A short object of length $L$ is placed along the principal axis of a concave mirror away from focus. The object distance is $u$. If the mirror has a focal length $f$, what will be the length of the image? You may take $L \ll |v - f|$.

A short object of length L is placed along the principal axis of a concave mirror away from focus. The object

NCERT Exemplar · Q22

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Concept understanding — Spherical Mirror Equation

The Spherical Mirror Equation: From Intuition to Formula

Imagine you're standing in front of a concave mirror — the kind that makes your face look bigger when you're close, but flips everything upside down when you step far back. That change isn't magic; it's geometry. The spherical mirror equation is the single relationship that predicts exactly where an image will form, and whether it's real or virtual, for any spherical mirror.

The Core Idea

Every point on an object sends out light rays in all directions. A mirror redirects those rays. The mirror equation tells you: given the mirror's curvature and the object's distance, where will those rays meet again (or appear to meet)?

There are only three quantities you need:

$u$ — object distance (from the mirror's pole)
$v$ — image distance (from the mirror's pole)
$f$ — focal length (a property of the mirror's curvature)

The equation is:

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

The power is in the sign convention, because every distance can point in one of two directions.

The Sign Convention (New Cartesian Sign Convention)

This is where most students slip. The equation works for all spherical mirrors — concave and convex — only if you follow the convention used throughout NCERT and CBSE:

All distances are measured from the mirror's pole.
The incident light is taken to travel left to right, so distances measured in that same direction (to the right) are positive, and distances measured against it (to the left) are negative.
Heights above the principal axis are positive; heights below are negative.

Because a real object is always placed in front of the mirror (to the left, where the incident light originates), its distance $u$ is always negative.

Important

Under this convention, the focal length of a concave mirror is negative (its focus $F$ sits in front of the mirror, on the same side as the object), and the focal length of a convex mirror is positive (its focus lies behind the mirror). This is one of the most frequently tested facts in CBSE board exams.

Watch out

A very common mistake is writing $f$ as positive for a concave mirror because "it converges light." Convergence tells you the type of mirror, not the sign — the sign comes purely from where the focus physically sits relative to the pole, under the convention above.

Where Does the Formula Come From?

For a concave mirror, parallel rays from a distant object converge at the focus, a point at (signed) distance $f$ from the mirror. The derivation uses similar triangles from a ray diagram.

›Proof

Consider an object of height $h_o$ in front of a concave mirror. Draw the ray parallel to the axis: it reflects through the focus $F$ . Draw the ray through the centre of curvature $C$ : it strikes the mirror normally and reflects straight back on itself. These two reflected rays cross to form the image, of height $h_i$ .

From similar triangles formed by the ray through $C$ :

$\frac{h_o}{h_i} = \frac{u - R}{R - v}$

where $R = 2f$ is the radius of curvature (with the same sign convention as $f$ ).

From similar triangles formed by the ray through $F$ :

$\frac{h_o}{h_i} = \frac{u - f}{f}$

Equating the two ratios and simplifying (using $R = 2f$ ) gives:

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

What the Equation Tells You

Rearranging for $v$ :

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

Because $u$ is negative for a real object, and $v$ takes the sign the geometry dictates:

$v$ negative → the image forms in front of the mirror → real image (can be projected on a screen).
$v$ positive → the image forms behind the mirror → virtual image.

For a concave mirror ( $f$ negative), using the magnitude of the object distance $|u|$ measured from the pole:

$|u| > 2|f|$ → real, inverted, diminished image between $f$ and $2f$
$|u| = 2|f|$ → real, inverted, same-size image at $2f$
$|f| < |u| < 2|f|$ → real, inverted, magnified image beyond $2f$
$|u| = |f|$ → image at infinity …

Why this formula?

Spherical Mirror Equation: Why the Formula Holds

The spherical mirror equation — also called the mirror formula — relates the object distance ( $u$ ), image distance ( $v$ ), and focal length ( $f$ ) of a spherical mirror. Let's build the reasoning step by step.

1. The Key Formula

For a spherical mirror (concave or convex):

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

Where:

$f$ = focal length (positive for concave, negative for convex)
$u$ = object distance from pole (always negative by sign convention)
$v$ = image distance from pole (sign depends on image location)

2. Why This Formula Holds — The Derivation

Step 1: Start with a ray diagram

Consider a concave mirror with:

Pole $P$
Centre of curvature $C$ (radius $R$ )
Focus $F$ (midpoint of $PC$ , so $f = R/2$ )

Take an object placed beyond $C$ . Draw two rays from the object's tip:

A ray parallel to the principal axis → reflects through $F$
A ray through $C$ → reflects back along itself

These rays meet at the image point.

Step 2: Use similar triangles

Let the object height be $h_o$ and image height be $h_i$ .

From the geometry of the ray through $C$ :

Triangle formed by object, $C$ , and axis is similar to triangle formed by image, $C$ , and axis.

This gives:

\frac{h_o}{h_i} = \frac{u - R}{R - v}

(Here $u$ and $v$ are distances from $P$ , with sign conventions applied later.)

Step 3: Use the parallel ray

From the ray parallel to the axis:

Triangle formed by object, $F$ , and axis is similar to triangle formed by image, $F$ , and axis.

This gives:

\frac{h_o}{h_i} = \frac{u - f}{f}

Step 4: Equate the two ratios

Since both ratios equal $h_o/h_i$ :

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

Step 5: Substitute $R = 2f$

For a spherical mirror, the focal length is half the radius of curvature:

R = 2f

Substitute:

\frac{u - 2f}{2f - v} = \frac{u - f}{f}

Step 6: Cross-multiply and simplify

Cross-multiply:

f(u - 2f) = (u - f)(2f - v)

Expand:

fu - 2f^2 = 2fu - uv - 2f^2 + fv

Cancel $-2f^2$ on both sides:

fu = 2fu - uv + fv

Bring all terms to one side:

0 = fu - uv + fv

Rearrange:

uv = fu + fv

Step 7: Divide by $uvf$

Divide both sides by $uvf$ :

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

This is the mirror formula.

3. Why the Sign Convention Matters

The derivation above used distances as positive magnitudes. In actual problem-solving, we use the Cartesian sign convention:

Distances measured against incident light are negative
Distances measured along incident light are positive

For a concave mirror:

$u$ is negative (object in front)
$f$ is negative (focus in front)
$v$ is negative for real images (in front) …

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For a small object placed along the principal axis of a concave mirror, the image length is found by differentiating the mirror equation. The result is $L' = \left( \frac{f}{u-f} \right)^2 L$ .

The key idea is that when an object of small length $L$ is placed along the principal axis, its two ends are at slightly different distances from the mirror. The image of each end forms at a slightly different distance, and the difference between those image distances gives the image length. Because $L$ is very small compared to the distance from the object to the focus, we can use calculus (differentiation) to find this difference accurately.

The spherical mirror equation relates object distance $u$ and image distance $v$ :

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

For a concave mirror, $f$ is positive. The object is placed away from the focus, so $u \neq f$ .

Why differentiation works here: If the object length $L$ is a small change in $u$ (i.e., $L = \Delta u$ ), then the corresponding change in $v$ (the image length $L' = \Delta v$ ) can be found from the derivative $dv/du$ . This is valid because $L \ll |v - f|$ , meaning the object is small enough that the mirror equation remains linear over that range.

Let’s work through it step by step.

Start with the mirror equation and differentiate implicitly.

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

Differentiate both sides with respect to $u$ :

-\frac{1}{u^2} - \frac{1}{v^2} \frac{dv}{du} = 0

(Remember: $f$ is constant, so its derivative is zero.)

Solve for $dv/du$ .

