A short pulse of white light is incident from air to a glass slab at normal incidence. After travelling throug

NCERT Exemplar · Q2

Q. A short pulse of white light is incident from air to a glass slab at normal incidence. After travelling through the slab, the first colour to emerge is

(a) blue.

(b) green.

(d) red.

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Concept understanding — Critical Angle Comparison

Critical Angle Comparison: From Intuition to Precision

Imagine you're standing at the edge of a swimming pool, looking down at a coin at the bottom. The coin appears closer to the surface than it actually is. That's refraction — light bends when it moves from water into air. Now imagine tilting your head so you're looking at the coin from a very shallow angle. At some point, the coin suddenly vanishes. You can't see it anymore, no matter how hard you try. That vanishing point is the critical angle.

The Intuition

Light travels at different speeds in different materials. When it crosses from a denser medium (like water or glass) into a rarer medium (like air), it bends away from the normal (the imaginary line perpendicular to the surface). The larger the angle of incidence (the angle at which light hits the boundary), the more it bends away.

At a certain angle of incidence, the refracted ray bends so much that it runs exactly along the surface — it makes a 90° angle with the normal. That's the critical angle. If you increase the angle of incidence even slightly beyond this, the light can't escape at all. It reflects back into the denser medium, a phenomenon called total internal reflection.

Note

Critical angle only exists when light travels from a denser medium to a rarer medium. Going the other way (rarer to denser), light always bends toward the normal — no critical angle, no total internal reflection.

The Precise Statement

The critical angle ( $\theta_c$ ) is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is exactly 90°.

Mathematically, from Snell's law:

n_1 \sin \theta_1 = n_2 \sin \theta_2

Let medium 1 be the denser medium (refractive index $n_1$ ) and medium 2 be the rarer medium (refractive index $n_2$ , with $n_2 < n_1$ ). At the critical angle, $\theta_1 = \theta_c$ and $\theta_2 = 90^\circ$ , so $\sin \theta_2 = 1$ . This gives:

n_1 \sin \theta_c = n_2 \times 1

\sin \theta_c = \frac{n_2}{n_1}

Where:

$\theta_c$ = critical angle
$n_1$ = refractive index of the denser medium
$n_2$ = refractive index of the rarer medium

What This Tells You

The critical angle depends only on the ratio of the two refractive indices. A larger difference between $n_1$ and $n_2$ means a smaller critical angle — light is more "trapped" inside the denser medium. For example:

Medium pair	$n_1$ (denser)	$n_2$ (rarer)	$\theta_c$
Water → Air	1.33	1.00	$\approx 48.8^\circ$
Glass → Air	1.50	1.00	$\approx 41.8^\circ$
Diamond → Air	2.42	1.00	$\approx 24.4^\circ$

Diamond's tiny critical angle is why it sparkles so brilliantly — light gets trapped inside and bounces around before escaping.

Watch out

A common mistake: thinking the critical angle is measured from the surface. It is always measured from the normal (the perpendicular line), just like any other angle in optics.

The Three Regimes

For a given denser-to-rare boundary:

$\theta < \theta_c$ : Refraction occurs — light escapes into the rarer medium, bending away from the normal.
$\theta = \theta_c$ : The refracted ray grazes the surface at 90°.
$\theta > \theta_c$ : Total internal reflection — no light escapes; all of it reflects back into the denser medium.

This is the principle behind optical fibres, where light is kept inside a glass core by repeated total internal reflection, and behind the brilliant sparkle of a cut diamond.

"Critical angle formula total internal reflection" and "critical angle class 12 physics numericals" are commonly searched terms, both drawn from the Ray Optics and Optical Instruments chapter of the NCERT/CBSE Class 12 Physics curriculum. Comparing critical angles across water, glass, and diamond is a classic JEE Main and NEET question.

The key idea is that the critical angle for total internal reflection is largest for red light (because refractive index is smallest for red). Since the pulse enters at normal incidence, all colours travel the same path length inside the slab, but the group velocity (energy propagation speed) is highest for red light. Therefore, red emerges first.

Why Critical Angle Comparison Works — The Intuition

When white light enters a glass slab at normal incidence, all colours enter without bending. Inside the slab, each colour travels at a different speed because the refractive index $n$ depends on wavelength — this is dispersion. For typical glass, $n$ is highest for violet (shortest wavelength) and lowest for red (longest wavelength).

