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Maximum Angle for Total Internal Reflection in Block

Learn how to find the maximum incident angle for total internal reflection in a rectangular block using refractive index and critical angle concepts..

 

❓ Question

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

(Assume the top surface is horizontal and the vertical face is flat — standard rectangular block geometry.)


đź–Ľ️ Question Image

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


✍️ Short Explanation

This problem is based on:

👉 Snell’s law
👉 Critical angle
👉 Total internal reflection (TIR).

Main idea:

For TIR:

iC\boxed{ i \ge C }

where critical angle satisfies:

sinC=ÎĽ1ÎĽ2\sin C=\frac{\mu_1}{\mu_2}

Maximum Angle for Total Internal Reflection in Block


đź”· Step 1 — Find Critical Angle đź’Ż

Given:

ÎĽ2=1.25=54\mu_2=1.25=\frac54 ÎĽ1=1\mu_1=1

Critical angle:

sinC=ÎĽ1ÎĽ2\sin C=\frac{\mu_1}{\mu_2}
=15/4=\frac{1}{5/4}
=45=\frac45

Thus:

C=sin1(45)\boxed{ C=\sin^{-1}\left(\frac45\right) }


đź”· Step 2 — Relation Between Internal Angles

Let refracted angle inside block at first face be rr.

At top surface, angle of incidence becomes:

90r90^\circ-r

For just TIR:

90r=C90^\circ-r=C
r=90Cr=90^\circ-C

Using:

sinC=45\sin C=\frac45

triangle gives:

cosC=35\cos C=\frac35

Hence:

sinr=sin(90C)\sin r=\sin(90^\circ-C)
=cosC=\cos C
=35=\frac35


đź”· Step 3 — Apply Snell’s Law at Entry

At first surface:

μ1sinθ=μ2sinr\mu_1\sin\theta=\mu_2\sin r
1sinθ=54351\cdot\sin\theta = \frac54\cdot\frac35
=34=\frac34

Thus:

sinθ=34\boxed{ \sin\theta=\frac34 }

Therefore:

θ=sin1(34)\boxed{ \theta=\sin^{-1}\left(\frac34\right) }


đź”· Step 4 — JEE Trap Alert 🚨

❌ Directly critical angle ko answer maan lena

❌ Top surface angle =r=r assume kar lena

❌ Complementary angle relation bhool jaana

Remember:

Top incidence angle=90r\boxed{ \text{Top incidence angle}=90^\circ-r }


✅ Final Answer

sin1(34)\boxed{ \sin^{-1}\left(\frac34\right) }

(Option 2)


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