Q.A $4.5\ \text{cm}$ needle is placed $12\ \text{cm}$ away from a convex mirror of focal length $15\ \text{cm}$. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Concept & Intuition

A fully silvered concave mirror is just a concave mirror — it reflects light. The plane mirror is placed behind it, so after the first reflection from the concave mirror, the light travels toward the plane mirror, reflects again, and the final image location is found by applying the mirror formula twice, carefully tracking sign conventions. The trick: the plane mirror simply creates a virtual image at the same distance behind it as the object is in front of it. We must keep distances measured from the concave mirror consistent.

Step-by-step solution

1. First reflection from the concave mirror
   • Object distance from concave mirror: $u_1 = -25$ cm (negative because object is in front, per Cartesian sign convention).
   • Focal length: $f = -15$ cm (concave mirror, negative focal length).
   • Mirror formula: $\frac{1}{v_1} + \frac{1}{u_1} = \frac{1}{f}$.

 So $v_1 = -37.5$ cm.  

• The negative sign means the image is in front of the concave mirror (real image), 37.5 cm from it.

2. Position of this image relative to the plane mirror

   • The plane mirror is 35 cm behind the concave mirror. So the distance from the concave mirror to the plane mirror is $+35$ cm (taking direction from concave toward plane as positive).
   • The first image is at $-37.5$ cm from the concave mirror (i.e., 37.5 cm in front).
   • Distance from this image to the plane mirror = $35 - (-37.5) = 72.5$ cm.
   • So the image is 72.5 cm in front of the plane mirror.

3. Second reflection from the plane mirror

   • For a plane mirror, the image is formed at the same distance behind the mirror as the object is in front.
   • Object (first image) is 72.5 cm in front of the plane mirror → second image is 72.5 cm behind the plane mirror.

4. Locate the final image relative to the concave mirror

   • The plane mirror is 35 cm behind the concave mirror.
   • The final image is 72.5 cm behind the plane mirror → total distance from concave mirror = $35 + 72.5 = 107.5$ cm behind the concave mirror.

Step 1 — Recall the reflection rule for a ray parallel to the axis

For a concave mirror, a ray parallel to the principal axis strikes the mirror and is reflected such that the reflected ray passes through the focus F, which lies midway between the pole and the centre of curvature C on the principal axis. This follows directly from the law of reflection applied to the mirror's curved surface at the point of incidence Q.

Step 2 — Identify the correct ray among the four options

Among the four candidate directions A, B, C and D shown emerging from Q, only the ray labelled C is directed so as to pass through F on the principal axis, consistent with this rule; the other three rays do not pass through F.

Step 1 — Write Snell's law for the vacuum $\to$ medium refraction.

Light goes from vacuum ($n_1 = 1$) into a medium of refractive index $n$:

Step 2 — Impose the given condition.

We are told the angle of incidence is twice the angle of refraction:

Substituting:

Step 3 — Use the double-angle identity.

so

Step 4 — Cancel $\sin r$.

Since the ray actually refracts, $r \neq 0$, so $\sin r \neq 0$ and we may divide both sides by it:

Step 5 — Return to the angle of incidence.

The question asks for $i$, not $r$:

Step 6 — Sanity check with a number.

Take $n = \sqrt{3}$. Then $\cos r = \sqrt3/2 \Rightarrow r = 30^\circ$, so $i = 60^\circ$. Check in Snell's law: $\sin 60^\circ = 0.866$ and $n\sin r = \sqrt3 \times 0.5 = 0.866$. ✓ Consistent, and indeed $i = 2r$.

Option (D) forgets the factor of 2 (it gives $r$, not $i$); options (A) and (C) wrongly use $\sin^{-1}$ instead of $\cos^{-1}$.

ANSWER: B

Step 1 — The lens-maker's formula is the concept.

The power depends only on the refractive index and the two radii of curvature — not on the aperture (the size/height of the lens).

Step 2 — Cut ALONG the principal axis.

This cut is made in a plane containing the principal axis, so each half still has both curved surfaces with the same $R_1$ and $R_2$; only the aperture is halved. Therefore

The piece $L_1$ (the half that is not cut again) has power $P$. (Only its light-gathering area, and hence image brightness, is halved.)

Step 3 — Cut PERPENDICULAR to the principal axis.

The second half is now sliced through the middle perpendicular to the axis. Each resulting piece is plano-convex: one surface keeps its curvature, the other becomes flat ($R \to \infty$).

For the original biconvex lens with $R_1 = +R$, $R_2 = -R$:

For each plano-convex piece:

So $L_2$ and $L_3$ each have power $\dfrac{P}{2}$.

Step 4 — Test the four statements.

• (A) Power of $L_2$ is $P/2$ — correct
• (B) Power of $L_3$ is $P/2$ — correct
• (C) Power of $L_1$ is $P$ — correct
• (D) Power of $L_1$ is $P/2$ — INCORRECT

The question asks for the incorrect option.

ANSWER: D

Concept and intuition

When a convex lens forms a real image, both object distance $u$ and image distance $v$ are positive (using the real-is-positive sign convention). The magnification for a real image is $m = v/u$. If we move the lens toward the object, $u$ decreases, so $v$ must increase to keep the lens equation $1/f = 1/u + 1/v$ satisfied. The problem tells us that the image distance increases by $x$, and gives two magnifications $m_1$ and $m_2$. The trick is to express $u$ and $v$ in terms of $m$ and $f$, then use the change in $v$ to solve for $f$.

Step-by-step solution

1. Relate magnification to object and image distances For a real image formed by a convex lens,

So we can write $v = m u$.

2. Use the lens formula The lens equation is

Substitute $v = m u$:

Hence

Then the image distance is

3. Apply to the two positions For the first position:

For the second position (after moving the lens toward the object):

4. Use the given change in image distance The image distance increases by $x$, so

Substituting:

which simplifies to

5. Solve for focal length

Concept and Intuition

The key idea here is to recall what determines the focal length of a mirror. For any spherical mirror — concave or convex — the focal length $f$ is given by $f = \frac{R}{2}$, where $R$ is the radius of curvature of the mirror. This radius is a purely geometric property: it is the radius of the sphere from which the mirror’s surface is cut. It does not depend on the color, wavelength, or frequency of the light used.

A common confusion arises because lenses do have wavelength-dependent focal lengths (chromatic aberration), but mirrors work by reflection, not refraction. Reflection obeys the law of reflection for all wavelengths equally — there is no dispersion. So, no matter what light you shine on a mirror, its focal length stays fixed.

Let’s walk through the reasoning step by step.

1. Recall the mirror formula for focal length For any spherical mirror (concave or convex), the focal length is

where $R$ is the radius of curvature. This is derived from geometry and the law of reflection, and it contains no term involving wavelength.

2. Identify what changes when wavelength is doubled

   The problem states that the wavelength of incident light is doubled, but the mirror’s area (and therefore its shape and radius) is kept constant. Changing the wavelength does not alter the physical shape of the mirror — the mirror remains the same piece of glass with the same silvered surface.

3. Check for any possible effect of wavelength on reflection

   In reflection, the angle of incidence equals the angle of reflection for all wavelengths. There is no refraction, no dispersion, and no change in the path of rays due to wavelength. The only way wavelength could affect a mirror is if the mirror were so small that diffraction became significant — but here the mirror’s area is constant and presumably much larger than the wavelength, so diffraction effects are negligible.

4. Conclude that the focal length remains unchanged

   Since $R$ is unchanged and the mirror’s behavior is independent of wavelength, the focal length stays exactly the same: