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MHT CET 2026 (19 May, Shift 2) · Total Internal Reflection

Total Internal Reflection question from MHT CET 2026 (19 May, Shift 2)

Q3MHT CET 2026 · 19 MAY · SHIFT 2Medium

Total Internal Reflection

A ray of light is incident at a glass-water interface at an angle (i1)(i_1), it emerges finally parallel to the surface of water, then the value of the refractive index of glass (μg)(\mu_g) is

{μw=\{\mu_w = refractive index of water =4/3,sin⁡(90)∘=1,cos⁡(90)∘=0}= 4/3, \sin(90)^\circ = 1, \cos(90)^\circ = 0\}

Figure

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Answer: (c)

  1. Snell's law gives μsin⁡θ=\mu \sin\theta = constant along the whole path.
  2. Glass to water to air: μgsin⁡i1=μwsin⁡r=μasin⁡90∘\mu_g \sin i_1 = \mu_w \sin r = \mu_a \sin 90^\circ.
  3. The ray finally grazes the water surface, so μgsin⁡i1=1×1=1\mu_g \sin i_1 = 1 \times 1 = 1. The value of μw\mu_w cancels out.
  4. μg=1sin⁡i\mu_g = \frac{1}{\sin i}. So the answer is (c).
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