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Q29(ii):
Draw a ray diagram showing the path of rays of light when it enters with oblique incidence (ii) from water into air.

Solution :

Given Variables & Initial Setup

In this scenario, a light ray travels from a denser medium (water) into a rarer medium (air). The refractive index of water ($n_1 \approx 1.33$) is greater than the refractive index of air ($n_2 \approx 1.00$). According to Snell’s Law:

$n_1 \sin(i) = n_2 \sin(r)$

Since $n_1 > n_2$, it follows that $\sin(r) > \sin(i)$, which implies that the angle of refraction ($r$) must be greater than the angle of incidence ($i$). Consequently, the refracted ray bends away from the normal. [Per Snell’s Law of Refraction]

Ray Diagram Representation

Interface Normal Incident Ray (i) Refracted Ray (r) Water (n₁) Air (n₂)

Step-by-Step Analytical Breakdown

  • Step 1: Establishing the Interface: The horizontal line represents the boundary separating water (bottom) and air (top).
  • Step 2: Drawing the Normal: A vertical dashed line is drawn perpendicular to the interface at the point of incidence. [By geometric construction of the normal]
  • Step 3: Incident Path: The incident ray is drawn from the water medium toward the point of incidence on the interface.
  • Step 4: Application of Optical Principles: As light transitions from a higher refractive index (denser) to a lower refractive index (rarer), the velocity of light increases. To satisfy the conservation of wavefronts, the ray must deviate from the normal path. [Per the Law of Refraction/Snell's Law]
  • Step 5: Refracted Path: The refracted ray is drawn emerging into the air, tilted at an angle $r > i$ relative to the normal, demonstrating refraction away from the normal.

Final Solution: When light travels from a denser medium (water) to a rarer medium (air), the refracted ray bends away from the normal, such that the angle of refraction is greater than the angle of incidence ($r > i$).


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