact only when \lvert u\rvert \to \infty. At finite \lvert u\rvert the true image is at v = \lvert u\rvert f / (\lvert u\rvert - f), slightly larger than f. For \lvert u\rvert = 250 f, the correction is f/(250-1) \approx 0.4\%.

So the measurement reads 20.08 cm for a true f of 20 cm — the method systematically overestimates f by about 0.4%. The observers averaged 19.95 cm, which is already 0.1% below 20 cm, so the two effects roughly cancel.

Step 3. Report.

f \;=\; 20.0 \pm 0.2\ \text{cm}.

Why: the \pm 0.2 cm is a conservative estimate — larger than either the random observer disagreement (0.15 cm) or the distance-approximation bias (0.08 cm), because either could be real and they could add.

Result. The focal length is 20.0 \pm 0.2 cm, or roughly 20 \pm 0.2 cm, a 1% measurement.

What this shows. The distant-object method gives a 1% answer in thirty seconds — good enough to plan the u–v method but not good enough for a lab report submission. The next step is to set up on the optical bench and take five u–v pairs for the plot.

Example 2: Convex lens by $1/v$-vs-$1/u$ plot

A student in a Pune ISC school sets up a convex lens on an optical bench. The object pin is fixed at x_O = 20.0 cm. Table 1 shows five measured lens-position/screen-position pairs. Find the focal length by plotting 1/v vs 1/u and reading the intercepts.

x_L (cm) x_I (cm) \lvert u\rvert = x_L - x_O (cm) v = x_I - x_L (cm) 1/u (cm⁻¹) 1/v (cm⁻¹)
47.0 100.2 27.0 53.2 0.0370 0.0188
52.0 92.5 32.0 40.5 0.0313 0.0247
58.0 88.8 38.0 30.8 0.0263 0.0325
65.0 87.3 45.0 22.3 0.0222 0.0448

(She collects a fourth row too, oops — check whether the fourth row is an outlier.)

1/v vs 1/u data points and best-fit lineFour data points plotted on a 1/u vs 1/v axes in inverse centimetres. The first three points lie close to a straight line with slope -1; the fourth point sits noticeably below, suggesting a misread or a misaligned measurement.1/u (×10⁻² cm⁻¹)1/v (×10⁻² cm⁻¹) 12345 12345 best-fit (slope −1)intercept = 0.05 = 1/f (1) (2) (3) (4) outlier? 1/f1/f
Four data points from the lab table. Points 1–3 lie tightly on the line $1/v = 1/f - 1/u$ with $1/f = 0.05\ \text{cm}^{-1}$, so $f = 20$ cm. Point 4 sits above the line — checking the original reading, $u = 45$ cm with $v = 22.3$ cm gives $1/v = 0.0448$, which does not fit; the student mis-read the screen position. Reject this point and report $f = 20$ cm from the other three.

Step 1. Compute 1/u and 1/v (already done in the table).

Step 2. Plot the four points on the (1/u, 1/v) plane.

Step 3. Points 1, 2, 3 satisfy 1/u + 1/v \approx 0.0558, 0.0560, 0.0588 — close but not identical. Point 4 gives 1/u + 1/v = 0.0222 + 0.0448 = 0.0670. This sum should equal 1/f and should be constant across rows. Point 4 is 20% off — an outlier.

Step 4. Discard point 4 and fit a line to the remaining three. The mean 1/f from the three is (0.0558 + 0.0560 + 0.0588)/3 = 0.0569, so f = 17.6 cm.

Why: the sum 1/u + 1/v is the working invariant of the experiment — it should equal 1/f at every measurement. A row whose sum disagrees is either a misread or a physical problem (misaligned lens, loose upright, twinkled eye). Discarding the outlier is the right move, and the lab book should note which row was discarded and why.

Step 5. Alternative: a least-squares fit forced to slope -1 gives y-intercept = \overline{1/u + 1/v} = 1/f, same answer.

Result. f = 17.6 \pm 0.3 cm from three consistent rows; point 4 discarded as a misread (the student's lab book should explain this).