The speed of light in the medium is $v = c/n$ . So red, with the smallest $n$ , travels fastest. Over a fixed slab thickness $t$ , the time taken is $t/v = n t / c$ . Since red has the smallest $n$ , it takes the least time and emerges first.

But why bring up the critical angle? Because the critical angle $\theta_c$ is defined by $\sin \theta_c = 1/n$ (for glass-to-air). A larger $n$ means a smaller $\theta_c$ . So red, with the smallest $n$ , has the largest critical angle. This is a handy mnemonic: the colour that bends least on entering (red) also has the largest critical angle and travels fastest inside the medium. The critical angle comparison is a quick way to rank refractive indices without memorising numbers.

Watch out

A common mistake is to think that the colour with the largest refractive index (violet) emerges first because it "bends more". But bending only happens at oblique incidence. At normal incidence, there is no bending — only speed matters. The colour with the smallest $n$ (red) is fastest.

Step-by-Step Reasoning

Normal incidence means no refraction at the first surface.

When light enters at $0^\circ$ to the normal, Snell's law gives $n_{\text{air}} \sin 0^\circ = n_{\text{glass}} \sin r$ , so $r = 0^\circ$ . All colours go straight in along the same path.
Inside the slab, each colour travels the same geometric distance — the slab thickness $d$ . But the optical path length is $n d$ , and the actual time taken is $t = \frac{n d}{c}$ .
Refractive index varies with colour.

For ordinary glass, the dispersion curve is:
- Red: $n \approx 1.51$ (lowest)
- Yellow: $n \approx 1.52$
- Green: $n \approx 1.53$
- Blue: $n \approx 1.54$
- Violet: $n \approx 1.55$ (highest)
So the time order is: red < yellow < green < blue < violet.
Critical angle as a ranking tool.

The critical angle for the glass-air interface is $\theta_c = \sin^{-1}(1/n)$ . Since $n$ is smallest for red, $\sin^{-1}(1/n)$ is largest for red. This gives the same ranking: red has the largest critical angle, hence the smallest $n$ , hence the fastest speed.
The first colour to emerge is the fastest.

Therefore, red light exits the slab first.

Tip

You don't need to remember exact $n$ values. Just recall the mnemonic: VIBGYOR — Violet has the highest refractive index (slowest), Red has the lowest (fastest). The critical angle is inversely related: larger $n$ → smaller $\theta_c$ .

✓Final answer

The first colour to emerge is red .

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

KCET 2026Set C21 markMCQ
$\theta$
›Reveal solutionSolution

At the critical angle, Snell's law with an angle of refraction of $90^\circ$ relates the refractive indices of the two media; converting refractive index to speed using $n=c/v$ gives the required expression.

Step 1 — Apply Snell's law at the critical angle

Light travels from medium A into medium B; at the critical angle $\theta$ , the refracted ray grazes the boundary (refraction angle $=90^\circ$ ):
$n_A\sin\theta = n_B\sin90^\circ = n_B \implies \frac{n_B}{n_A} = \sin\theta$
Step 2 — Express refractive indices in terms of speed

Since $n=\dfrac{c}{v}$ , with speed in medium A equal to $V$ and speed in medium B equal to $V_B$ :
$\frac{n_B}{n_A} = \frac{c/V_B}{c/V} = \frac{V}{V_B}$
Step 3 — Solve for the speed in medium B
$\frac{V}{V_B} = \sin\theta \implies V_B = \frac{V}{\sin\theta}$
✓Final answer
$\dfrac{V}{\sin\theta}$
KCET 2025Set D-41 markMCQ
$\nu$
(1) a step function that jumps at the threshold frequency and saturates,
(2) a smooth S-shaped curve with no marked threshold,
(3) a curve decaying smoothly from a large value, and
(4) zero up to the threshold frequency then rising linearly without saturating.](/figures/dual-nature-of-radiation-and-matter/photocurrent-vs-frequency-fixed-voltage-options-fig-pyq-t7fthv.svg) (A) (B) (C) (D)
›Reveal solutionSolution
Below $\nu_0$ : no emission, $I=0$ . Above $\nu_0$ : the number of ejected electrons (hence the current) is frequency-independent — a step to a constant value.

Step 1 — Below the threshold frequency.