-\frac{1}{v^2} \frac{dv}{du} = \frac{1}{u^2}

Multiply both sides by $-v^2$ :

\frac{dv}{du} = -\frac{v^2}{u^2}

The negative sign tells us that as $u$ increases, $v$ decreases — which makes sense for a concave mirror: moving the object away from the mirror moves the image closer.

Relate $v$ to $u$ using the mirror equation. From $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{u - f}{uf}$ , we get:

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

Substitute $v$ into the derivative. $\frac{dv}{du} = -\frac{1}{u^2} \left( \frac{uf}{u - f} \right)^2 = -\frac{f^2}{(u - f)^2}$ …

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

TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
$21.4\ \text{cm}$
›Reveal solutionSolution
This is the classic “displacement method” for finding focal length: when object and screen are fixed, two lens positions give a sharp image; the focal length is given by $f = \frac{D^2 - d^2}{4D}$ , where $D = 90\ \text{cm}$ and $d = 30\ \text{cm}$ . The result is $f = 20\ \text{cm}$ , so option (B) is correct.

Concept and intuition

When an object and a screen are a fixed distance $D$ apart, a convex lens can form a sharp image on the screen in two distinct positions (provided $D > 4f$ ). At one position the image is magnified; at the other it is diminished. The distance between these two lens positions is $d$ . This is the displacement method for measuring focal length. The key insight: the lens formula and symmetry imply that the two object distances are $u$ and $D - u$ , and the difference between them is $d$ . Solving the lens equation yields a neat formula that avoids solving for $u$ directly.

Step-by-step solution

Set up the variables

Let the fixed distance between object and screen be $D = 90\ \text{cm}$ .

Let the distance between the two lens positions be $d = 30\ \text{cm}$ .

For the first position, let the object distance be $u$ ; then the image distance is $v = D - u$ .

For the second position, the roles swap: the object distance becomes $v$ and the image distance becomes $u$ .

The separation between the two lens positions is $|u - v| = d$ .

Relate $u$ and $v$ to $D$ and $d$

Since $u + v = D$ and $|u - v| = d$ , we can solve:

$u = \frac{D + d}{2}, \quad v = \frac{D - d}{2}$
(or vice versa; the absolute difference is $d$ ).

Here:
$u = \frac{90 + 30}{2} = 60\ \text{cm}, \quad v = \frac{90 - 30}{2} = 30\ \text{cm}.$

Apply the lens formula The thin lens equation:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$
Substitute $u = 60\ \text{cm}$ , $v = 30\ \text{cm}$ :
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TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
$Q. The focal lengths of the objective and the eyepiece of a compound microscope are 2 cm and 3 cm respectively and the distance between them is 15 cm. The final image formed by the eyepiece is at infinity. The distances of the object and the image produced by the object from the objective lens are respectively (A) 2.4 cm, 12 cm (B) 2.4 cm, 15 cm (C) 2.3 cm, 12 cm (D) 2.3 cm, 3 cm$
›Reveal solutionSolution
For a compound microscope with the final image at infinity, the eyepiece must receive parallel rays, so the objective’s image lies exactly at the eyepiece’s focal point. Using the lens formula for the objective, we find the object distance and the image distance from the objective, leading to option (A).

Concept & Intuition

In a compound microscope, the objective lens forms a real, inverted, and magnified image of the object. This image then acts as the object for the eyepiece (the second lens). When the final image is at infinity, the eyepiece is being used as a simple magnifier in its “normal adjustment” — meaning the intermediate image must be placed exactly at the eyepiece’s focal point. That gives us the distance from the eyepiece to the intermediate image. Since we know the separation between the two lenses, we can then find the distance from the objective to that intermediate image. Finally, using the lens formula for the objective, we solve for the object distance.

