JEE Challenger
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PhyMedAdv 2008 • Paper 2

Angle of Incidence for Light Beam to Miss Entering Region IV

A light beam is traveling from Region I to Region IV (Refer Figure). The refractive index in Regions I, II, III and IV are n0n_0, n02\frac{n_0}{2}, n06\frac{n_0}{6} and n08\frac{n_0}{8}, respectively. The angle of incidence θ\theta for which the beam just misses entering Region IV is

Diagram 0

Options

A

sin⁡−1(34)\sin^{-1}\left(\frac{3}{4}\right)

B

sin⁡−1(18)\sin^{-1}\left(\frac{1}{8}\right)

Correct
C

sin⁡−1(14)\sin^{-1}\left(\frac{1}{4}\right)

D

sin⁡−1(13)\sin^{-1}\left(\frac{1}{3}\right)

Topics & Concepts

Ray Optics and Optical InstrumentsReflection and Refraction

Step-by-Step Solution

To find the angle of incidence θ\theta in Region I for which the light beam just misses entering Region IV, we analyze the propagation of light across parallel interfaces.

1. Principle of Refraction Across Parallel Media

According to Snell's Law, for a set of parallel planar interfaces, the product of the refractive index nn of a medium and the sine of the angle of refraction θ\theta inside that medium relative to the normal remains constant throughout all layers:

n1sin⁡θ1=n2sin⁡θ2=n3sin⁡θ3=n4sin⁡θ4=constantn_1 \sin \theta_1 = n_2 \sin \theta_2 = n_3 \sin \theta_3 = n_4 \sin \theta_4 = \text{constant}

2. Condition for "Just Missing" Region IV

For the beam to just miss entering Region IV, total internal reflection (TIR) must occur at the interface between Region III and Region IV. This occurs when the angle of refraction at the boundary becomes θ4=90∘\theta_4 = 90^\circ (grazing refraction).

Applying Snell's Law directly between Region I and Region IV:
n1sin⁡θ=n4sin⁡θ4n_1 \sin \theta = n_4 \sin \theta_4

Given:

  • Refractive index of Region I, n1=n0n_1 = n_0
  • Refractive index of Region IV, n4=n08n_4 = \frac{n_0}{8}
  • Critical condition angle in Region IV, θ4=90∘\theta_4 = 90^\circ

Substitute these values into the equation:
n0sin⁡θ=(n08)sin⁡90∘n_0 \sin \theta = \left(\frac{n_0}{8}\right) \sin 90^\circ

Since sin⁡90∘=1\sin 90^\circ = 1:
n0sin⁡θ=n08n_0 \sin \theta = \frac{n_0}{8}

sin⁡θ=18\sin \theta = \frac{1}{8}

θ=sin⁡−1(18)\theta = \sin^{-1}\left(\frac{1}{8}\right)

3. Verification for Intermediate Regions

To ensure that the beam actually reaches the Region III–IV interface without experiencing total internal reflection at earlier boundaries:

  1. At Region I – Region II Interface:
    sin⁡θ2=n1n2sin⁡θ=n0n02⋅18=28=14≤1(Light enters Region II)\sin \theta_2 = \frac{n_1}{n_2} \sin \theta = \frac{n_0}{\frac{n_0}{2}} \cdot \frac{1}{8} = \frac{2}{8} = \frac{1}{4} \le 1 \quad (\text{Light enters Region II})

  2. At Region II – Region III Interface:
    sin⁡θ3=n2n3sin⁡θ2=n02n06⋅14=3⋅14=34≤1(Light enters Region III)\sin \theta_3 = \frac{n_2}{n_3} \sin \theta_2 = \frac{\frac{n_0}{2}}{\frac{n_0}{6}} \cdot \frac{1}{4} = 3 \cdot \frac{1}{4} = \frac{3}{4} \le 1 \quad (\text{Light enters Region III})

  3. At Region III – Region IV Interface:
    sin⁡θ4=n3n4sin⁡θ3=n06n08⋅34=86⋅34=1  ⟹  θ4=90∘\sin \theta_4 = \frac{n_3}{n_4} \sin \theta_3 = \frac{\frac{n_0}{6}}{\frac{n_0}{8}} \cdot \frac{3}{4} = \frac{8}{6} \cdot \frac{3}{4} = 1 \implies \theta_4 = 90^\circ

Thus, total internal reflection occurs precisely at the boundary of Region IV.

Correct Option: (B) sin⁡−1(18)\sin^{-1}\left(\frac{1}{8}\right)

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