Q.Consider sunlight incident on a slit of width $10^4\ \text{\AA}$. The image seen through the slit shall

Step 1 — Decode the geometry

The equiconvex lens is cut by a vertical plane through its centre, perpendicular to the principal axis. So:

• The left piece is bounded by the front convex surface ($R_1=+14$ cm) and the flat cut ($R_2=\infty$) → a plano-convex lens of $n_1=1.5$.
• The right piece is bounded by the flat cut ($R_1=\infty$) and the rear convex surface ($R_2=-14$ cm) → a plano-convex lens of $n_2=1.2$.

They are cemented together, i.e. two thin lenses in contact, sharing the same axis. ("Equiconvex, radius 14 cm" fixes both magnitudes at 14 cm; the sign convention makes the second surface negative because its centre of curvature lies to the left.)

Step 2 — Focal length of each half (lens-maker's formula)

Left lens ($n=1.5$, $R_1=+14$, $R_2=\infty$):

Right lens ($n=1.2$, $R_1=\infty$, $R_2=-14$):

Step 3 — Combine (lenses in contact ⇒ powers add)

For thin lenses in contact, $P=P_1+P_2$, i.e. $\dfrac{1}{f}=\dfrac{1}{f_1}+\dfrac{1}{f_2}$:

Step 4 — Thin-lens equation

Object at 40 cm on the incident side: $u=-40$ cm.

Step 5 — Sanity check

The object sits at $2f$ ($=40$ cm) of the combined lens, and the textbook result for an object at $2f$ is a real image at $2f$ on the other side, of unit magnification — exactly what we got. The image is 40 cm from the lens.

ANSWER: D

The key concept is that frequency of an electromagnetic wave is determined by the source and does not change when the wave moves from one medium to another. What changes are the wave’s speed and wavelength, because the medium’s permittivity (and possibly permeability) alters the phase velocity. For a non-magnetic dielectric ($\mu_r = 1$), the refractive index is $n = \sqrt{\varepsilon_r}$. The wave slows down by a factor $n$, and since $v = f\lambda$, the wavelength must shrink by the same factor.

1. Identify the given data

   • Frequency in vacuum: $f = 3\ \text{MHz}$ (this will remain the same in the dielectric).
   • Relative permittivity: $\varepsilon_r = 4.0$.
   • Relative permeability: $\mu_r = 1$ (non-magnetic).

2. Find the refractive index

   For a non-magnetic medium, $n = \sqrt{\mu_r \varepsilon_r} = \sqrt{1 \times 4} = 2$.

   This means the speed of the wave in the dielectric is $v = \frac{c}{n} = \frac{c}{2}$.

3. Relate speed, frequency, and wavelength

   In vacuum: $c = f \lambda_0$.

   In the dielectric: $v = f \lambda$.

   Since $f$ is unchanged, we have

So the wavelength is halved.

4. Check the options
   • (A) Wavelength halved, frequency doubled → wrong (frequency doesn’t double).
   • (B) Wavelength doubled, frequency unchanged → wrong (wavelength is halved, not doubled).
   • (C) Both unchanged → wrong (wavelength changes).
   • (D) Wavelength halved, frequency unchanged → correct.

Concept: when light enters a denser medium its FREQUENCY is unchanged (it is set by the source), while its speed drops to v = c/n; hence the wavelength shortens:

lambda_medium = lambda_air / n.

Given lambda_air = 600 nm and n = 1.5:

lambda_medium = 600 / 1.5 = 400 nm.

So while travelling through the medium the light has a wavelength of 400 nm.

ANSWER: D

1. What stays the same and what changes

When light passes from medium 1 into medium 2:

2. Why a change of speed makes the ray bend

Consider a plane wavefront striking the boundary obliquely. The edge of the wavefront that enters the denser medium first immediately slows down, while the rest of the wavefront is still travelling at the old (faster) speed. In a given time the two ends advance by different distances, so the wavefront swivels — and the ray, being perpendicular to the wavefront, bends.

Huygens' construction turns this straight into Snell's law. In time $t$, the incident wavefront covers $v_1 t$ in medium 1 while the refracted one covers $v_2 t$ in medium 2 across the same interface segment $AB$:

Notice the speed ratio is the whole content of Snell's law. If $v_1 = v_2$, then $\sin i = \sin r$ and there is no bending at all — which proves the change in speed is the cause.

3. Rejecting the other options

1. (A) Frequency is different — false. Frequency is invariant across a boundary; if it changed, the wave would have to "pile up" or "vanish" at the interface (crests would not be conserved). ✗
2. (C) Coefficient of elasticity is different — this is a property of a mechanical/elastic medium; light is an electromagnetic wave and needs no elastic medium at all (it travels through vacuum). ✗
3. (D) Amplitude is smaller — the amplitude does change (some energy is reflected and the medium may absorb), but amplitude affects only the intensity/brightness, never the direction of propagation. A dim ray and a bright ray refract through exactly the same angle. ✗

ANSWER: B

The key here is that the image is formed on a screen. A screen can only capture a real image — one where light rays actually converge at a point. That immediately tells you the lens must be convex (converging), because a concave lens always produces a virtual image that cannot be projected onto a screen.

Now, the lens formula is the standard tool for relating object distance, image distance, and focal length. But the sign convention matters. In the Cartesian sign convention (used in most Indian board exams):

• Distances measured from the optical centre toward the incident light are negative.
• Distances measured opposite to the incident light are positive.

The candle (object) is on one side of the lens, and the screen (image) is on the other side. So the object distance $u$ is negative, and the image distance $v$ is positive.

1. Assign signs correctly

   Object distance: $u = -25~\mathrm{cm}$ (since object is on the incident side)

   Image distance: $v = +75~\mathrm{cm}$ (since image is on the opposite side, real)

2. Apply the lens formula

   The lens formula is:

Substitute the values:

3. Simplify

Therefore:

4. Interpret the sign The focal length is positive, which confirms a convex lens. A concave lens would have a negative focal length.