Q.A magician during a show makes a glass lens with $n = 1.47$ disappear in a trough of liquid. What is the refractive index of the liquid? Could the liquid be water?

Concept & Intuition

In a series RC circuit, the voltage across the resistor is in phase with the current, but the voltage across the capacitor lags the current by $90^\circ$. The total voltage is the vector sum of these two, so the current leads the voltage by some angle $\phi$ between $0^\circ$ and $90^\circ$. That angle is determined by the ratio of capacitive reactance $X_C$ to resistance $R$: $\tan\phi = X_C / R$. So the problem reduces to computing $X_C$ from the given capacitance and frequency, then finding $\phi$.

Step-by-step solution

1. Find the capacitive reactance The formula is $X_C = \frac{1}{2\pi f C}$. Given: $f = 100\ \text{Hz}$, $C = \frac{40}{\pi}\ \mu\text{F} = \frac{40}{\pi} \times 10^{-6}\ \text{F}$. So

2. Identify the resistance

   Given $R = \frac{125}{\sqrt{3}}\ \Omega$.

3. Compute the phase difference

   For a series RC circuit, the phase angle by which current leads voltage is

   $\phi = \tan^{-1}\left(\frac{X_C}{R}\right).$ …

Concept

A lens refracts light because of the difference in refractive index between the lens material and its surroundings. In a medium, the lens maker's formula uses the relative index $n_{\text{lens}}/n_{\text{medium}}$:

Solution

Curvature term. For the biconvex lens, $R_1 = +6\text{ cm}$ and $R_2 = -12\text{ cm}$:

Relative index term. With $n_{\text{lens}} = 1.5$ and $n_{\text{medium}} = 1.3$: …

The key concept here is Brewster’s law: when light reflects off a surface and the reflected ray is completely polarised, the angle of incidence is such that the reflected and refracted rays are perpendicular. This happens at a specific angle called Brewster’s angle ($i_B$), where $\tan i_B = n$ (the refractive index of the medium, assuming the first medium is air with $n=1$). The beauty of this condition is that it also forces the angle of refraction to be the complement of the angle of incidence — that is, $i_B + r = 90^\circ$. So once we find $i_B$, we immediately get $r$.

Let’s work through it step by step.

1. Apply Brewster’s law For total polarisation by reflection, the angle of incidence $i_B$ satisfies

Since $\tan 60^\circ = \sqrt{3}$, we have

2. Use the perpendicular condition At Brewster’s angle, the reflected and refracted rays are at right angles:

This is a direct consequence of Snell’s law combined with $\tan i_B = n$ — a neat shortcut that saves recalculating. …

Concept & Intuition

The lens maker’s formula for a thin lens in air is

where $n_{\text{lens}}$ is the refractive index of the lens material relative to air. When the lens is immersed in a liquid, the “bending” of light depends on the relative refractive index of the lens material with respect to the liquid:

So the formula becomes

The key insight: the lens bends light less when surrounded by a medium closer to its own refractive index, so the focal length increases.

Step-by-step solution

1. Identify the given data

   • Radii of curvature: $R_1 = 6\ \text{cm}$, $R_2 = -12\ \text{cm}$ (sign convention: convex surface facing incident light is positive; the second surface is concave relative to the first, so negative).
   • Refractive index of lens material: $n_{\text{lens}} = 1.5$
   • Refractive index of liquid: $n_{\text{liquid}} = 1.3$

2. Compute the relative refractive index

3. Apply the lens maker’s formula for the immersed lens

Substitute the values:

Simplify the bracket:

Simplify the curvature term:

So:

Therefore:

4. Check the sign and interpretation …

The key concept is Brewster’s law: when unpolarised light reflects from a surface, the reflected light is completely polarised if the angle of incidence equals Brewster’s angle $i_B$, satisfying $\tan i_B = \frac{n_2}{n_1}$, where $n_1$ and $n_2$ are the refractive indices of the incident and transmitting media. At this special angle, the reflected and refracted rays are perpendicular. So the problem reduces to finding $i_B$, then using Snell’s law to get the angle of refraction.

1. Identify the media. Light travels from air ($n_1 = 1$) into a medium with $n_2 = \sqrt{3}$. The reflected light is totally polarised, so the angle of incidence is Brewster’s angle $i_B$.

2. Apply Brewster’s law.

Hence $i_B = \tan^{-1}(\sqrt{3}) = 60^\circ$.

3. Use the perpendicular-ray condition (or Snell’s law). At Brewster’s angle, the reflected and refracted rays are at $90^\circ$ to each other. If $r$ is the angle of refraction, then $i_B + r = 90^\circ$, so

Alternatively, apply Snell’s law: $n_1 \sin i_B = n_2 \sin r$, i.e.

The key idea here is that when two soap bubbles merge, the air inside them is not simply added up in volume. Instead, the process is governed by the balance of surface tension forces under isothermal conditions. For a soap bubble, the excess pressure inside is inversely proportional to its radius, and the surface energy is proportional to the surface area. When they coalesce, the total surface energy of the system remains constant (since no work is done by an external agent and the process is isothermal), and the final bubble has a single radius that satisfies this energy conservation.

Let’s work through it step by step.

1. Relate volume to radius. The volume of a spherical bubble is $V = \frac{4}{3}\pi r^3$. So for the two bubbles, with volumes $27V$ and $64V$, we have:

Taking cube roots, the radii are proportional to the cube roots of the volumes:

So we can write $r_1 = 3k$ and $r_2 = 4k$, where $k$ is a constant that depends on $V$ (specifically $k = \left(\frac{3V}{4\pi}\right)^{1/3}$).

