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 and Intuition

Total internal reflection occurs when light travels from a denser to a rarer medium and the angle of incidence exceeds the critical angle $\theta_c$, where $\sin\theta_c = \dfrac{n_{rarer}}{n_{denser}}$ (this comes directly from Snell's law with the refracted ray at $90°$). The glass's critical angle in air tells us $\mu_{glass}$ relative to air ($n=1$). Once we know $\mu_{glass}$, we can find the critical angle for glass in contact with any other medium (here, a liquid of index $6/5$) using the same Snell's law relation between glass and that medium — air is no longer involved.

Step-by-Step Solution

1. Critical angle in air is $30°$, and air has $n=1$, so

2. Now the glass is placed in a liquid of refractive index $\mu_{liquid} = 6/5 = 1.2$. The new critical angle $\theta_c'$ for light going from glass into this liquid satisfies

3. We are given $\sin53° = \dfrac45$. Since $37°$ and $53°$ are complementary ($37°+53°=90°$), $\sin37° = \cos53° = \sqrt{1-\left(\frac45\right)^2} = \sqrt{1-\frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac35$.
4. So $\sin\theta_c' = \dfrac35 = \sin37°$, giving $\theta_c' = 37°$.

Common Mistakes

• Re-using air's refractive index in the second step instead of switching to the liquid's index — the critical angle for glass-to-liquid has nothing to do with air once the glass is submerged.
• Mixing up which angle corresponds to $\sin=3/5$ versus $4/5$ (i.e. picking $53°$ instead of $37°$).

ANSWER: B

Concept and Intuition

When light enters the top face of the slab and refracts, the geometry of the rectangular (perpendicular) faces means the angle the ray makes with the vertical-face normal is the complement of the angle it makes with the top-face normal. If total internal reflection just occurs at the side face, the angle of incidence there equals the critical angle. This links the entry refraction angle directly to the critical angle, and hence to the refractive index through both Snell's law and the critical-angle relation $\sin C = 1/n$.

Step-by-Step Solution

1. At the top face: $\sin45^\circ = n\sin r$, where $r$ is the refraction angle (measured from the vertical normal).
2. Since the top and side faces are perpendicular, the ray's angle from the horizontal normal at the side face is $90^\circ - r$.
3. TIR occurs at B, so this angle equals the critical angle: $90^\circ - r = C \Rightarrow r = 90^\circ - C$.
4. So $\sin r = \sin(90^\circ-C) = \cos C$.
5. Substitute into Snell's law: $\sin45^\circ = n\cos C$.
6. Using $\sin C = 1/n \Rightarrow \cos C = \sqrt{1-1/n^2}$, so $\sin45^\circ = n\sqrt{1-1/n^2} = \sqrt{n^2-1}$.
7. $\sin45^\circ = 1/\sqrt2$, so $n^2 - 1 = 1/2 \Rightarrow n^2 = 3/2 \Rightarrow n = \sqrt{3/2}$.

Common Mistakes

• Forgetting that the angle at the side face is the complement of the refraction angle at the top face, not the same angle.
• Confusing the condition "TIR just occurs" (angle = critical angle) with "angle greater than critical angle".

ANSWER: A

Concept and Intuition

A ray going from denser to rarer medium bends away from the normal. As the incidence angle $i$ increases toward the critical angle $C$, the refraction angle approaches $90^\circ$ and the deviation $\delta_{refr}=(90^\circ-i)$ increases, topping out at $90^\circ-C$ right at $i=C$. Beyond $C$, refraction is no longer possible and the ray totally internally reflects, with deviation $\delta_{TIR}=180^\circ-2i$ — this is a decreasing function of $i$, so it is largest at $i$ just above $C$, i.e. effectively $180^\circ-2C$. Since TIR deviations are generically much larger than refraction deviations near the critical angle, the true maximum over all possible incidence angles is the TIR value.

Step-by-Step Solution

1. Refraction deviation at incidence $i\le C$: $\delta = 90^\circ - i$ (since refraction angle $\to 90^\circ$ as $i\to C$). Maximum at $i=C$: $\delta_{max,refr}=90^\circ-C$.
2. TIR deviation for $i>C$: reflection makes angle $i$ with normal on both sides, so deviation $= 180^\circ - 2i$. This decreases as $i$ increases, so its largest value (as $i\to C^+$) is $180^\circ-2C$.
3. Compare: for any $C<90^\circ$, $180^\circ-2C > 90^\circ - C$ (equivalent to $90^\circ > C$, always true).
4. Hence overall maximum deviation $=\pi-2C$.

