Step-by-step Solution:

The problem asks us to compare the speed of light in two different media, medium-1 and medium-2, based on a diagram showing the refraction of wavefronts at the interface XY.

There are two main ways to analyze the provided figure to determine the relationship between the speeds.

Method 1: Analysis using Wavelength

1. Identify Wavefronts and Wavelength: The lines ab and cd represent wavefronts in medium-1, and ef and gh represent wavefronts in medium-2. A wavefront is a surface of constant phase. The perpendicular distance between two consecutive wavefronts (like cd and ab, or gh and ef) is the wavelength (λ) of the wave in that medium.

2. Compare Wavelengths from the Figure: Let λ₁ be the wavelength in medium-1 and λ₂ be the wavelength in medium-2. By visually inspecting the diagram, we can see that the separation between the incident wavefronts (ab and cd) is greater than the separation between the refracted wavefronts (ef and gh). Therefore, we can conclude that λ₁ > λ₂.

3. Relate Wavelength and Speed: The speed of a wave (v), its frequency (f), and its wavelength (λ) are related by the equation v = fλ. When light passes from one medium to another, its frequency f remains constant.

4. Compare Speeds: Since f is constant, the speed v is directly proportional to the wavelength λ (v ∝ λ). Because we observed that λ₁ > λ₂, it follows that the speed of light in medium-1 (v₁) must be greater than the speed of light in medium-2 (v₂). $v₁ > v₂$

Method 2: Analysis using Snell's Law and Angles

1. Define Angles of Incidence and Refraction: The angle of incidence i is the angle between the incident ray and the normal to the interface. It is also equal to the angle between the incident wavefront (e.g., cd) and the interface XY. Similarly, the angle of refraction r is the angle between the refracted ray and the normal, which is equal to the angle between the refracted wavefront (e.g., ef) and the interface XY.

2. Compare Angles from the Figure: From the diagram, we can see that the incident wavefront cd makes a larger angle with the interface XY than the refracted wavefront ef. Therefore, i > r.

3. Apply Snell's Law: Snell's law relates the angles of incidence and refraction to the speeds of light in the two media (v₁ and v₂) or the refractive indices (n₁ and n₂): $\frac{\sin i}{\sin r} = \frac{v₁}{v₂} = \frac{n₂}{n₁}$

4. Compare Speeds: Since we determined from the figure that i > r, and for angles between 0° and 90°, the sine function is an increasing function, it follows that sin i > sin r. Therefore, the ratio (sin i) / (sin r) is greater than 1. $\frac{\sin i}{\sin r} > 1$ Using Snell's law, this implies: $\frac{v₁}{v₂} > 1 \implies v₁ > v₂$

Conclusion:

Both methods lead to the same conclusion: the speed of light is larger in medium-1 than in medium-2.

Evaluating the Options:

• A: the same in medium-1 and medium-2: Incorrect. This would mean no refraction (i=r), which contradicts the figure.
• B: larger in medium-1 than in medium-2: Correct. Our analysis shows v₁ > v₂.
• C: larger in medium-2 than in medium-1: Incorrect. This would imply v₂ > v₁ and r > i, which contradicts the figure.
• D: different at b and d: Incorrect. The speed of light in a homogeneous medium (like medium-1) is constant everywhere within that medium. So, the speed at point b is the same as at point d.