2. Surface energy of a soap bubble. A soap bubble has two surfaces (inner and outer), so its total surface area is $2 \times 4\pi r^2 = 8\pi r^2$. The surface energy is surface tension $\sigma$ times area: $E = 8\pi \sigma r^2$. For the two initial bubbles:

Total initial surface energy:

3. Final bubble after coalescence. Let the final bubble have radius $R$. Its surface energy is $E_{\text{final}} = 8\pi \sigma R^2$. Under isothermal conditions, the total surface energy is conserved (the air inside does work against surface tension, but no heat exchange occurs, and the process is slow enough that energy is just redistributed). So: $8\pi \sigma R^2 = 8\pi \sigma (25k^2) \quad \Rightarrow \quad R^2 = 25k^2 \quad \Rightarrow \quad R = 5k.$ …

The key idea is that in a common-emitter amplifier, the voltage gain $A_V$ is not just the current gain $\beta$ — it also depends on how the resistances in the collector and base circuits transform that current gain into a voltage swing. The transistor gives you current amplification, but the resistors determine how much voltage change you get at the output for a given input voltage change.

For a CE configuration, the voltage gain is given by:

where $\beta$ is the current gain (also called $h_{fe}$), $R_C$ is the collector resistance, and $R_B$ is the base resistance. This formula comes from the fact that a small change in base current $\Delta I_B$ produces a change $\Delta I_C = \beta \Delta I_B$ in the collector. The output voltage change is $\Delta V_C = -\Delta I_C \cdot R_C$, and the input voltage change is $\Delta V_B = \Delta I_B \cdot R_B$. The ratio gives the magnitude of voltage gain.

Let's work through it step by step.

1. Identify the given values.

   Current gain $\beta = 45$

   Collector resistance $R_C = 4.5~\text{k}\Omega = 4500~\Omega$

   Base resistance $R_B = 900~\Omega$

2. Apply the voltage gain formula.

3. Simplify the fraction. $\frac{4500}{900} = 5$ So $A_V = 45 \times 5 = 225$ …

The Earth's magnetic field at any location can be resolved into two perpendicular components: a horizontal component $B_H$ that points toward magnetic north, and a vertical component $B_V$ that points into or out of the ground. The angle of dip (or magnetic inclination) $\delta$ is the angle the total field vector makes with the horizontal plane.

Think of it this way: if you hold a freely suspended magnetic needle, it won't stay horizontal except at the magnetic equator. The angle it makes below the horizontal is precisely the angle of dip, determined by the ratio of vertical to horizontal components.

The geometry gives us:

This follows directly from the right triangle formed by $B_H$ (adjacent side), $B_V$ (opposite side), and the total field $B$ (hypotenuse).

Now let's apply the given condition:

1. Set up the relationship. We're told that $B_H = \sqrt{3} \cdot B_V$, which means the horizontal component is larger than the vertical component.

2. Express the tangent of the dip angle. Substituting into our formula: …

Concept and Intuition

The key idea is that when two thin lenses are placed in contact, the overall bending of light (the power) adds up. Think of it like two magnifying glasses stacked together — they bend light more strongly than either alone, so the combined focal length is shorter than either individual focal length. The power of a lens is defined as $P = \frac{1}{f}$ (with $f$ in meters), and for lenses in contact, $P_{\text{total}} = P_1 + P_2$. This works because the first lens bends the light by a certain amount, and the second lens bends it further, so the net effect is the sum of the two bends.

Step-by-step solution

1. Write the formula for lenses in contact For two thin lenses of focal lengths $f_1$ and $f_2$ placed coaxially in contact, the combined focal length $F$ satisfies:

This is derived from the fact that the image formed by the first lens acts as the object for the second, and when the separation is negligible, the equations combine neatly.

2. Substitute the given values Here, $f_1 = 20$ cm and $f_2 = 30$ cm. Note that both are convex (converging) lenses, so their focal lengths are positive.

3. Find a common denominator and add The least common multiple of 20 and 30 is 60. $\frac{1}{F} = \frac{3}{60} + \frac{2}{60} = \frac{5}{60}$ …

The critical angle only exists when light travels from a denser to a rarer medium — that is, from a higher refractive index to a lower one. Here $n_1 = 2$ (denser) and $n_2 = \sqrt{3} \approx 1.732$ (rarer), so the condition is satisfied. The concept is simple: at the critical angle $C$, the refracted ray grazes the boundary, making the angle of refraction exactly $90^\circ$.

1. Apply Snell’s law at the critical condition. Snell’s law: $n_1 \sin \theta_1 = n_2 \sin \theta_2$. At the critical angle, $\theta_1 = C$ and $\theta_2 = 90^\circ$. So:

2. Solve for $\sin C$.

3. Find $C$ from the known sine value. Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$, we get: $C = 60^\circ$ …

The key idea is that when you ask for the velocity of B as measured in car A, you want the relative velocity of B with respect to A. That means you subtract A's velocity from B's velocity, treating both as vectors. The result tells you how B appears to move from A's frame — its speed and its direction.

A classic trap is to add the speeds or to forget that direction matters. Here, A moves east and B moves north at the same speed. Their relative velocity is not simply the difference in magnitudes; it's the vector difference, which gives a larger magnitude and a diagonal direction.

Let’s work it through.

1. Set up the velocities as vectors.

   Take east as the positive $x$-direction and north as the positive $y$-direction.

   Car A: $\vec{v}_A = 30\ \hat{i}$ km/hr (east).

   Car B: $\vec{v}_B = 30\ \hat{j}$ km/hr (north).

2. Write the relative velocity formula.

   The velocity of B as seen from A is

This is the vector you would measure if you were sitting in car A and watching car B.

3. Subtract the vectors.

The $-30\ \hat{i}$ means the relative motion has a component pointing west (opposite to east), and the $+30\ \hat{j}$ means it has a component pointing north.

4. Find the magnitude.

5. Find the direction. …