A transparent cube of side $d$, made of a material of refractive index ${\mu }_{2}$, is immersed in a liquid of refractive index ${\mu }_{1}({\mu }_{1}<{\mu }_{2})$. A ray is incident on the face $AB$ at an angle $\theta$ (shown in the figure). Total internal reflection takes place at the point $E$ on the face $BC.$ ![JEE Main 2019 Physics, Ray Optics — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Ray_Optics/648b5a67417cc3fb48d66851/question_1__q_648b5a67417cc3fb48d66851__cdn-question-pool.getmarks.app__0459979_que__88f33f22a3_final_ppt_sync.jpg) Then, $\theta$ must satisfy

JEE Main 2019 — Physics Optics

2019mcqmedium

A transparent cube of side $d$ , made of a material of refractive index ${\mu }_{2}$ , is immersed in a liquid of refractive index ${\mu }_{1}({\mu }_{1}<{\mu }_{2})$ . A ray is incident on the face $AB$ at an angle $\theta$ (shown in the figure). Total internal reflection takes place at the point $E$ on the face $BC.$

JEE Main 2019 Physics, Ray Optics — question figure

Then, $\theta$ must satisfy

Official previous-year question

Held on 12 Apr 2019 · Verified 6 Jul 2026.

Options

A
$\theta <{\mathrm{sin}}^{-1}\sqrt{\frac{{\mu }_{2}^{2}}{{\mu }_{1}^{2}}-1}$
B
$\theta >{\mathrm{sin}}^{-1}\frac{{\mu }_{1}}{{\mu }_{2}}$
C
$\theta <{\mathrm{sin}}^{-1}\frac{{\mu }_{1}}{{\mu }_{2}}$
D
$\theta >{\mathrm{sin}}^{-1}\sqrt{\frac{{\mu }_{2}^{2}}{{\mu }_{1}^{2}}-1}$

Did you get this right?

Solution

JEE Main 2019 Physics, Ray Optics — solution figure

At surface (1)

${\mu }_{1}\mathrm{sin}\theta ={\mu }_{2}\mathrm{sin}r \ldots (1)$

At surface (2)

${\mu }_{1}={\mu }_{2}\mathrm{sin}(90-r)$

${\mu }_{1}={\mu }_{2}\mathrm{cos}r...(2)$

From equation $(1)$ and $(2)$

$\Rightarrow \mathrm{sin}\theta =\frac{\mathrm{sin}r}{\mathrm{cos}r}$

$\Rightarrow \mathrm{sin}\theta =\frac{\frac{(\sqrt{{\mu }_{2}^{2}-{\mu }_{1}^{2}})}{{\mu }_{2}}}{(\frac{{\mu }_{1}}{{\mu }_{2}})}$

$\Rightarrow \mathrm{sin}\theta =\sqrt{\frac{{\mu }_{2}^{2}}{{\mu }_{1}^{2}}-1 }$

$\theta$ must satisfy,

$\Rightarrow \theta <{\mathrm{sin}}^{-1}(\sqrt{\frac{{\mu }_{2}^{2}}{{\mu }_{1}^{2}}-1})$

Your note

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Options

$\theta <{\mathrm{sin}}^{-1}\sqrt{\frac{{\mu }_{2}^{2}}{{\mu }_{1}^{2}}-1}$

$\theta >{\mathrm{sin}}^{-1}\frac{{\mu }_{1}}{{\mu }_{2}}$

$\theta <{\mathrm{sin}}^{-1}\frac{{\mu }_{1}}{{\mu }_{2}}$

$\theta >{\mathrm{sin}}^{-1}\sqrt{\frac{{\mu }_{2}^{2}}{{\mu }_{1}^{2}}-1}$

Solution

JEE Main 2019 Physics, Ray Optics — solution figure

At surface (1)

${\mu }_{1}\mathrm{sin}\theta ={\mu }_{2}\mathrm{sin}r \ldots (1)$

At surface (2)

${\mu }_{1}={\mu }_{2}\mathrm{sin}(90-r)$

${\mu }_{1}={\mu }_{2}\mathrm{cos}r...(2)$

From equation $(1)$ and $(2)$

$\Rightarrow \mathrm{sin}\theta =\frac{\mathrm{sin}r}{\mathrm{cos}r}$

$\Rightarrow \mathrm{sin}\theta =\frac{\frac{(\sqrt{{\mu }_{2}^{2}-{\mu }_{1}^{2}})}{{\mu }_{2}}}{(\frac{{\mu }_{1}}{{\mu }_{2}})}$

$\Rightarrow \mathrm{sin}\theta =\sqrt{\frac{{\mu }_{2}^{2}}{{\mu }_{1}^{2}}-1 }$

$\theta$ must satisfy,

$\Rightarrow \theta <{\mathrm{sin}}^{-1}(\sqrt{\frac{{\mu }_{2}^{2}}{{\mu }_{1}^{2}}-1})$

JEE Main 2019 — Physics Optics

Options

Did you get this right?

Solution

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JEE Main 2019 — Physics Optics

Options

Did you get this right?

Solution

Your note

JEE Main Optics in other years

More JEE Main Optics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Thermodynamics

Waves & Oscillations

Keep practicing Optics