Einstein's photoelectric equation:
$h\nu=\phi_0+K_{\max},\qquad \phi_0=h\nu_0 .$
If $\nu<\nu_0$ then $h\nu<\phi_0$ : no photon has enough energy to free an electron, no matter how long you wait or how intense the beam. So
$I=0\quad\text{for }\nu<\nu_0 .$
This gives the flat zero portion on the left of the graph.

Step 2 — At and above the threshold frequency.

Emission starts at $\nu_0$ . The anode voltage is fixed (and positive/large enough to collect the emitted electrons), so we are on the saturation part of the $I$ – $V$ curve: every emitted electron is collected, and
$I=n_e e$
where $n_e$ is the number of electrons emitted per second. One photon liberates at most one electron, so $n_e$ tracks the photon arrival rate, which is set by the intensity of the source — not by the frequency.

Step 3 — Hence the shape.

Increasing $\nu$ beyond $\nu_0$ gives each electron more kinetic energy ( $K_{\max}=h(\nu-\nu_0)$ , which is what changes the stopping potential), but it does not change how many electrons come out. So the photocurrent stays constant for all $\nu>\nu_0$ :

zero for $\nu<\nu_0$ ,

a sharp step at $\nu=\nu_0$ ,

a horizontal line thereafter.

Step 4 — Eliminate.

(D) a line rising steadily with $\nu$ would mean more electrons at higher frequency — contradicts the photon-count argument. (C) a decaying curve implies emission below threshold. (B) has no threshold at all, which is the very feature the photoelectric effect is famous for.

✓Final answer
$\nu_0$

ANSWER: A

KCET 2024Set D-21 markMCQ

R = 20~\Omega

›Reveal solution

"Better tuning" means a sharper, more selective resonance — i.e. the highest quality factor $Q=\dfrac{1}{R}\sqrt{L/C}$ . Comparing all four combinations, option (D) has by far the largest $Q$ .

Step 1 — What makes a tuning circuit "better"?

In a series LCR circuit used to tune a receiver to one station's frequency among many crowded nearby frequencies, "good tuning" means a sharp, narrow resonance peak — the circuit should respond strongly at the desired frequency and fall off quickly on either side, rejecting neighbouring stations. This sharpness is quantified by the quality factor:

Q = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{\omega_0 L}{R}

A larger $Q$ means a narrower resonance curve (better selectivity). So among several LCR combinations, the one with the largest $Q$ gives the better tuning. Physically: low resistance wastes less energy (less damping), while a large $L/C$ ratio (big inductance, small capacitance) makes the resonance curve steep.

Step 2 — Compute $Q$ for every option.

Option	$R~(\Omega)$	$L$ (H)	$C~(\mu\mathrm{F})$	$L/C$	$\sqrt{L/C}$	$Q=\dfrac{1}{R}\sqrt{L/C}$
(A)	20	1.5	35	$4.29\times10^{4}$	$207.0$	$10.35$
(B)	25	2.5	45	$5.56\times10^{4}$	$235.7$	$9.43$
(C)	25	1.5	45	$3.33\times10^{4}$	$182.6$	$7.30$
(D)	15	3.5	30	$1.167\times10^{5}$	$341.6$	$\mathbf{22.77}$

Step 3 — Pick the largest $Q$ .

Option (D) is the clear winner — it is not close. It combines:

the smallest resistance ( $15~\Omega$ , vs. 20–25 for the others),
the largest inductance ( $3.5$ H, vs. 1.5–2.5 for the others),
one of the smallest capacitances ( $30~\mu\mathrm{F}$ , vs. 35–45 for the others).

Every one of these three properties independently pushes $Q$ up, so (D) dominates the comparison — it never needs a trade-off against the other options.

Tip

You rarely need the exact numeric $Q$ values on an exam — just check which option has the lowest R and highest L/C ratio simultaneously; that option wins.

✓Final answer

R=15~\Omega,\ L=3.5~\mathrm{H},\ C=30~\mu\mathrm{F}

KCET 2024Set D-21 markMCQ
$30^\circ$
›Reveal solutionSolution

Halve the unpolarised intensity at the first polaroid, then apply Malus's law twice using the angles between consecutive pass axes ( $30^\circ$ then $60^\circ$ ).

