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JEE Main 20257 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Evening Shift
Subject
Physics
Chapter
Ray Optics

JEE Main 2025 · 7 April · Evening Shift · Physics · Ray Optics

A transparent block A having refractive index μ=1.25 is surrounded...

A transparent block A having refractive index μ=1.25\mu=1.25 is surrounded by another medium of refractive index μ=1.0\mu=1.0 as shown in figure. A light ray is incident on the flat face of the block with incident angle θ\theta as shown in figure. What is the maximum value of θ\theta for which light suffers total internal reflection at the top surface of the block?
A transparent block A having refractive index μ=1.25 is surrounded by another medium of refractive index μ=1.0 as shown

Options

  1. Atan⁡−1(4/3)\tan ^{-1}(4 / 3)
  2. Btan⁡−1(3/4)\tan ^{-1}(3 / 4)
  3. Csin⁡−1(3/4)\sin ^{-1}(3 / 4)
  4. Dcos⁡−1(3/4)\cos ^{-1}(3 / 4)

Correct answer

C. sin⁡−1(3/4)\sin ^{-1}(3 / 4)

Step-by-step solution

A transparent block A having refractive index μ=1.25 is surrounded by another medium of refractive index μ=1.0 as shown
r+θC=90∘μ1sin⁡θ=μ2sin⁡rsin⁡θ=μ2μ1sin⁡(90−θC)sin⁡θ=μ2μ1cos⁡θCsin⁡θC=μ1μ2\begin{aligned} & \mathrm{r}+\theta_{\mathrm{C}}=90^{\circ} \\ & \mu_1 \sin \theta=\mu_2 \sin \mathrm{r} \\ & \sin \theta=\frac{\mu_2}{\mu_1} \sin \left(90-\theta_{\mathrm{C}}\right) \\ & \sin \theta=\frac{\mu_2}{\mu_1} \cos \theta_{\mathrm{C}} \\ & \sin \theta_{\mathrm{C}}=\frac{\mu_1}{\mu_2}\end{aligned}
sin⁡θ=μ2μ11−μ12μ22sin⁡θ=μ22−μ12μ12=2516−11sin⁡θ=34θ=sin⁡−1(34)\begin{aligned} & \sin \theta=\frac{\mu_2}{\mu_1} \sqrt{1-\frac{\mu_1^2}{\mu_2^2}} \\ & \sin \theta=\sqrt{\frac{\mu_2^2-\mu_1^2}{\mu_1^2}}=\sqrt{\frac{\frac{25}{16}-1}{1}} \\ & \sin \theta=\frac{3}{4} \\ & \theta=\sin ^{-1}\left(\frac{3}{4}\right)\end{aligned}
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