Q.Figures 9.27(a) and

Refractive index is a property of the material, so we read $n_{\text{glass}}$ and $n_{\text{water}}$ from the two given air interfaces and reuse them at the water-glass interface (Snell's law).

From Fig. 9.27(a), air$\to$glass, $i = 60^\circ$, $r \approx 35^\circ$:

From Fig. 9.27(b), air$\to$water, $i = 60^\circ$, $r \approx 41^\circ$:

At the water$\to$glass interface, $i = 45^\circ$: …

Step 1 — refractive index of glass

In Fig. 9.27(a) light passes from air into glass with $i = 60^\circ$ and refraction angle $r_g \approx 35^\circ$. By Snell's law ($n_{\text{air}} = 1$):

Step 2 — refractive index of water

In Fig. 9.27(b) light passes from air into water with $i = 60^\circ$ and $r_w \approx 41^\circ$:

Step 3 — water to glass

At the water-glass interface [Fig. 9.27(c)] the ray travels from water into glass with $i = 45^\circ$. Snell's law across this boundary:

$n_{\text{water}}\sin 45^\circ = n_{\text{glass}}\sin\theta.$ …

Method: Finding an Unknown Refractive Index by Chaining Snell's Law

This method applies whenever a medium's refractive index isn't given directly, but can be worked out from a refraction event you ARE given (usually against a known medium like air), and then reused at a different interface.

Steps

Step 1: Recognise that refractive index is a fixed material property

Because $n$ belongs to the material, not to any one ray, a single measured (angle of incidence, angle of refraction) pair against a KNOWN medium is enough to pin it down — and that same value can then be reused for any other interface involving that material.

Step 2: Compute each unknown medium's index from its given air-interface data

using $n_{\text{air}}=1$. Do this once for every medium whose index you need but aren't told directly.

Step 3: Write Snell's law at the NEW interface, between the two now-known media

$n_1 \sin i_1 = n_2 \sin i_2$ …