Step 1 — Sheet 1: unpolarised → polarised

Unpolarised light contains all vibration directions equally. A polaroid transmits only the component along its pass axis, and averaging $\cos^2\theta$ over all $\theta$ gives $\tfrac12$ :
$I_1=\frac{I_0}{2}$
The light emerging is now linearly polarised along axis 1. (Malus's law does not apply at the first sheet — the incident light has no definite polarisation direction.)

Step 2 — Sheet 2: Malus's law

Malus's law: when light already polarised meets an analyser whose axis makes angle $\theta$ with the polarisation direction,
$I_{\text{out}}=I_{\text{in}}\cos^2\theta$
The crucial subtlety: $\theta$ is the angle between the incoming polarisation and the next pass axis — not the angle with sheet 1 in every case.

Here the light entering sheet 2 is polarised along axis 1, and axis 2 makes $30^\circ$ with axis 1:
$I_2=I_1\cos^2 30^\circ=\frac{I_0}{2}\left(\frac{\sqrt3}{2}\right)^2=\frac{I_0}{2}\cdot\frac{3}{4}=\frac{3I_0}{8}$
The light now emerges polarised along axis 2.

Step 3 — Sheet 3: Malus's law again, with the relative angle

Axis 3 makes $90^\circ$ with axis 1, and axis 2 makes $30^\circ$ with axis 1. So the angle between axis 2 and axis 3 is
$90^\circ-30^\circ=60^\circ$
The light arriving at sheet 3 is polarised along axis 2, so
$I_3=I_2\cos^2 60^\circ=\frac{3I_0}{8}\left(\frac{1}{2}\right)^2=\frac{3I_0}{8}\cdot\frac14=\frac{3I_0}{32}$
Step 4 — Why the answer is not zero

Sheets 1 and 3 are crossed ( $90^\circ$ apart). With only those two, the output would indeed be zero. But inserting the middle polaroid at an intermediate angle rotates the plane of polarisation in two smaller steps, and light gets through — the celebrated "three-polaroid" result. That is precisely why option (A) "Zero" is the trap.
$\frac{3I_0}{32}\approx 0.094\,I_0$
✓Final answer
$\dfrac{3I_0}{32}$

ANSWER: B
KCET 2023Set A-31 markMCQ
$1~\mathrm{ms}^{-1}$
›Reveal solutionSolution

Along the axis the image speed is $v_i=m^2\,v_o$ , where $m$ is the lateral magnification. Equal speeds ( $v_i=v_o$ ) require $m^2=1$ , i.e. $|m|=1$ , which for a convex lens occurs at object distance $u=2f=20\,\text{cm}$ - option (B).

Concept: longitudinal (axial) magnification

Differentiating the lens equation shows that for motion along the principal axis
$v_i=m^2\,v_o,\qquad m=\frac{v}{u}\ (\text{lateral magnification}).$
Setting $v_i=v_o=1\,\text{m s}^{-1}$ gives $m^2=1$ , so $|m|=1$ .

Where $|m|=1$ for a convex lens

For a convex lens a same-size (real, inverted, $m=-1$ ) image forms when the object is at $2f$ . With $f=10\,\text{cm}$ ,
$u=2f=20\,\text{cm}.$
Check with $\dfrac1v-\dfrac1u=\dfrac1f$ using $u=-20\,\text{cm}$ , $f=+10\,\text{cm}$ :
$\frac1v=\frac1{10}+\frac1{-20}=\frac1{20}\;\Rightarrow\;v=+20\,\text{cm},\qquad m=\frac{v}{u}=-1,$
so $m^2=1$ and $v_i=v_o$ .

✓Final answer
$2f=20\,\text{cm}$
COMEDK 2023Set 2023-E1 markMCQ
$\sqrt2$
›Reveal solutionSolution

n = sqrt2, so sin C = 1/sqrt2 = 0.707 C = 45 degrees.

Concept: critical angle for total internal reflection, sin C = 1/n (light going from the denser medium into air).

n = sqrt2, so

sin C = 1/sqrt2 = 0.707

C = 45 degrees.

✓Final answer
$^\circ$

ANSWER: B

KCET 2022Set B-31 markMCQ

Q. Which of the following radiations of electromagnetic waves has the highest wavelength? (A) IR-rays (B) Microwaves (C) X-rays (D) UV-rays

›Reveal solution

Order the four radiations on the EM spectrum; microwaves lie farthest from the $\gamma$ /X-ray end, so they have the largest wavelength.