Step-by-step solution

Understand the setup

Objective focal length: $f_o = 2\ \text{cm}$

Eyepiece focal length: $f_e = 3\ \text{cm}$

Distance between objective and eyepiece (tube length, but not exactly the standard “tube length” definition here): $L = 15\ \text{cm}$

Final image is at infinity → rays entering the eyepiece are parallel → the intermediate image must lie at the first focal point of the eyepiece.

Locate the intermediate image

For the eyepiece, if the object (intermediate image) is at its focal point, the image is at infinity. So the intermediate image is at a distance $f_e = 3\ \text{cm}$ in front of the eyepiece.

Since the lenses are 15 cm apart, the distance from the objective to this intermediate image is:

$v_o = L - f_e = 15\ \text{cm} - 3\ \text{cm} = 12\ \text{cm}$
This is the image distance for the objective lens.

Apply the lens formula to the objective The lens formula:

$\frac{1}{f_o} = \frac{1}{u_o} + \frac{1}{v_o}$
where $u_o$ is the object distance from the objective (positive if real object is on the same side as incoming light, but we treat distances as positive in the formula with sign conventions; here we use the real-is-positive convention for simplicity). …
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TG EAPCET 2021Set ap-2021-08-09-AN1 markMCQ
$+50$
›Reveal solutionSolution
The problem asks for the power of a lens that shifts the near point from 50 cm to the standard 25 cm. Using the lens formula with object at 25 cm and image at 50 cm (virtual, on the same side as the object), the required power is +2 D. The correct option is (B).

Concept and intuition

A person whose least distance of distinct vision (near point) is 50 cm cannot see objects clearly closer than that. To read at the normal near point of 25 cm, we need a lens that makes an object at 25 cm appear to be at 50 cm (the person’s actual near point). This is a classic case of a converging lens used as a simple magnifier for a hypermetropic (farsighted) eye. The lens creates a virtual image at the person’s near point, which the eye can then focus on.

Step-by-step reasoning

Identify the object and image distances

The object is placed at the normal near point: $u = -25$  cm (negative by sign convention: object is on the same side as incoming light).

The image must be formed at the person’s actual near point: $v = -50$  cm (negative because the image is virtual and on the same side as the object).

Apply the lens formula

The thin lens equation is

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
Substituting the values (all in cm):
$\frac{1}{f} = \frac{1}{-50} - \frac{1}{-25} = -\frac{1}{50} + \frac{1}{25}$ $\frac{1}{f} = -\frac{1}{50} + \frac{2}{50} = \frac{1}{50}$
So the focal length $f = +50$  cm. The positive sign confirms a converging lens.

Convert focal length to power Power in dioptres is $P = \frac{1}{f(\text{in metres})}$ .
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TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
$21.38\ \text{cm}$
›Reveal solutionSolution
The problem describes the displacement method for finding the focal length of a convex lens. Using the given distances, the focal length is calculated as $\boxed{21.38\ \text{cm}}$ .

Concept and Intuition

This problem uses a common experimental technique called the displacement method (or conjugate foci method) to determine the focal length of a convex lens. The core idea is that for a fixed distance $D$ between an object and a screen, a convex lens can form a sharp, real image on the screen at two different positions, provided $D$ is greater than or equal to four times the focal length ( $D \ge 4f$ ).

This phenomenon arises from the principle of conjugate foci. If an object at position $O$ forms an image at position $I$ through a lens, then if the object is placed at $I$ , its image will be formed at $O$ . In the displacement method, the object and screen are fixed. When the lens is at the first position ( $L_1$ ), it forms an image of the object on the screen. If we then move the lens to a second position ( $L_2$ ), another sharp image is formed on the screen. The key insight is that the object distance for the first position becomes the image distance for the second position, and vice-versa.

Let $u_1$ be the object distance and $v_1$ be the image distance for the first lens position.

Then, for the second lens position, the object distance will be $v_1$ and the image distance will be $u_1$ .

The total distance between the object and the screen is $D = u_1 + v_1$ .

