Q.A small bulb is placed at the bottom of a tank containing water to a depth of $80\ \text{cm}$. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is $1.33$. (Consider the bulb to be a point source.)

The key here is to connect the prism angle, the path of the ray, and the critical angle condition. When a ray enters a prism normally (perpendicular to the surface), it does not bend at the first face. So the ray inside the prism travels straight toward the second face. The angle at which it meets that second face is simply the prism angle itself. If that angle is larger than the critical angle for the glass-air interface, the ray cannot escape — it reflects internally.

Why this approach works:

We don’t need Snell’s law at the first face because normal incidence means zero refraction. The only question is whether the ray can exit the second face. That depends on comparing the angle of incidence on the second face to the critical angle. The prism angle is given as $45^\circ$, and the critical angle is $42^\circ$. Since $45^\circ > 42^\circ$, total internal reflection occurs.

1. Ray enters normally — The ray is perpendicular to the first face, so it passes straight into the glass without bending. Its direction inside the prism is exactly along the normal to the first face.

2. Geometry inside the prism — In a prism, the angle between the two faces is the prism angle $A = 45^\circ$. The ray, traveling straight from the first face, will hit the second face at an angle equal to $A$. This is because the ray’s path is perpendicular to the first face, and the second face is tilted by $A$ relative to the first. So the angle of incidence on the second face is $i = 45^\circ$.

3. Compare with the critical angle — The critical angle for glass-air is given as $42^\circ$. Total internal reflection occurs when the angle of incidence inside the denser medium exceeds the critical angle. Here $45^\circ > 42^\circ$, so the ray cannot refract out — it must reflect internally. …

Condition for total internal reflection. TIR can occur only when light travels from the optically denser medium into the rarer one, at an angle of incidence greater than or equal to the critical angle $\theta_c$.

Here the solid ($n_2=2.9$) is denser than the liquid ($n_1=1.45$), so the ray must go solid $\to$ liquid. This immediately rules out any choice describing "liquid $\to$ solid" (rarer to denser can never give TIR).

Critical angle:

$\sin\theta_c=\frac{n_{\text{rarer}}}{n_{\text{denser}}}=\frac{1.45}{2.9}=0.5\ \Rightarrow\ \theta_c=30^{\circ}$ …

Step 1 — Account for all the energy at an interface.

When light meets a boundary between two media, the incident energy in general splits three ways:

Step 2 — Apply the stated conditions.

The question says the second medium is non-reflecting and non-absorbing. Therefore

Substituting:

So the total energy carried by the wave is unchanged. There is simply no physical channel left through which it could grow or shrink — and energy cannot be created, so "increases" is ruled out on principle.

Step 3 — What does change on refraction, and what does not.

It is worth being precise, because this is where the distractors live:

• Frequency $\nu$ — unchanged. The oscillating fields at the boundary drive the second medium at the same rate; the frequency is set by the source. This is why the photon energy $E = h\nu$ of each photon is also unchanged.
• Speed $v$ — changes: $v = c/n$.
• Wavelength $\lambda$ — changes: $\lambda = v/\nu = \lambda_{vac}/n$.
• Intensity (energy per unit area per unit time) — can change, because the beam cross-section and speed change — but the total energy carried does not. …

Concept and Intuition

Total internal reflection occurs when light travels from a denser medium (higher refractive index) to a rarer medium (lower refractive index) and the angle of incidence exceeds a certain threshold called the critical angle. At the critical angle, the refracted ray grazes the boundary (angle of refraction = 90°). For angles larger than this, no light enters the second medium — it is all reflected back. Here, glass (n = 1.5) is denser than water (n ≈ 1.333), so total internal reflection is possible. The minimum angle for which no light enters water is exactly this critical angle.

Step-by-step solution

1. Identify the refractive indices

   Glass: $n_g = \frac{3}{2}$

   Water: $n_w = \frac{4}{3}$

   Since $n_g > n_w$, light travels from denser to rarer medium.

2. Recall Snell’s law at the critical angle

   Snell’s law: $n_g \sin i = n_w \sin r$

   At the critical angle $i_c$, the angle of refraction $r = 90^\circ$, so $\sin r = 1$.

   Thus:

3. Solve for $\sin i_c$ $$$$

Total internal reflection occurs when light travels from a denser to a rarer medium at an angle of incidence exceeding the critical angle $\theta_c$, where $\sin\theta_c=\dfrac{n_{\text{air}}}{n_{\text{medium}}}$.

1. Speed of light in the medium. It covers $3\text{ cm}=0.03\text{ m}$ in $0.2\text{ ns}=0.2\times10^{-9}\text{ s}$:

2. Refractive index.

3. Critical angle for the medium-to-air interface: …

Light enters the flat end face at angle $i$ and refracts at $r$, then strikes the lateral face at angle $(90^\circ-r)$. For it never to escape the side, this must exceed the critical angle $\theta_c$ for all $i$:

$90^\circ-r\ge\theta_c\ \Rightarrow\ \cos r\ge\sin\theta_c=\frac{1}{n}.$ …

1. First polaroid — the factor of one half.

Unpolarised light contains all vibration directions with equal probability. A polaroid transmits only the component along its pass axis, and averaging $\cos^2\theta$ over all angles gives $\langle\cos^2\theta\rangle = \tfrac12$. Hence, for any orientation of the first polaroid:

and this light is now plane-polarised along the first pass axis.

2. Second polaroid — Malus' law.

The light reaching the second polaroid is polarised, so Malus' law applies. Only the component of the electric field along the second pass axis gets through, $E_2 = E_1\cos\theta$, and since intensity $\propto E^2$:

where $\theta$ is the angle between the pass axes.

3. Impose the given condition. …

1. The governing formula

where $\mu_{\text{rel}} = \dfrac{\mu_{\text{lens}}}{\mu_{\text{medium}}}$ is the refractive index of the lens with respect to the medium it sits in. This is the crux of the problem: the radii are fixed by the glass, but the relative index changes when you move the lens from air to water.

Let the (constant, purely geometric) factor be

2. Extract the absolute indices from the data

• Air → glass: ${}_{a}\mu_{g} = \dfrac{3}{2} = 1.5 \quad\Rightarrow\quad \mu_{\text{glass}} = 1.5$ (taking $\mu_{\text{air}} = 1$)
• Air → water: ${}_{a}\mu_{w} = \dfrac{4}{3} \quad\Rightarrow\quad \mu_{\text{water}} = \dfrac{4}{3}$

3. Focal length in AIR

4. Focal length in WATER

5. Take the ratio

Divide (1) by (2):

$\Longrightarrow \quad \frac{f_w}{f_a} = 4 \quad\Longrightarrow\quad \boxed{f_w : f_a = 4 : 1}$ …

The critical angle is set by $\sin\theta_c = 1/\mu$. Diamond's refractive index is very high ($\mu \approx 2.42$), giving a low critical angle of about $24^\circ$. Because so many internal rays exceed this angle, they undergo repeated total internal reflection and finally emerge concentrated, producing the characteristic sparkle. Thus brilliance requi …