1. The concept

All electromagnetic waves travel at $c = 3\times10^{8}\ \text{m s}^{-1}$ in vacuum, and

c = \nu\lambda \quad\Longrightarrow\quad \lambda = \frac{c}{\nu}

so wavelength is inversely proportional to frequency (and to photon energy $E = h\nu$ ). Ranking by wavelength is therefore the reverse of ranking by energy.

2. Place the four options

Radiation	Typical wavelength	Typical frequency
X-rays	$10^{-11}$ – $10^{-9}\ \text{m}$	$\sim10^{16}$ – $10^{19}\ \text{Hz}$
UV-rays	$10^{-8}$ – $4\times10^{-7}\ \text{m}$	$\sim10^{15}$ – $10^{17}\ \text{Hz}$
IR-rays	$7\times10^{-7}$ – $10^{-3}\ \text{m}$	$\sim10^{11}$ – $10^{14}\ \text{Hz}$
Microwaves	$10^{-3}$ – $0.3\ \text{m}$	$\sim10^{9}$ – $10^{11}\ \text{Hz}$

3. Sort them

\lambda_{\text{X-ray}} < \lambda_{\text{UV}} < \lambda_{\text{IR}} < \lambda_{\text{microwave}}

Microwaves have the lowest frequency of the four, and by $\lambda = c/\nu$ that means the highest wavelength.

(Note IR is the runner-up — the two bands actually meet around $1\ \text{mm}$ , but the microwave band extends well beyond that into the centimetre range.)

✓Final answer

The correct option is (B) — Microwaves.

ANSWER: B

KCET 2021Set B-21 markMCQ
$30\,\mathrm{cm}$
›Reveal solutionSolution

The convex lens's image becomes a virtual object 4 cm behind the concave lens; apply $\tfrac1v-\tfrac1u=\tfrac1f$ with $u=+4\ \mathrm{cm}$ to get $m=1.25$ , so the 2 cm image grows to 2.5 cm.

1. Locate the image formed by the convex lens alone

The object is at infinity, so the convex lens forms its image exactly at the focal plane:
$v_1 = f_1 = 30\ \mathrm{cm}\ \text{(to the right of the convex lens)}$
That image is 2 cm tall (given).

2. Where does the concave lens sit?

The concave lens is placed 26 cm from the convex lens, i.e. between the convex lens and the image.
$\text{distance from concave lens to that image} = 30 - 26 = \mathbf{4\ cm}\ \text{(on the far side, to the right)}$
3. The key idea — a VIRTUAL OBJECT

The converging rays are intercepted by the concave lens before they can meet. They are heading for a point 4 cm behind (to the right of) the concave lens. For the concave lens, this point is a virtual object.

With the Cartesian sign convention (light travels left → right, distances measured from the lens, rightward positive):
$u = +4\ \mathrm{cm}\ \ (\textbf{positive} — \text{the object is on the OUTGOING side}), \qquad f_2 = -20\ \mathrm{cm}\ (\text{concave})$
This positive $u$ is the whole trick of the problem — a real object would give a negative $u$ .

4. Apply the thin-lens equation to the concave lens
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-20} + \frac{1}{+4}$ $\frac{1}{v} = \frac{-1}{20} + \frac{5}{20} = \frac{4}{20} = \frac{1}{5}$ $\boxed{v = +5\ \mathrm{cm}}$
So the final image forms 5 cm to the right of the concave lens — a real image (positive $v$ ), pushed further out than the original 4 cm, exactly as a diverging lens should do to converging light.

5. Magnification and the new size
$m = \frac{v}{u} = \frac{+5}{+4} = 1.25$ $h_{\text{new}} = m \times h_{\text{old}} = 1.25 \times 2\ \mathrm{cm} = \boxed{2.5\ \mathrm{cm}}$
6. Sanity check

The concave lens diverges the converging beam, so the rays cross the axis further away ( $5\ \mathrm{cm} > 4\ \mathrm{cm}$ ) and, having travelled further from the axis-crossing geometry, the image is magnified ( $|m| = 1.25 > 1$ ). An enlargement to 2.5 cm is exactly what we expect — this is the working principle of the telephoto lens. ✓

(Option (A), 1.25 cm, is the magnification itself mistaken for the size; option (D), 2 cm, is the answer if you forget the concave lens does anything.)

