Physics Q41 – JEE-MAIN 2026

One side of an equilateral prism is painted by a transparent material of refractive index

n_{2}

. The refractive index of prism is

1.6

. The minimum value of

n_{2}

required for total internal reflection from painted face is _______.

Clue 1 of 3: Total Internal Reflection

For total internal reflection (TIR) to occur, light must travel from a denser medium to a rarer medium, and the angle of incidence must be greater than or equal to the critical angle.

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Step 1: Condition for Total Internal Reflection (TIR)✦ Active

For total internal reflection to occur at the painted face, the angle of incidence inside the prism ( $r_{2}$ ) must be greater than or equal to the critical angle $C$ . The critical angle is defined by $\sin C = \frac{n_{2}}{n_{p}}$ , where $n_{p}$ is the refractive index of the prism and $n_{2}$ is the refractive index of the painted material. Thus, for TIR, $n_{p} \sin r_{2} \geq n_{2}$ . The maximum value of $n_{2}$ for which TIR is possible (i.e., the threshold value) is when $n_{2} = n_{p} \sin r_{2}$ .

Step 2: Maximizing the Angle of Incidence at the Second Face○ Expand

To find the maximum possible $n_{2}$ , we need to maximize the angle of incidence $r_{2}$ at the painted face. For an equilateral prism, the prism angle is $A = 60^{\circ}$ . The relation between angles inside the prism is $A = r_{1} + r_{2}$ . To maximize $r_{2}$ , $r_{1}$ must be minimized. The minimum value of $r_{1}$ occurs when the incident ray strikes the first face normally ( $i_{1} = 0^{\circ}$ ), which results in $r_{1} = 0^{\circ}$ . Therefore, the maximum possible angle of incidence at the second face is $r_{2} = A - 0^{\circ} = 60^{\circ}$ .

💡 Teacher's Secret Hint

The diagram showing an incident ray at an angle is a general illustration, not specific to the condition for 'minimum $n_{2}$ required for TIR'.

Step 3: Calculating the Minimum

n_{2}

○ Expand

Using the maximum possible $r_{2} = 60^{\circ}$ and the refractive index of the prism $n_{p} = 1.6$ , the maximum $n_{2}$ for which TIR can occur (which is the 'minimum $n_{2}$ required for TIR' in this context) is:

n_{2} = n_{p} \sin r_{2}

\Rightarrow n_{2} = 1.6 \sin 60^{\circ}

\Rightarrow n_{2} = 1.6 \times \frac{\sqrt{3}}{2}

\Rightarrow n_{2} = 0.8 \sqrt{3}

\Rightarrow n_{2} = \frac{8}{10} \sqrt{3} = \frac{4 \sqrt{3}}{5}

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Physics Question 41 – JEE-MAIN 2026

Clue 1 of 3: Total Internal Reflection

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