CBSE 2025, Set 1

Question

A rectangular glass slab ABCD (refractive index 1·5) is surrounded by a transparent liquid (refractive index 1·25) as shown in the figure. A ray of light is incident on face AB at an angle $i$ such that it is refracted out grazing the face AD. Find the value of angle $i$.

![Glass slab ABCD in liquid, ray incident on AB](https://www.edvaya.com/images/cbse/physics/2025/set1/q27-glass-slab-abcd-liquid-refraction.webp)

Edvaya · Accepted Answer

The correct answer is Subjective. **Step 1: Find the critical angle at the glass-liquid interface (face AD).** The ray grazes face AD, which means it undergoes **total internal reflection** just at the critical angle $i_c$ at the glass-liquid interface. For the glass-liquid interface: $$\sin i_c = \frac{n_{	ext{liquid}}}{n_{	ext{glass}}} = \frac{1\cdot25}{1\cdot5} = \frac{5}{6}$$ $$\cos i_c = \sqrt{1 - \sin^2 i_c} = \sqrt{1 - \frac{25}{36}} = \sqrt{\frac{11}{36}} = \frac{\sqrt{11}}{6}$$ **Step 2: Find the angle of refraction $r$ at face AB.** Face AB is perpendicular to face AD. The ray travels inside the glass slab and hits face AD at angle $i_c$. Since AB ⊥ AD: $$i_c + r = 90° \implies r = 90° - i_c$$ $$\sin r = \sin(90° - i_c) = \cos i_c = \frac{\sqrt{11}}{6}$$ **Step 3: Apply Snell's law at face AB (air-to-glass interface).** At face AB, the ray goes from air (effectively, the upper medium is air as shown in diagram) into glass: $$\frac{\sin i}{\sin r} = n_{	ext{glass}} = 1\cdot5$$ $$\sin i = 1\cdot5 	imes \frac{\sqrt{11}}{6} = \frac{\sqrt{11}}{4}$$ $$\boxed{i = \sin^{-1}\!\left(\frac{\sqrt{11}}{4}ight) \approx 55\cdot7°}$$ > **Method summary:** (1) Find critical angle at glass-liquid interface → (2) geometry gives $r = 90° - i_c$ → (3) Snell's law at AB gives $i$. This is a classic problem combining total internal reflection and Snell's law.

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