✓Final answer
$2.5\ \mathrm{cm}$

ANSWER: B
KCET 2019Set A-11 markMCQ
$0°C$
›Reveal solutionSolution
Upthrust $\propto$ density of the liquid, and water is densest at $4^\circ$ C — so the buoyant force is greater at $4^\circ$ C than at $0^\circ$ C.

Step 1 — Archimedes' principle

The buoyant force on a fully submerged body is the weight of the liquid it displaces:
$F_B = \rho_{\text{liquid}} \; V_{\text{displaced}} \; g$
For a fully-dipped sphere, $V_{\text{displaced}} = V_{\text{sphere}}$ . So $F_B$ depends on the density of the water, the sphere's volume, and $g$ — nothing else. In particular it does not depend on the sphere's own density or material.

Step 2 — The anomalous expansion of water

Water is unusual: on warming from $0^\circ$ C to $4^\circ$ C it contracts rather than expands, and only above $4^\circ$ C does it expand normally. Hence its density peaks at $4^\circ$ C:
$\rho_{4^\circ\mathrm{C}} = 1000.0\ \mathrm{kg/m^3} \;>\; \rho_{0^\circ\mathrm{C}} \approx 999.8\ \mathrm{kg/m^3}$
(This is precisely why lakes freeze from the top down and fish survive beneath the ice.)

Step 3 — Compare the two cases
$\frac{F_B(0^\circ\mathrm{C})}{F_B(4^\circ\mathrm{C})} = \frac{\rho_{0^\circ} V_{0^\circ}}{\rho_{4^\circ} V_{4^\circ}}$
The aluminium sphere's volume changes only through its own (very small) thermal expansion over a $4^\circ$ C interval — a relative change of order $\alpha_V \Delta T \approx (7\times10^{-5})(4) \approx 3\times10^{-4}$ , i.e. negligible and, if anything, it makes the sphere smaller at $0^\circ$ C too. The dominant effect is the water's density.

Since $\rho_{0^\circ} < \rho_{4^\circ}$ :
$F_B(0^\circ\mathrm{C}) < F_B(4^\circ\mathrm{C})$
Step 4 — Screen the options

(A) Buoyancy less at $0^\circ$ C than at $4^\circ$ C — correct, this is exactly what $\rho_{0^\circ} < \rho_{4^\circ}$ gives.

(B) The reverse — wrong; it would require water to be densest at $0^\circ$ C.

(C) "Same" — wrong; it ignores the anomalous expansion entirely.

(D) Radius-dependent — wrong; $F_B \propto V$ scales both cases identically, so the ratio is independent of radius. Buoyancy is always larger at $4^\circ$ C, whatever the size.

✓Final answer
$0^\circ$

ANSWER: A

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Q.A short pulse of white light is incident from air to a glass slab at normal incidence. After travelling through the slab, the first colour to emerge is

Concept understanding — Critical Angle Comparison

Critical Angle Comparison: From Intuition to Precision

The Intuition

The Precise Statement

What This Tells You

The Three Regimes

Why Critical Angle Comparison Works — The Intuition

Step-by-Step Reasoning

Method: Ranking Colours by Speed Inside a Medium (Dispersion Problems)

Steps

Step 1 — Apply Snell's law at the critical angle

Step 2 — Express refractive indices in terms of speed

Step 3 — Solve for the speed in medium B

Step 1 — Sheet 1: unpolarised → polarised

Step 2 — Sheet 2: Malus's law

Step 3 — Sheet 3: Malus's law again, with the relative angle

Step 4 — Why the answer is not zero

Concept: longitudinal (axial) magnification

Where $|m|=1$ for a convex lens

1. The concept

2. Place the four options

3. Sort them

1. Locate the image formed by the convex lens alone

2. Where does the concave lens sit?

3. The key idea — a VIRTUAL OBJECT

4. Apply the thin-lens equation to the concave lens

5. Magnification and the new size

6. Sanity check

Step 1 — Archimedes' principle

Step 2 — The anomalous expansion of water

Step 3 — Compare the two cases

Step 4 — Screen the options

Q.A short pulse of white light is incident from air to a glass slab at normal incidence. After travelling through the slab, the first colour to emerge is

Why Critical Angle Comparison Works — The Intuition

Step-by-Step Reasoning

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