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 top surface of the block [2025]

Question

Smart Achievers · Accepted Answer

(3)

$r + θ_{C} = 90 ° \Rightarrow r = 90 ° - θ_{C}$

At $1^{st}$ refraction : $1 \times \sin θ = \frac{5}{4} \sin r$

$\Rightarrow \sin r = \frac{4}{5} \sin θ$ ... (i)

At point TIR:

$\frac{5}{4} \sin θ_{C} = 1 . \sin 90 ° \Rightarrow \frac{5}{4} \sin θ_{C} = 1$

$\sin (90 - r) = \frac{4}{5} \Rightarrow \cos r = \frac{4}{5}$

$\Rightarrow \sin r = \frac{3}{5}$ ... (ii)

From (i) and (ii), $θ = \sin^{- 1} (\frac{3}{4})$

Solution

1 $\tan^{- 1} (4 / 3)$

2 $\tan^{- 1} (3 / 4)$

3 $\sin^{- 1} (3 / 4)$

4 $\cos^{- 1} (3 / 4)$

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Solution

1 tan–1(4/3)

2 tan–1(3/4)

3 sin–1(3/4)

4 cos–1(3/4)

Ans. (3) r+θC=90° ⇒ r=90°–θC At 1st refraction : 1×sin θ=54sin r ⇒ sin r=45sin θ ... (i) At point TIR: 54sin θC=1.sin 90° ⇒ 54sin θC=1 sin(90–r)=45 ⇒ cos r=45 ⇒ sin r=35 ... (ii) From (i) and (ii), θ=sin–1(34)

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1 $\tan^{- 1} (4 / 3)$

2 $\tan^{- 1} (3 / 4)$

3 $\sin^{- 1} (3 / 4)$

4 $\cos^{- 1} (3 / 4)$