Common Mistakes

• Stopping at the refraction-only answer $90^\circ-C$ and missing that TIR gives a larger deviation.
• Confusing the TIR deviation formula (which is $180^\circ-2i$, using the reflection law) with the refraction deviation formula.

ANSWER: C

Concept and Intuition

At the polarising (Brewster) angle, the reflected and refracted rays are perpendicular to each other, which leads to the simple relation $n=\tan i_p$ where $n$ is the refractive index of the medium relative to the incident (air) side. The critical angle, relevant when light tries to go the other way (medium to air), is defined by $\sin C = 1/n$. Both are properties of the same pair of media, so one determines the other.

Step-by-Step Solution

1. Brewster's law: $n = \tan(i_p) = \tan 60^\circ = \sqrt3$.
2. Critical angle relation (denser medium to air): $\sin C = \dfrac{1}{n} = \dfrac{1}{\sqrt3}$.
3. Therefore $C = \sin^{-1}\left(\dfrac{1}{\sqrt3}\right)$.

Common Mistakes

• Using $\cos$ or $\sin$ in Brewster's law instead of $\tan$.
• Writing the critical angle as $\sin^{-1}\sqrt3$ (impossible, since $\sqrt3>1$) instead of its reciprocal.

ANSWER: D

Concept and Intuition

Critical angle for total internal reflection is given by $\sin\theta_c = \dfrac{1}{n}$, where $n$ is the refractive index of the medium for that colour of light. Because a medium's refractive index normally increases as wavelength decreases (normal dispersion), violet (shortest visible wavelength) has the highest $n$, and red (longest visible wavelength) has the lowest $n$. Since $\theta_c$ varies inversely with $n$, red — having the smallest $n$ — has the largest critical angle.

Step-by-Step Solution

1. Write $\theta_c=\sin^{-1}(1/n)$.
2. Order refractive indices (normal dispersion): $n_{violet}>n_{green}>n_{yellow}>n_{red}$.
3. Since $\theta_c$ decreases as $n$ increases, the ordering of critical angles is reversed: $\theta_{c,red}>\theta_{c,yellow}>\theta_{c,green}>\theta_{c,violet}$.
4. So red has the highest critical angle among the given options.

Common Mistakes

• Assuming the colour with the highest refractive index (violet) has the highest critical angle — it's actually the opposite, an inverse relationship.

ANSWER: B

Concept and Intuition

Total internal reflection occurs beyond the critical angle, and by Snell's law applied at the critical angle (refracted ray grazing the surface at $90^\circ$): $n\sin C = 1\times\sin 90^\circ \Rightarrow \sin C = 1/n$.

Step-by-Step Solution

1. Refractive index of diamond, $n\approx2.42$.
2. $\sin C = 1/2.42 \approx 0.4132$.
3. $C = \sin^{-1}(0.4132) \approx 24.4^\circ$.

Common Mistakes

• Using the refractive index of glass ($\approx1.5$, giving $C\approx41.8^\circ$, close to a distractor option) instead of diamond's much higher value.

ANSWER: D

Concept and Intuition

The critical angle for total internal reflection depends on the refractive index via

$\sin\theta_c=1/n$: a larger refractive index means a smaller critical angle. In

normal dispersion, refractive index increases as wavelength decreases — violet/blue

bends more than red. Yellow light has a shorter wavelength than red, so it has a

slightly larger refractive index, and hence a slightly smaller critical angle.

Step-by-Step Solution

1. Critical angle relation: $\sin\theta_c=\dfrac{1}{n}$, so $\theta_c$ decreases as $n$ increases.
2. Red light ($\lambda_1$, longer wavelength) has critical angle $\theta$, i.e. $\sin\theta=1/n_{red}$.
3. Yellow light ($\lambda_2$, shorter wavelength than red) has a slightly higher refractive index: $n_{yellow}>n_{red}$ (normal dispersion).
4. Since $n_{yellow}>n_{red}$, we get $\theta_{c,yellow}<\theta$.