What this shows. The plot exposes the outlier that a simple average would have hidden. If the student had reported f = \text{mean of }uv/(u+v)\text{ over all four rows} she would have got (17.95 + 17.85 + 17.00 + 14.91)/4 = 16.9 cm — contaminated by 1 cm (6%) by the one misread row. The plot caught it.

Example 3: Concave lens by the lens-combination method

A Bangalore JEE coaching lab has a convex lens of focal length f_1 = 25.0 cm and an unknown concave lens. When the two are placed in contact and treated by the u–v method on an optical bench, the equivalent focal length is F = 42.0 cm. Find the concave lens's focal length.

Step 1. Apply the combination formula.

\frac{1}{F} \;=\; \frac{1}{f_1} + \frac{1}{f_2} \quad\Longrightarrow\quad \frac{1}{f_2} \;=\; \frac{1}{F} - \frac{1}{f_1}.

Step 2. Substitute.

\frac{1}{f_2} \;=\; \frac{1}{42.0} - \frac{1}{25.0} \;=\; 0.02381 - 0.04000 \;=\; -0.01619\ \text{cm}^{-1}.

Why: the convex lens alone has 1/f_1 = 0.04\ \text{cm}^{-1}. When the combination is weaker (larger F, smaller 1/F = 0.0238\ \text{cm}^{-1}), the second lens must be pulling down the sum — so 1/f_2 is negative, confirming a concave lens.

Step 3. Invert.

f_2 \;=\; \frac{1}{-0.01619} \;=\; -61.8\ \text{cm}.

Step 4. Convert to dioptres.

P_2 \;=\; \frac{1}{f_2 \text{ (in m)}} \;=\; \frac{1}{-0.618} \;=\; -1.62\ \text{D}.

Why: opticians prescribe in dioptres, not centimetres. A concave lens of power -1.62 D is a mild myopia-correction lens — could easily belong to a student with mild short-sightedness.

Result. The concave lens has f_2 = -61.8 cm or P_2 = -1.62 D.

What this shows. A concave lens alone forms no real image — a screen placed behind it catches nothing. The combination method is the only bench technique that measures a concave lens. In its pure form it requires only one extra measurement beyond the usual u–v (namely, f_1 of the convex lens alone), and gives the concave lens's focal length to the same precision as the convex lens's — typically about 1%. This is how every pair of spectacles in an optician's shop gets its power verified before the student leaves the counter.

Common confusions

If you came here to set up the optical bench and submit the CBSE or ISC practical, you now have everything you need. What follows is the error-propagation algebra, the principal-plane analysis for a thick lens, and the aberration corrections that separate the school-lab answer from a research-grade focal length.

Error propagation through f = uv/(u+v)

Take the logarithmic derivative:

\ln f \;=\; \ln u + \ln v - \ln(u+v).
\frac{df}{f} \;=\; \frac{du}{u} + \frac{dv}{v} - \frac{du + dv}{u+v}.

Why: differentiate \ln f and use the rule d\ln x = dx/x. The subtraction of the third term comes from the -\ln(u+v), and d(u+v) = du + dv.

For \lvert u\rvert = v (the symmetric case, minimising the combined error):

\frac{df}{f} \;=\; \frac{du}{u} + \frac{dv}{v} - \frac{du+dv}{2u} \;=\; \frac{1}{2}\,\frac{du}{u} + \frac{1}{2}\,\frac{dv}{v}.

Why: at u = v, the third term becomes (du+dv)/(2u) = du/(2u) + dv/(2v), halving the contributions from du/u and dv/v. This is a textbook example of the error halving at a specific geometric configuration.

With du = dv = 0.1 cm (parallax residual) and u = v = 40 cm, df/f = (1/2)(0.0025) + (1/2)(0.0025) = 0.0025 = 0.25\%. A single row at the symmetric point gives f to 0.25% — already better than chromatic aberration (2%). Five such rows drive the random component down to 0.25/\sqrt{5} \approx 0.1\%; the systematics (lens thickness, aberrations) then dominate.

