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MCQEasyJEE 2026Prisms & Total Internal Reflection

JEE Physics 2026 Question with Solution

Consider light travelling from a medium A to medium B separated by a plane interface. If the light undergoes total internal reflection during its travel from medium A to medium B and the speed of light in media A and B are 2.4×108 m/s2.4 \times 10^8 \, \text{m/s} and 2.7×108 m/s2.7 \times 10^8 \, \text{m/s}, respectively, then the value of critical angle is :

  • A

    sin⁡−1(98)\sin^{-1}\left(\frac{9}{8}\right)

  • B

    cos⁡−1(89)\cos^{-1}\left(\frac{8}{9}\right)

  • C

    tan⁡−1(817)\tan^{-1}\left(\frac{8}{\sqrt{17}}\right)

  • D

    cot⁡−1(315)\cot^{-1}\left(\frac{3}{\sqrt{15}}\right)

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Light travels from medium A to medium B with speeds vA=2.4×108 m/sv_A = 2.4 \times 10^8 \, \text{m/s} and vB=2.7×108 m/sv_B = 2.7 \times 10^8 \, \text{m/s}.

Find: The critical angle for total internal reflection.

For total internal reflection, light must travel from a denser medium to a rarer medium. Since vA<vBv_A < v_B, medium A is denser and medium B is rarer.

Using refractive index relation μ=cv\mu = \frac{c}{v}, the critical angle satisfies

sin⁡θc=μrarerμdenser=vdenservrarer\sin \theta_c = \frac{\mu_{\text{rarer}}}{\mu_{\text{denser}}} = \frac{v_{\text{denser}}}{v_{\text{rarer}}}

Substituting the given values,

sin⁡θc=vAvB=2.4×1082.7×108=2427=89\sin \theta_c = \frac{v_A}{v_B} = \frac{2.4 \times 10^8}{2.7 \times 10^8} = \frac{24}{27} = \frac{8}{9}

Now convert this into the form of the given options. If

sin⁡θc=89\sin \theta_c = \frac{8}{9}

then in a right triangle, opposite side =8= 8 and hypotenuse =9= 9. Therefore, the adjacent side is

92−82=81−64=17\sqrt{9^2 - 8^2} = \sqrt{81 - 64} = \sqrt{17}

Hence,

tan⁡θc=oppositeadjacent=817\tan \theta_c = \frac{\text{opposite}}{\text{adjacent}} = \frac{8}{\sqrt{17}}

So,

θc=tan⁡−1(817)\theta_c = \tan^{-1}\left(\frac{8}{\sqrt{17}}\right)

Therefore, the value of the critical angle is tan⁡−1(817)\tan^{-1}\left(\frac{8}{\sqrt{17}}\right). The correct option is C.

Common mistakes

  • Using vBvA\frac{v_B}{v_A} instead of vAvB\frac{v_A}{v_B} for sin⁡θc\sin \theta_c. This is wrong because the critical-angle relation uses denser-to-rarer ordering. First identify which medium is denser from the smaller speed, then apply the ratio correctly.

  • Assuming the denser medium has greater speed of light. This is wrong because refractive index is inversely proportional to speed, so lower speed means higher refractive index. Compare vAv_A and vBv_B before deciding the media.

  • Stopping at sin⁡−1(89)\sin^{-1}\left(\frac{8}{9}\right) without matching the equivalent trigonometric form in the options. This is incomplete because the correct option is expressed using tan⁡−1\tan^{-1}. Draw a right triangle and convert the ratio carefully.

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