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JEE Physics 2025 Question with Solution

Two light beams fall on a transparent material block at point 11 and 22 with angle θ1\theta_1 and θ2\theta_2, respectively, as shown in the figure. After refraction, the beams intersect at point 33 which is exactly on the interface at the other end of the block. Given: the distance between 11 and 22, d=43 cmd = \frac{4}{3} \, \text{cm} and θ1=θ2=cos⁡−1(n22n1)\theta_1 = \theta_2 = \cos^{-1} \left( \frac{n_2}{2n_1} \right), where n2n_2 is the refractive index of the block and n1n_1 is the refractive index of the outside medium, then the thickness of the block is _____ cm.

A rectangular transparent block with upper and lower parallel faces, points 1 and 2 on the top face, point 3 on the bottom face, incident rays making angles theta1 and theta2 outside, and refracted rays inside meeting at point 3; regions labeled n1 outside and n2 inside.
  • A

    1 cm1 \, \text{cm}

  • B

    2 cm2 \, \text{cm}

  • C

    3 cm3 \, \text{cm}

  • D

    4 cm4 \, \text{cm}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Two rays are incident on the top surface at points 11 and 22 with θ1=θ2=cos⁡−1(n22n1)\theta_1 = \theta_2 = \cos^{-1}\left(\frac{n_2}{2n_1}\right) and d=43 cmd = \frac{4}{3} \, \text{cm}. The refracted rays meet at point 33 on the lower face.

Find: The thickness of the block.

Use Snell's law at the upper surface:

n1sin⁡θ=n2sin⁡rn_1 \sin\theta = n_2 \sin r

where θ=cos⁡−1(n22n1)\theta = \cos^{-1}\left(\frac{n_2}{2n_1}\right).

So,

cos⁡θ=n22n1\cos\theta = \frac{n_2}{2n_1}

and hence

sin⁡θ=1−cos⁡2θ=1−(n22n1)2\sin\theta = \sqrt{1-\cos^2\theta} = \sqrt{1-\left(\frac{n_2}{2n_1}\right)^2}

the solution states that using Snell's law together with the geometry of the refracted beams gives the thickness of the block as 3 cm3 \, \text{cm}.

Thus, the correct option is C.

Common mistakes

  • Using only Snell's law and ignoring the geometry of the two refracted rays. Snell's law gives the refracted angle, but the thickness comes from how the two rays meet at point 33. Always combine refraction with the ray-path geometry.

  • Confusing the angle of incidence with the angle of refraction. The given θ1\theta_1 and θ2\theta_2 are outside the block, whereas the internal angles are different. First apply Snell's law, then use the internal geometry.

  • Substituting cos⁡θ\cos\theta directly into Snell's law instead of using sin⁡θ\sin\theta. Since Snell's law involves sine of the angle, convert the given cosine form carefully before proceeding.

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