The distance the lens is displaced between the two positions is $x = |v_1 - u_1|$ .

These two equations allow us to find $u_1$ and $v_1$ in terms of $D$ and $x$ , which can then be substituted into the lens formula to find the focal length $f$ .
$f$
›Proof

Let $u_1$ be the object distance and $v_1$ be the image distance for the first position of the lens.

The total distance between the object and the screen is $D$ . So, $u_1 + v_1 = D$ .

According to the lens formula (using magnitudes for $u$ and $v$ for a real object and real image):

$\frac{1}{v_1} + \frac{1}{u_1} = \frac{1}{f} \quad (*)$

For the second position of the lens, due to the principle of conjugate foci, the object distance becomes $v_1$ and the image distance becomes $u_1$ .

The distance between the two lens positions is $x$ . From the geometry, this displacement is the difference between the image and object distances: $x = |v_1 - u_1|$ .

Assuming $v_1 > u_1$ (the lens is moved away from the object), we have:

$u_1 + v_1 = D$

$v_1 - u_1 = x$ Adding (1) and (2): $(u_1 + v_1) + (v_1 - u_1) = D + x$ $2v_1 = D + x \implies v_1 = \frac{D+x}{2}$ Subtracting (2) from (1): $(u_1 + v_1) - (v_1 - u_1) = D - x$ $2u_1 = D - x \implies u_1 = \frac{D-x}{2}$ Now, substitute these expressions for $u_1$ and $v_1$ into the lens formula $(*)$ : $\frac{1}{f} = \frac{1}{\frac{D+x}{2}} + \frac{1}{\frac{D-x}{2}}$ $\frac{1}{f} = \frac{2}{D+x} + \frac{2}{D-x}$ $\frac{1}{f} = 2 \left( \frac{1}{D+x} + \frac{1}{D-x} \right)$ $\frac{1}{f} = 2 \left( \frac{(D-x) + (D+x)}{(D+x)(D-x)} \right)$ $\frac{1}{f} = 2 \left( \frac{2D}{D^2 - x^2} \right)$ …
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TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
$\dfrac{20}{3}$
›Reveal solutionSolution
Use the lens maker’s formula to find the focal length, then apply the thin lens equation to locate the image from the lens center, and finally subtract the lens half-thickness to get the distance from the rear surface.

Concept and intuition

A symmetrical biconvex lens with equal radii means the two surfaces are identical in curvature. The lens maker’s formula gives the focal length directly from the radii and refractive index. Once we know the focal length, the thin lens equation tells us where the image forms relative to the optical center. But the question asks for the distance from the rear surface, not from the center. So we must account for the lens thickness — here the lens is thin enough that we can treat its center as the midpoint, but the problem expects us to realize the image distance from the rear surface is simply the image distance from the center minus half the (negligible) thickness? Actually, careful: For a thin lens, the distance from the rear surface is essentially the same as from the optical center, because the lens thickness is small compared to other distances. However, the problem gives radii of 5 cm, so the lens is not extremely thin — but in standard optics problems, “thin lens” approximation is used unless thickness is given. Here no thickness is given, so we assume the lens is thin and the optical center coincides with the lens center. Then the image distance from the rear surface equals the image distance from the optical center.

Let’s work it through.

Find the focal length using the lens maker’s formula For a lens in air (refractive index of air = 1), the formula is

$\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
For a biconvex lens, the first surface (front) has $R_1 = +5$ cm (positive because convex toward the object), and the second surface (rear) has $R_2 = -5$ cm (negative because convex toward the object means the center of curvature is on the opposite side).

