Q.A small candle, $2.5\ \text{cm}$ in size is placed at $27\ \text{cm}$ in front of a concave mirror of radius of curvature $36\ \text{cm}$. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

Concept and Intuition

If the final image coincides with the original object, light must retrace its exact path after reflecting off the mirror. A ray retraces its path off a spherical mirror only if it strikes the mirror normally, i.e. it is travelling along a radius of the mirror — which means the point the rays are converging toward (before the mirror intervenes) must be the mirror's centre of curvature $C$. So the strategy is: (1) find where the lens alone would have focused the light, (2) find how far that point lies from the mirror, and (3) equate that distance to $R$, giving $f_{mirror}=R/2$.

Step-by-Step Solution

1. Lens formula (sign convention: distances measured from the lens, light travelling left to right): $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$, with $u=-15$ cm, $f=+10$ cm (convex lens).
2. $\dfrac{1}{v} = \dfrac{1}{10} + \dfrac{1}{-15}\cdot(-1)$... more carefully: $\dfrac1v = \dfrac1f + \dfrac1u = \dfrac{1}{10} - \dfrac{1}{15} = \dfrac{3-2}{30}=\dfrac{1}{30}$, so $v=30$ cm.
3. So without the mirror, the lens would converge light to a real image 30 cm from the lens, on the far side.
4. The convex mirror is placed at the lens's focus on that side, i.e. at 10 cm from the lens. The converging beam is intercepted 10 cm before it would have converged, so relative to the mirror the rays are aiming at a point $30 - 10 = 20$ cm behind the mirror. This is a virtual object for the mirror.
5. For the mirror to send the rays straight back the way they came (so that after passing back through the lens they reconverge exactly on the original object), the rays must hit the mirror along its normal — i.e. they must be directed at the mirror's centre of curvature $C$.
6. Hence $R = 20$ cm, and for a spherical mirror $f = R/2 = 10$ cm.

Common Mistakes

• Confusing the focus of the mirror with the point where the rays converge (that point is $C$, not $F$).
• Forgetting to subtract the mirror's own position (10 cm) from the lens's image distance (30 cm) before treating it as the mirror's object/virtual-object distance.
• Assuming the image coincides with the object automatically just because the mirror is at the lens's focus — that condition alone doesn't guarantee retro-reflection; the normal-incidence condition is what does.

ANSWER: B

Concept and Intuition

For a mirror, the image and object distances are related by the mirror formula $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$. As the object moves slowly along the axis, the image also moves — but not at the same speed, because the relationship between $v$ and $u$ is nonlinear. Differentiating the mirror equation with respect to time shows that the instantaneous ratio of image speed to object speed equals the square of the ratio $v/u$ (this ratio is also, incidentally, the linear magnification's magnitude). This squared dependence is why images near a focal point can appear to move dramatically faster or slower than the object itself.

Step-by-Step Solution

1. Find the image distance using the mirror formula (magnitude form, real object and real image, concave mirror): $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$, or equivalently $v=\dfrac{uf}{u-f}$.
2. Substitute $u=20\ cm$, $f=15\ cm$: $v = \dfrac{20\times15}{20-15} = \dfrac{300}{5}=60\ cm$.
3. Differentiate $\dfrac1v+\dfrac1u=\dfrac1f$ with respect to time (treating $f$ as constant):

4. Take magnitudes: $\left|\dfrac{dv}{dt}\right| = \left(\dfrac{v}{u}\right)^2\left|\dfrac{du}{dt}\right| = \left(\dfrac{60}{20}\right)^2\times 5 = 9\times5 = 45\ m/s$.

Common Mistakes

• Assuming the image moves at the same speed as the object (ignoring the nonlinear $v$-$u$ relationship).
• Using the ratio $(v/u)$ instead of its square for the speed relation — a very common slip.

ANSWER: A

Concept and Intuition

A dentist uses a concave mirror held close to the tooth (object distance less than the focal length) so that the mirror produces a virtual, erect, and magnified image — letting the dentist see an enlarged view of the cavity. This is a standard application of the mirror formula in the regime $u < f$.

