Q.Let us list some of the factors, which could possibly influence the speed of wave propagation:

The key concept is that the deviation produced by a thin prism is proportional to the difference between the refractive index of the prism material relative to the surrounding medium and 1. For a thin prism with small angle $A$, the deviation $\delta$ is given by $\delta = (\mu_{\text{relative}} - 1)A$. When the prism is in air, the relative index is simply the prism's own refractive index $\mu$. When immersed in a liquid, the relative index becomes $\mu / \mu_{\text{liquid}}$. The ratio of deviations then reduces to a ratio of these relative indices minus one.

1. Write the deviation formula for a thin prism in air. Let the refractive index of the prism material be $\mu$ (typically around 1.5 for glass). In air, the surrounding refractive index is 1, so the relative refractive index is $\mu$. The deviation is:

where $A$ is the prism angle (small).

2. Write the deviation formula when the prism is immersed in water. Water has refractive index $\mu_w = \frac{4}{3}$. The relative refractive index of the prism with respect to water is $\frac{\mu}{\mu_w}$. Hence the deviation in water is:

3. Find the ratio $\delta_{\text{air}} : \delta_{\text{water}}$. The prism angle $A$ cancels out:

Simplify the denominator:

So the ratio becomes:

4. Substitute the known value $\mu_w = \frac{4}{3}$. We still need $\mu$. For a typical glass prism, $\mu = \frac{3}{2}$ (a common value used in such problems). Substituting: $\frac{\delta_{\text{air}}}{\delta_{\text{water}}} = \frac{\left(\frac{3}{2} - 1\right) \cdot \frac{4}{3}}{\frac{3}{2} - \frac{4}{3}}$ …

Step 1 — Check the true statements

The refractive index $n_{21} = \dfrac{v_1}{v_2} = \dfrac{n_2}{n_1}$ is a ratio that depends on the optical density (nature) of the first medium and the second medium — so statements (A) and (B) are correct.

Refractive index also varies with wavelength (this is the basis of dispersion, e.g. in a prism), so statement (C) is also correct.

Step 2 — Identify the false statement …

Step 1 — Read off $A$ and $D_m$ from the graph

The prism is equilateral, so its apex angle is $A = 60°$. From the graph, the minimum point of the deviation-versus-incidence curve occurs at a deviation of $D_m = 60°$ (at an incidence angle of $60°$).

Step 2 — Apply the prism formula …

The key idea here is the definition of refractive index. It’s not a mysterious property — it simply tells you how much slower light travels in a material compared to empty space. If the refractive index is 1.2, light in that medium moves at $1/1.2$ times the vacuum speed. So to go from the medium speed back to the vacuum speed, you multiply by 1.2.

Let’s walk through it step by step.

1. Recall the definition The absolute refractive index $n$ of a medium is given by

where $c$ is the speed of light in vacuum and $v$ is the speed of light in the medium. This is the fundamental relationship.

2. Plug in the given numbers We are told $n = 1.2$ and the speed in the medium is $v$ (the same symbol, but careful — here $v$ is the speed in the medium, not vacuum). So:

3. Solve for $c$ Multiply both sides by $v$:

That’s it — the vacuum speed is simply $1.2$ times the medium speed.

4. Check the options
   • (A) $\frac{v}{1.2}$ would be the speed in the medium if the vacuum speed were $v$ — backwards.
   • (B) $2.4v$ is double the correct factor — no reason for that.
   • (C) $1.2v$ matches exactly. …

Step 1 — Recognise the phenomenon.

When a light source moves away from an observer, the wavelength of light it emits appears stretched (increased) — this is the optical Doppler effect, and an increase in wavelength is called a redshift. Here the spectral line shifts from its rest value $\lambda_0 = 600$ nm to an observed value $\lambda = 601$ nm — a shift toward longer wavelength, consistent with the galaxy moving away from Earth (matching the problem statement).

Step 2 — Find the wavelength shift.

Step 3 — Apply the (non-relativistic) Doppler formula.

For speeds much smaller than $c$ (true here, as the answer itself will confirm), the fractional wavelength shift equals the fractional speed: …

Concept & Intuition

The problem is a classic prism refraction scenario where the ray emerges perpendicular to the second face. That means the angle of emergence is $0^\circ$ relative to the normal, so the ray inside the prism hits the second face at exactly the critical angle for that interface? No—actually, if it emerges normal, the angle of refraction at the second face is $0^\circ$, so by Snell’s law the angle of incidence inside the prism at that face must also be $0^\circ$. That’s a huge clue: the ray inside the prism travels parallel to the base? Let’s check carefully.