So
$\frac{1}{f} = (1.5 - 1)\left(\frac{1}{5} - \frac{1}{-5}\right) = 0.5 \times \left(\frac{1}{5} + \frac{1}{5}\right) = 0.5 \times \frac{2}{5} = \frac{1}{5}$
Thus $f = 5$ cm.
$f = 5 \text{ cm}$

Apply the thin lens equation The object is placed 20 cm from the front surface. For a thin lens, the object distance $u$ is measured from the optical center. Since the lens is thin, the front surface is essentially at the optical center, so $u = -20$ cm (negative by sign convention: object on the incident side). The lens equation:

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Substituting: …
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TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
$Q. An object is placed in front of a spherical concave mirror between the focal point and the radius of curvature. Its image is (A) Inverted, real, farther than radius of curvature from mirror (B) Inverted, virtual, closer than focal point to mirror (C) Upright, real, farther than radius of curvature from mirror (D) Inverted, Real, closer than radius of curvature to mirror$
›Reveal solutionSolution
For an object between the focus and centre of curvature of a concave mirror, the image is real, inverted, and located beyond the centre of curvature — matching option (A).

The key to this problem is the mirror equation and the behaviour of rays from a concave mirror. When you place an object between the focal point $F$ and the centre of curvature $C$ , the reflected rays converge to form a real image on the same side as the object. The exact position and nature of that image follow directly from the mirror formula and a quick ray diagram.

Let’s walk through it.

Recall the mirror equation For a spherical mirror,

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$
where $f$ is the focal length (negative for concave), $u$ is the object distance (negative, by sign convention), and $v$ is the image distance (negative for real images).

The radius of curvature $R = 2f$ , so $C$ is at distance $2f$ from the mirror.

Set up the object position The object lies between $F$ and $C$ , so

$f < |u| < 2f$
(Remember: $u$ is negative, but we work with magnitudes for clarity.)

Find the image distance From the mirror equation:

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$
Since $|u| > f$ , the term $1/|u|$ is smaller than $1/|f|$ , so $1/v$ is negative — meaning $v$ is negative, hence a real image.

Moreover, because $|u| < 2f$ , we have $1/|u| > 1/(2f)$ , so
$\frac{1}{|v|} = \frac{1}{|f|} - \frac{1}{|u|} < \frac{1}{|f|} - \frac{1}{2|f|} = \frac{1}{2|f|}$
This gives $|v| > 2|f|$ , i.e. the image lies beyond the centre of curvature.

Determine magnification and orientation Magnification $m = -v/u$ . Since both $v$ and $u$ are negative, $m$ is negative — meaning the image is inverted relative to the object. Also $|v| > |u|$ , so $|m| > 1$ — the image is enlarged. …
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Q.A short object of length $L$ is placed along the principal axis of a concave mirror away from focus. The object distance is $u$ . If the mirror has a focal length $f$ , what will be the length of the image? You may take $L \ll |v - f|$ .

Concept understanding — Spherical Mirror Equation

The Spherical Mirror Equation: From Intuition to Formula

The Core Idea

The Sign Convention (New Cartesian Sign Convention)

Where Does the Formula Come From?

What the Equation Tells You

Why this formula?

Spherical Mirror Equation: Why the Formula Holds

1. The Key Formula

2. Why This Formula Holds — The Derivation

Step 1: Start with a ray diagram

Step 2: Use similar triangles

Step 3: Use the parallel ray

Step 4: Equate the two ratios

Step 5: Substitute $R = 2f$

Step 6: Cross-multiply and simplify

Step 7: Divide by $uvf$

3. Why the Sign Convention Matters

Method: Finding the Length of an Image Along the Axis (Longitudinal Magnification)

Steps

Concept and Intuition

Q.A short object of length LLL is placed along the principal axis of a concave mirror away from focus. The object distance is uuu. If the mirror has a focal length fff, what will be the length of the image? You may take L≪∣v−f∣L \ll |v - f|L≪∣v−f∣.

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Q.A short object of length $L$ is placed along the principal axis of a concave mirror away from focus. The object distance is $u$ . If the mirror has a focal length $f$ , what will be the length of the image? You may take $L \ll |v - f|$ .

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