Step-by-Step Solution

1. Using the Cartesian sign convention (distances measured from the pole, incident light direction positive, real objects/distances to the left are negative): for a concave mirror, $f = -24$ mm; the object is at $u = -16$ mm.
2. Mirror formula: $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$.
3. $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{-24} - \dfrac{1}{-16} = -\dfrac{1}{24} + \dfrac{1}{16}$.
4. Common denominator 48: $-\dfrac{2}{48} + \dfrac{3}{48} = \dfrac{1}{48}$, so $v = 48$ mm.
5. Since $v$ is positive (opposite sign convention region from a real image), this indicates a virtual image formed behind the mirror.
6. Magnification: $m = -\dfrac{v}{u} = -\dfrac{48}{-16} = 3$.
7. So the image is magnified 3 times, virtual and erect — exactly what lets the dentist see the cavity clearly enlarged.

Common Mistakes

• Treating this as a case with $u > f$, which would give a real, inverted, diminished/magnified image on the same side — but here $u=16 < f=24$, so the image is virtual and behind the mirror.
• Sign errors in the mirror formula, leading to a negative or fractional magnification instead of a clean integer value.

ANSWER: B

Concept and Intuition

For a concave mirror forming a real (inverted) image, the magnification and object distance are linked through the focal length by $m=\dfrac{f}{u-f}$ (using magnitudes, with $u>f$ so the object sits beyond the focus). Two different object positions with two different (given) magnifications give two independent equations in the two unknowns $x$ (the first object distance) and $f$ (focal length) — solving them together pins down both.

Step-by-Step Solution

1. First position, $u=x$, $m=3$: $3=\dfrac{f}{x-f} \Rightarrow 3(x-f)=f \Rightarrow 3x=4f \Rightarrow f=\dfrac{3x}{4}$.
2. Second position, $u=x+5$, $m=2$: $2=\dfrac{f}{(x+5)-f} \Rightarrow 2[(x+5)-f]=f \Rightarrow 2x+10=3f \Rightarrow f=\dfrac{2x+10}{3}$.
3. Equate the two expressions for $f$: $\dfrac{3x}{4}=\dfrac{2x+10}{3} \Rightarrow 9x = 8x+40 \Rightarrow x=40$.
4. $f = \dfrac{3(40)}{4} = 30$ cm (check with the other equation: $f=\dfrac{2(40)+10}{3}=\dfrac{90}{3}=30$ cm — consistent).

Common Mistakes

• Using the virtual-image magnification sign convention ($m=f/(f-u)$) instead of the real-image form — mixing conventions gives an inconsistent pair of equations.
• Forgetting to move the object farther (adding 5, not subtracting) as stated, which is why the magnification decreases from 3 to 2.

ANSWER: D

We have a convex mirror with radius of curvature $R = 20\ \text{cm}$, so its focal length is $f = \frac{R}{2} = 10\ \text{cm}$. For a convex mirror, the focal length is taken as positive in the standard sign convention (since the focus lies behind the mirror). The rod AB lies along the principal axis, with end B at 20 cm from the pole (so $u_B = -20\ \text{cm}$ using the Cartesian sign convention: distances measured from the pole, positive in the direction of incident light, which is toward the mirror; here the object is on the left, so object distances are negative). End A is 5 cm farther from the mirror than B, so $u_A = -25\ \text{cm}$.

The mirror formula is:

where $v$ is the image distance (positive for virtual images behind a convex mirror).

1. Image of end B For $u_B = -20\ \text{cm}$ and $f = +10\ \text{cm}$:

This is positive, so the image of B is virtual, behind the mirror, at $\frac{20}{3}$ cm from the pole.

2. Image of end A For $u_A = -25\ \text{cm}$:

Again positive, so the image of A is also virtual, behind the mirror.

3. Length of the image Since both images lie on the same side of the mirror (behind it), the image of the rod is the segment between these two image points. The length is the absolute difference of their distances from the pole:

Compute:

Concept and Intuition

For mirrors (New Cartesian sign convention: distances against the incident-light direction are negative; for a convex mirror $f$ is positive; a real object always has $u$ negative), the governing relations are:

A convex mirror always forms a virtual, erect, diminished image of a real object, so its magnification is positive and less than 1 — exactly consistent with a size ratio $n<1$. The goal is to eliminate $v$ between the two equations to express $u$ purely in terms of $f$ and $n$.

