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MCQMediumJEE 2025Prisms & Total Internal Reflection

JEE Physics 2025 Question with Solution

A transparent block A having refractive index μ2=1.25\mu_2 = 1.25 is surrounded by another medium of refractive index μ1=1.0\mu_1 = 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 ?

Ray entering rectangular block A from surrounding medium with mu1 equals 1.0, block refractive index mu2 equals 1.25, incident angle theta at left face, refracted ray reaches top face where critical angle theta_C is marked, then emerges along the top surface.
  • A

    tan⁡−1(4/3)\tan^{-1}(4/3)

  • B

    tan⁡−1(3/4)\tan^{-1}(3/4)

  • C

    sin⁡−1(3/4)\sin^{-1}(3/4)

  • D

    cos⁡−1(3/4)\cos^{-1}(3/4)

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: A block has refractive index μ2=1.25=54\mu_2 = 1.25 = \frac{5}{4} and the surrounding medium has refractive index μ1=1.0\mu_1 = 1.0. A ray is incident on the side face with angle θ\theta, and after refraction inside the block it strikes the top surface.

Find: The maximum value of θ\theta for which the ray undergoes total internal reflection at the top surface.

Let the angle of refraction inside the block at the first face be rr. Applying Snell's law at the first interface,

μ1sin⁡θ=μ2sin⁡r\mu_1 \sin \theta = \mu_2 \sin r

so,

sin⁡θ=1.25sin⁡r=54sin⁡r\sin \theta = 1.25 \sin r = \frac{5}{4} \sin r

Hence,

sin⁡r=45sin⁡θ\sin r = \frac{4}{5} \sin \theta

At the top surface, the angle of incidence inside the denser medium is

i=90∘−ri = 90^\circ - r

For total internal reflection to occur, this angle must be at least the critical angle θC\theta_C.

For the block-air interface, the critical angle satisfies

μ2sin⁡θC=μ1sin⁡90∘\mu_2 \sin \theta_C = \mu_1 \sin 90^\circ

Therefore,

sin⁡θC=μ1μ2=11.25=45\sin \theta_C = \frac{\mu_1}{\mu_2} = \frac{1}{1.25} = \frac{4}{5}

Now the condition for total internal reflection is

90∘−r≥θC90^\circ - r \ge \theta_C

which gives

cos⁡r≥45\cos r \ge \frac{4}{5}

Using cos⁡r=1−sin⁡2r\cos r = \sqrt{1-\sin^2 r} and substituting sin⁡r=45sin⁡θ\sin r = \frac{4}{5}\sin \theta,

1−(45sin⁡θ)2≥45\sqrt{1-\left(\frac{4}{5}\sin \theta\right)^2} \ge \frac{4}{5}

Squaring both sides,

1−1625sin⁡2θ≥16251 - \frac{16}{25}\sin^2 \theta \ge \frac{16}{25} 925≥1625sin⁡2θ\frac{9}{25} \ge \frac{16}{25}\sin^2 \theta 9≥16sin⁡2θ9 \ge 16\sin^2 \theta sin⁡2θ≤916\sin^2 \theta \le \frac{9}{16}

Since 0∘≤θ≤90∘0^\circ \le \theta \le 90^\circ,

sin⁡θ≤34\sin \theta \le \frac{3}{4}

The maximum value occurs when equality holds:

sin⁡θ=34\sin \theta = \frac{3}{4}

Therefore,

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

So, the correct option is C.

Geometry and critical angle shortcut

Given: The ray refracts into the block and then strikes the top surface. The critical angle at the top surface is determined by sin⁡θC=45\sin \theta_C = \frac{4}{5}.

Find: The largest incident angle θ\theta that still allows total internal reflection.

At the limiting case for total internal reflection, the angle of incidence at the top surface equals the critical angle. Hence,

r+θC=90∘r + \theta_C = 90^\circ

So,

sin⁡r=cos⁡θC=1−sin⁡2θC=1−(45)2=35\sin r = \cos \theta_C = \sqrt{1-\sin^2 \theta_C} = \sqrt{1-\left(\frac{4}{5}\right)^2} = \frac{3}{5}

Now apply Snell's law at the first face:

sin⁡θ=54sin⁡r=54⋅35=34\sin \theta = \frac{5}{4}\sin r = \frac{5}{4}\cdot\frac{3}{5} = \frac{3}{4}

Thus,

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

This shortcut works because the maximum allowed incident angle corresponds exactly to the limiting condition before refraction at the top surface changes from total internal reflection to transmission.

Common mistakes

  • Using the critical angle condition at the first interface is incorrect because total internal reflection occurs at the top surface, not where the ray enters. First apply Snell's law at the side face, then check the condition at the top face.

  • Taking the angle of incidence at the top surface as rr is wrong. The normal to the top face is vertical, so the actual angle of incidence there is 90∘−r90^\circ - r. Always identify the normal before writing the angle.

  • Writing sin⁡θC=μ2μ1\sin \theta_C = \frac{\mu_2}{\mu_1} reverses the ratio. For total internal reflection from denser to rarer medium, the correct expression is sin⁡θC=μ1μ2\sin \theta_C = \frac{\mu_1}{\mu_2}, which must be less than or equal to 11.

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