JEE Challenger
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PhyHardAdv 2019 • Paper 1

Maximum Time Taken by Light Ray Traversing Optical Slab via Total Internal Reflection

A planar structure of length LL and width WW is constructed from two distinct optical media having refractive indices n1=1.5n_1 = 1.5 and n2=1.44n_2 = 1.44, as depicted in the diagram. Given that L≫WL \gg W, a light ray entering from edge AB emerges from edge CD provided the condition for total internal reflection is satisfied inside the structure. For L=9.6 mL = 9.6\text{ m}, as the angle of incidence θ\theta is varied, the maximum duration taken by a ray to exit the boundary CD is t×10−9 st \times 10^{-9}\text{ s}. Find the value of tt. [Speed of light c=3×108 m/sc = 3 \times 10^8\text{ m/s}]

Question Diagram 1
Official Numerical Answer50

Topics & Concepts

Ray Optics and Optical InstrumentsReflection and Refraction

Step-by-Step Solution

To find the maximum time taken by a light ray to exit from the edge CDCD, we need to analyze the trajectory of the ray inside the medium of refractive index n1n_1.

  1. Geometry and Refraction at the Entrance Face (AB): Let θ\theta be the angle of incidence in air at the face ABAB, and rr be the angle of refraction inside the medium n1n_1 measured relative to the normal to ABAB (which is parallel to the length LL of the structure).

    By Snell's law at the entrance face ABAB: sin⁡θ=n1sin⁡r\sin\theta = n_1 \sin r

  2. Total Internal Reflection (TIR) Condition at the Interface: The angle of incidence of the ray at the n1−n2n_1 - n_2 boundary is i=90∘−ri = 90^\circ - r. For Total Internal Reflection to occur at this boundary: sin⁡i≥n2n1\sin i \ge \frac{n_2}{n_1} sin⁡(90∘−r)=cos⁡r≥n2n1\sin(90^\circ - r) = \cos r \ge \frac{n_2}{n_1}

    Substituting the given refractive indices n1=1.5n_1 = 1.5 and n2=1.44n_2 = 1.44: cos⁡r≥1.441.5=0.96\cos r \ge \frac{1.44}{1.5} = 0.96

    (Note: Since cos⁡r=0.96  ⟹  sin⁡r=1−0.962=0.28\cos r = 0.96 \implies \sin r = \sqrt{1 - 0.96^2} = 0.28, the required angle rr is achievable via refraction since sin⁡θ=n1sin⁡r=1.5×0.28=0.42≤1\sin\theta = n_1 \sin r = 1.5 \times 0.28 = 0.42 \le 1.)

  3. Time Taken to Cross the Slab: The total path length traveled by the ray inside medium n1n_1 to cover a horizontal distance LL is: s=Lcos⁡rs = \frac{L}{\cos r}

    The speed of light in medium n1n_1 is: v=cn1v = \frac{c}{n_1}

    Therefore, the time TT taken by the ray to exit the boundary CDCD is: T=sv=Lcos⁡rcn1=n1Lccos⁡rT = \frac{s}{v} = \frac{\frac{L}{\cos r}}{\frac{c}{n_1}} = \frac{n_1 L}{c \cos r}

  4. Maximum Time Calculation: To maximize the time TT, the term cos⁡r\cos r must take its minimum permissible value while still satisfying the TIR condition: cos⁡rmin=n2n1\cos r_{\text{min}} = \frac{n_2}{n_1}

    Substituting cos⁡rmin\cos r_{\text{min}} into the time expression gives: Tmax=n12Ln2cT_{\text{max}} = \frac{n_1^2 L}{n_2 c}

    Plugging in the given values (L=9.6 mL = 9.6\text{ m}, c=3×108 m/sc = 3 \times 10^8\text{ m/s}, n1=1.5n_1 = 1.5, and n2=1.44n_2 = 1.44): Tmax=(1.5)2×9.61.44×3×108T_{\text{max}} = \frac{(1.5)^2 \times 9.6}{1.44 \times 3 \times 10^8} Tmax=2.25×9.61.44×3×108=21.64.32×108=5×10−8 s=50×10−9 sT_{\text{max}} = \frac{2.25 \times 9.6}{1.44 \times 3 \times 10^8} = \frac{21.6}{4.32 \times 10^8} = 5 \times 10^{-8}\text{ s} = 50 \times 10^{-9}\text{ s}

Thus, the value of tt is 50.

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