The symmetric configuration u = v = 2f. This is a sweet spot for the u–v measurement. Always take at least one reading there, labelled in the table as "unit-magnification check" — the image on the screen should be the same size as the object.

Principal planes of a thick lens

A real lens, especially a thick one, has two principal planes H_1 (on the incident side) and H_2 (on the emergent side). These are conjugate planes of unit lateral magnification. The lens formula 1/v - 1/u = 1/f holds with u measured from H_1 and v measured from H_2 — not from the geometric centre.

For a biconvex lens of glass (n = 1.5) with both surfaces of radius 30 cm and a thickness d = 2 cm, the lensmaker's formula (with thickness correction) gives

\frac{1}{f} \;=\; (n-1)\left[\frac{1}{R_1} - \frac{1}{R_2} + \frac{(n-1)\,d}{n\,R_1 R_2}\right].

The principal-plane separations from the lens surfaces are of order d/n \approx 1.3 cm each. A student measuring u and v to the geometric centre of the lens makes a systematic error of about \pm 0.5 cm per measurement — about 1% on a 50 cm bench.

Practical remedy. For CBSE-level lenses, the thickness is typically < 5 mm and the correction is negligible (< 0.3%). For a thicker research-grade lens, measure u and v to each lens surface separately and interpolate; or acknowledge the \pm 0.5 cm uncertainty in the error bar.

Aberrations — the fundamental limit

A perfect spherical lens is not a perfect optical imaging device. Rays from an on-axis point at distance u focus at a distance that depends slightly on which part of the lens they went through — rays near the edge (marginal rays) focus closer than rays near the axis (paraxial rays). This is spherical aberration. The effect: the "sharpest" image on the screen is not a point but a small disk (the circle of least confusion), and f is defined only to within the width of this disk.

For a typical school convex lens of aperture 5 cm and focal length 20 cm, the spherical aberration contributes about 0.3% to the uncertainty in f. For a high-speed (low f-number) lens, the effect is much larger. Aspheric lenses, used in modern cameras, correct spherical aberration by carefully shaped surfaces but are far more expensive.

Chromatic aberration adds a wavelength-dependent contribution. With the usual white-light illumination, the refractive index varies by about 1% from red to violet, and f varies by a similar amount. The remedy used in all precision optical benches is a sodium lamp, which emits essentially monochromatic yellow light at \lambda = 589 nm — eliminating chromatic spread at the cost of a yellow-tinted experience.

The Bessel method — a slick alternative

There is an elegant method for the focal length of a convex lens that avoids measuring u and v directly. Fix the distance D between the object and the screen with D > 4f. There are then two positions of the lens between them at which a sharp image forms (one enlarged, one diminished). If d is the distance between these two lens positions, then

f \;=\; \frac{D^2 - d^2}{4D}.

Why: the two conjugate positions satisfy u_1 + v_1 = D and u_2 + v_2 = D, and symmetry demands u_1 = v_2 and u_2 = v_1, giving d = u_2 - u_1 = v_1 - v_2. The algebra reduces to the Bessel formula.

This method only uses two lens positions on a fixed bench and avoids the ambiguity of defining the lens's principal plane. It is particularly valuable for thick lenses. JEE Advanced occasionally asks students to derive the Bessel formula — a 5-minute algebra exercise.

Why focal length is an aperture-dependent quantity

In a perfectly spherical lens, different rays focus at different distances — so there is no single "focal length" when the aperture is large. The focal length quoted by the manufacturer is the paraxial focal length — the focus for rays arbitrarily close to the principal axis. A real-world measurement gives something between the paraxial focus and the marginal focus, depending on how the image sharpness was judged.

For an aspheric lens (corrected for spherical aberration), all rays focus at the same point and f is well-defined. For the school's cheap glass biconvex, f is well-defined to about 1% — the spread between paraxial and marginal. Quote a good school-lab answer as f = 20.0 \pm 0.2 cm; the extra decimal is meaningless without specifying the aperture used.

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