Step-by-Step Solution

1. From the mirror formula: $\dfrac{1}{v} = \dfrac{1}{f}-\dfrac{1}{u} = \dfrac{u-f}{fu}$, so $v = \dfrac{fu}{u-f}$.
2. Substitute into $m=-v/u$: $m = -\dfrac{1}{u}\cdot\dfrac{fu}{u-f} = \dfrac{-f}{u-f} = \dfrac{f}{f-u}$.
3. Set $m=n$ (given the image is $n$ times the object size): $n = \dfrac{f}{f-u}$.
4. Solve for $u$: $f-u = \dfrac{f}{n} \Rightarrow u = f - \dfrac{f}{n}$.
5. This matches option (C). (Sanity check: for $n\to1$, $u\to0$ — object at the pole, image same size, as expected; as $n\to0$, $u\to-\infty$ — object very far away, image tiny, as expected.)

Common Mistakes

• Mixing up magnitude-only formulas with signed Cartesian ones, leading to a sign-flipped final expression.
• Forgetting that for a convex mirror with a real object, $m$ is positive (erect image), not negative.

ANSWER: C

Concept and Intuition

A real image can only form in front of a mirror (same side as the object); anything found behind the mirror must be virtual. Only convex mirrors (and concave mirrors with the object between pole and focus) can produce a virtual image; here we let the sign of $f$ that comes out of the calculation tell us which mirror type is actually consistent with the data.

Step-by-Step Solution

1. Sign convention: distances of real objects/images (in front, same side as incident light) are negative; distances of virtual images (behind the mirror) are positive.
2. Object: real, in front, so $u = -18\ \text{cm}$.
3. Image: stated to form "on the other side," i.e., behind the mirror $\Rightarrow$ virtual, so $v = +4\ \text{cm}$.
4. Mirror formula: $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$.

5. $f = \dfrac{36}{7} \approx 5.14\ \text{cm}$, and it's positive.
6. In this convention, a positive focal length belongs to a convex mirror (focus behind the mirror); the image being at positive $v$ (behind the mirror) confirms it is virtual.

Common Mistakes

• Assuming a concave mirror by default — but a positive $f$ here rules that out; only a convex mirror (or a special concave case, which would give negative $f$) is consistent.
• Mixing up which side counts as positive; consistently using "real = negative, virtual = positive" for mirrors avoids the confusion.

ANSWER: C

Concept and Intuition

For rays close to the principal axis (paraxial approximation), a geometric analysis of reflection at a spherical mirror surface shows that the focal point lies exactly halfway between the pole and the centre of curvature.

Step-by-Step Solution

1. Consider a ray parallel to the principal axis striking the mirror close to the pole and reflecting through the focus.
2. Using the geometry of the mirror (radius $R$, centre of curvature $C$) and the law of reflection, for small angles the focal length works out to $f = \dfrac{R}{2}$.
3. This relation holds for both concave and convex spherical mirrors under the paraxial approximation.

Common Mistakes

• Writing $R=f/2$ (inverting the relation) instead of $f=R/2$.
• Confusing the mirror formula ($f=R/2$) with the analogous but different relation for thin lenses.

ANSWER: B

Concept and Intuition

A convex mirror always forms a virtual, erect, diminished image, so its magnification $m$ is positive and less than 1. Combining the mirror formula with the magnification definition gives a clean relation between object distance $x$, focal length $f$, and $m$.

Step-by-Step Solution

1. Let object distance (measured as a positive magnitude) be $x$; using the Cartesian sign convention, $u=-x$, and for a convex mirror $f=+30$ cm.
2. Mirror formula: $\dfrac1v+\dfrac1u=\dfrac1f \Rightarrow \dfrac1v = \dfrac1f-\dfrac1u = \dfrac1f+\dfrac1x = \dfrac{x+f}{fx} \Rightarrow v=\dfrac{fx}{x+f}$.
3. Magnification: $m=-\dfrac{v}{u} = -\dfrac{fx/(x+f)}{-x} = \dfrac{f}{x+f}$.
4. Given $m=\dfrac14$: $\dfrac{f}{x+f}=\dfrac14 \Rightarrow 4f=x+f \Rightarrow x=3f$.
5. Substitute $f=30$ cm: $x=90$ cm.

Common Mistakes

• Using the concave-mirror sign convention or forgetting that for a convex mirror $f$ is taken positive, which flips the algebra.
• Missing that the "quarter size" image being upright (positive $m$) is what confirms it's a convex mirror scenario, consistent with a straightforward positive-$f$ formula.

ANSWER: B