We have a prism with apex angle $A = 30^\circ$. Light enters at $i_1 = 60^\circ$ on the first face. It emerges normal to the second face, meaning the emergent ray makes $0^\circ$ with the normal, so $i_2 = 0^\circ$ (angle of incidence inside the prism at the second face). Using geometry of the prism, we can find the angle of refraction $r_1$ at the first face, then apply Snell’s law.

Step-by-step reasoning

1. Geometry of the prism In a prism, the sum of the two internal angles of refraction equals the apex angle:

Here $A = 30^\circ$. Since the emergent ray is normal to the second face, the angle of incidence inside the prism at that face is $r_2 = 0^\circ$ (because the ray is perpendicular to the surface).

Therefore:

2. Apply Snell’s law at the first face At the first face, light goes from air ($n=1$) into the prism ($n$). The angle of incidence is $i_1 = 60^\circ$, and the angle of refraction inside is $r_1 = 30^\circ$. Snell’s law gives: $1 \cdot \sin 60^\circ = n \cdot \sin 30^\circ$ …

The key idea is Snell’s law: when light passes from one medium to another, the ratio of the sines of the angles equals the inverse ratio of the refractive indices. The intuition: light bends toward the normal when entering a denser medium (higher refractive index). Here, glass (1.5) is denser than the first medium (1.25), so the refraction angle will be smaller than the incidence angle. We just need to compute exactly how much smaller.

1. Write Snell’s law

where $n_1 = \frac{5}{4}$ (incident medium), $n_2 = \frac{3}{2}$ (glass), $i = 30^\circ$, and $r$ is the unknown angle of refraction.

2. Substitute the known values

Since $\sin 30^\circ = \frac{1}{2}$, this becomes:

3. Solve for $\sin r$ Multiply both sides by $\frac{2}{3}$:

4. Interpret the result …

Concept: prism at minimum deviation - the ray passes symmetrically, so r1 = r2 = A/2, and n = sin i / sin(A/2).

Equilateral prism: A = 60 degrees.

Given i = (3/4) A = (3/4)(60) = 45 degrees.

At minimum deviation, r = A/2 = 30 degrees.

Refractive index: …

1. The formula and why it applies

In Young's double-slit experiment, bright fringes occur where the path difference is an integral number of wavelengths. The separation between consecutive bright (or dark) fringes — the fringe width — is

where $D$ = slit-to-screen distance and $d$ = slit separation.

The key structural point: $D$ and $d$ are properties of the apparatus, not of the light. If we simply change the colour of the source in the same set-up, $D$ and $d$ are unchanged, so

2. Wavelengths of the two colours

At the two ends of the visible spectrum:

3. Take the ratio

$\frac{\beta_{\text{red}}}{\beta_{\text{violet}}} = \frac{\lambda_{\text{red}}}{\lambda_{\text{violet}}} \approx \frac{700}{400} = 1.75 \;\approx\; 2$ …

For a thin prism the deviation is

Solving for the prism angle: …

Step 1 — The prism formula

At minimum deviation the ray passes symmetrically through the prism, and the refractive index is

where $A$ is the refracting (apex) angle and $\delta_m$ the angle of minimum deviation.

Step 2 — Substitute the given $\mu$

We are told $\mu = \cot\dfrac{A}{2}$. Writing the cotangent as a ratio:

So

Step 3 — Cancel and use the co-function identity

The common denominator $\sin\frac A2$ cancels (it is non-zero for a real prism, $0 < A < 180^\circ$):

Step 4 — Equate the angles and solve …

Step 1 — Extract the refractive index from the graph.

Snell's law for light entering the medium from vacuum:

So a plot of $\sin r$ (y-axis) against $\sin i$ (x-axis) is a straight line through the origin with slope $\dfrac{1}{n}$ — which is exactly what the figure shows.

Step 2 — Read the slope.

The line makes $30^\circ$ with the $\sin i$ axis, so

Therefore

(Sensible: a typical glass has $n\approx1.5$–$1.7$.)

Step 3 — Brewster's law.

At the Brewster (polarising) angle $\theta_B$, the reflected ray is completely plane-polarised, and the reflected and refracted rays are perpendicular. That geometric condition ($\theta_B + r = 90^\circ$) fed into Snell's law gives …