Common Mistakes

• Assuming refractive index is the same for all visible colours (ignoring dispersion).
• Getting the direction of dispersion backwards (thinking longer wavelength bends more).

ANSWER: C

Concept and Intuition

Total internal reflection can occur only when light travels from a denser medium into a rarer one. The critical angle $\theta_c$ is the angle of incidence (in the denser medium) for which the refracted ray just grazes the interface at $90^\circ$.

Step-by-Step Solution

1. Snell's law: $n_1\sin\theta_c=n_2\sin90^\circ=n_2$.
2. $\sin\theta_c=\dfrac{n_2}{n_1}=\dfrac12$.
3. $\theta_c=30^\circ$.

Common Mistakes

• Inverting the ratio (using $n_1/n_2$ instead of $n_2/n_1$), which would give an invalid $\sin\theta_c>1$.

ANSWER: A

Concept and Intuition

Total internal reflection (and hence a critical angle) only occurs when light travels from an optically denser medium toward a rarer one. The critical angle is defined by Snell's law applied at $90^\circ$ refraction: $n_1\sin\theta_c = n_2\sin90^\circ = n_2$.

Step-by-Step Solution

1. Here $n_1=2$ (denser, since $2>\sqrt3\approx1.732$) and $n_2=\sqrt3$ (rarer).
2. Snell's law at the critical angle: $n_1\sin\theta_c = n_2$.
3. $\sin\theta_c = \dfrac{n_2}{n_1} = \dfrac{\sqrt3}{2}$.
4. $\theta_c = \sin^{-1}\left(\dfrac{\sqrt3}{2}\right) = 60^\circ$.

Common Mistakes

• Computing $n_1/n_2$ instead of $n_2/n_1$ in the critical-angle formula.
• Forgetting that critical angle only applies going from denser to rarer medium (the direction matters).

ANSWER: C

Concept and Intuition

The refractive index of the liquid relative to air is fixed by Snell's law for light going from air (less dense) into the liquid (denser): $n_1\sin\theta = n_2\sin\alpha$, i.e. $\sin\theta = n_{liquid}\sin\alpha$ (since $n_{air}=1$). The critical angle is defined for light travelling the other way (liquid to air), at which the refracted ray grazes the surface ($90^\circ$): $n_{liquid}\sin\theta_c = n_{air}\sin90^\circ = 1$.

Step-by-Step Solution

1. Snell's law, air → liquid: $1\cdot\sin\theta = n_{liquid}\cdot\sin\alpha \implies n_{liquid} = \dfrac{\sin\theta}{\sin\alpha}$.
2. Critical angle condition, liquid → air: $n_{liquid}\sin\theta_c = 1 \implies \sin\theta_c = \dfrac{1}{n_{liquid}}$.
3. Substitute: $\sin\theta_c = \dfrac{1}{\sin\theta/\sin\alpha} = \dfrac{\sin\alpha}{\sin\theta}$.

Common Mistakes

• Inverting the refractive index formula (writing $n=\sin\alpha/\sin\theta$ instead of $\sin\theta/\sin\alpha$) since air is the first, less dense medium.
• Confusing $\theta$ (angle of incidence in air) with $\theta_c$ (critical angle, a different angle defined for the reverse path).

ANSWER: A

Concept and Intuition

Total internal reflection (and hence a critical angle) can only occur when light travels from an optically denser medium (higher refractive index) toward a rarer medium (lower refractive index). Here the slab ($n=5/3$) is denser than water ($n=4/3$), so light going from glass to water can undergo total internal reflection beyond the critical angle.

Step-by-Step Solution

1. Using Snell's law at the critical angle, the refracted ray in the rarer medium grazes along the interface ($90^\circ$):

2. Solve for $\sin\theta_c$:

3. Hence:

Common Mistakes

• Inverting the ratio (using $n_{glass}/n_{water}$ instead of $n_{water}/n_{glass}$), which would incorrectly give a value greater than 1 for $\sin\theta_c$ (impossible).
• Forgetting that critical angle formulas require going from denser to rarer medium — here that direction is glass$\to$water, exactly as stated in the question.

ANSWER: A