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PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364958 Critical angle of incidence for a pair of optical media is $45^{\circ}$ . The refractive indices of first and second media are in the ratio

1

1 : 2

2

1 : \sqrt{2}

3

\sqrt{2} : 1

4

2 : 1

Explanation:

$\sin c = \frac{n_{2}}{n_{1}} \Rightarrow \frac{n_{2}}{n_{1}} = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$
$\Rightarrow \frac{n_{1}}{n_{2}} = \sqrt{2}$

JEE - 2024

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364959 Assertion :
The images formed by total internal reflections are much brighter than those formed by mirrors or lenses.
Reason :
There is no loss of intensity in total internal refraction.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Total internal reflection (TIR) retains $100 %$ of incident light without intensity loss, creating
$\Rightarrow$ Brighter images are formed by TIR (compared to reflection from mirrors or refraction through lenses, which involve absorption and so their images are less intense). So correct option is (1).

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364960 A ray of light is incident at an angle $i$ from denser to rarer medium. The reflected and the refracted rays are mutually perpendicular. The angle of reflection and the angle of refraction are respectively $r$ and $r^{'}$ , then the critical angle will be
supporting img

1

\sin^{- 1} (\tan r^{'})

2

\sin^{- 1} (\sin r)

3

\tan^{- 1} (\sin i)

4

\sin^{- 1} (\tan i)

Explanation:

Let $μ_{1} & μ_{2}$ are the refractive indices of denser and rarer mediums respectively.
$μ_{1} \sin i = μ_{2} \sin r^{'} (1)$
The critical angle is
$μ_{1} \sin C = μ_{2} \sin 90^{\circ}$
$\Rightarrow \frac{μ_{1}}{μ_{2}} = \frac{1}{\sin C} (2)$
From (1) & (2)
$\frac{1}{\sin C} = \frac{\sin r^{'}}{\sin r} = \frac{\sin (90 - r)}{\sin r} = \frac{\cos r}{\sin r}$
$\sin C = \tan r = \tan i (i = r)$
$C = \sin^{- 1} (\tan i)$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364961 The critical angle of light going from medium $A$ to medium $B$ is $θ$ . The speed of light in medium $A$ is $v$ . The speed of light in medium $B$ is:

1

\frac{v}{\sin θ}

2

v \sin θ

3

v \cot θ

4

v \tan θ

Explanation:

$\sin θ = \frac{1}{μ} = \frac{C_{A}}{C_{B}} \Rightarrow C_{B} = \frac{V}{\sin θ}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364958 Critical angle of incidence for a pair of optical media is $45^{\circ}$ . The refractive indices of first and second media are in the ratio

1

1 : 2

2

1 : \sqrt{2}

3

\sqrt{2} : 1

4

2 : 1

Explanation:

$\sin c = \frac{n_{2}}{n_{1}} \Rightarrow \frac{n_{2}}{n_{1}} = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$
$\Rightarrow \frac{n_{1}}{n_{2}} = \sqrt{2}$

JEE - 2024

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364959 Assertion :
The images formed by total internal reflections are much brighter than those formed by mirrors or lenses.
Reason :
There is no loss of intensity in total internal refraction.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Total internal reflection (TIR) retains $100 %$ of incident light without intensity loss, creating
$\Rightarrow$ Brighter images are formed by TIR (compared to reflection from mirrors or refraction through lenses, which involve absorption and so their images are less intense). So correct option is (1).

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364960 A ray of light is incident at an angle $i$ from denser to rarer medium. The reflected and the refracted rays are mutually perpendicular. The angle of reflection and the angle of refraction are respectively $r$ and $r^{'}$ , then the critical angle will be
supporting img

1

\sin^{- 1} (\tan r^{'})

2

\sin^{- 1} (\sin r)

3

\tan^{- 1} (\sin i)

4

\sin^{- 1} (\tan i)

Explanation:

Let $μ_{1} & μ_{2}$ are the refractive indices of denser and rarer mediums respectively.
$μ_{1} \sin i = μ_{2} \sin r^{'} (1)$
The critical angle is
$μ_{1} \sin C = μ_{2} \sin 90^{\circ}$
$\Rightarrow \frac{μ_{1}}{μ_{2}} = \frac{1}{\sin C} (2)$
From (1) & (2)
$\frac{1}{\sin C} = \frac{\sin r^{'}}{\sin r} = \frac{\sin (90 - r)}{\sin r} = \frac{\cos r}{\sin r}$
$\sin C = \tan r = \tan i (i = r)$
$C = \sin^{- 1} (\tan i)$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364961 The critical angle of light going from medium $A$ to medium $B$ is $θ$ . The speed of light in medium $A$ is $v$ . The speed of light in medium $B$ is:

1

\frac{v}{\sin θ}

2

v \sin θ

3

v \cot θ

4

v \tan θ

Explanation:

$\sin θ = \frac{1}{μ} = \frac{C_{A}}{C_{B}} \Rightarrow C_{B} = \frac{V}{\sin θ}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364958 Critical angle of incidence for a pair of optical media is $45^{\circ}$ . The refractive indices of first and second media are in the ratio

1

1 : 2

2

1 : \sqrt{2}

3

\sqrt{2} : 1

4

2 : 1

Explanation:

$\sin c = \frac{n_{2}}{n_{1}} \Rightarrow \frac{n_{2}}{n_{1}} = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$
$\Rightarrow \frac{n_{1}}{n_{2}} = \sqrt{2}$

JEE - 2024

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364959 Assertion :
The images formed by total internal reflections are much brighter than those formed by mirrors or lenses.
Reason :
There is no loss of intensity in total internal refraction.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Total internal reflection (TIR) retains $100 %$ of incident light without intensity loss, creating
$\Rightarrow$ Brighter images are formed by TIR (compared to reflection from mirrors or refraction through lenses, which involve absorption and so their images are less intense). So correct option is (1).

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364960 A ray of light is incident at an angle $i$ from denser to rarer medium. The reflected and the refracted rays are mutually perpendicular. The angle of reflection and the angle of refraction are respectively $r$ and $r^{'}$ , then the critical angle will be
supporting img

1

\sin^{- 1} (\tan r^{'})

2

\sin^{- 1} (\sin r)

3

\tan^{- 1} (\sin i)

4

\sin^{- 1} (\tan i)

Explanation:

Let $μ_{1} & μ_{2}$ are the refractive indices of denser and rarer mediums respectively.
$μ_{1} \sin i = μ_{2} \sin r^{'} (1)$
The critical angle is
$μ_{1} \sin C = μ_{2} \sin 90^{\circ}$
$\Rightarrow \frac{μ_{1}}{μ_{2}} = \frac{1}{\sin C} (2)$
From (1) & (2)
$\frac{1}{\sin C} = \frac{\sin r^{'}}{\sin r} = \frac{\sin (90 - r)}{\sin r} = \frac{\cos r}{\sin r}$
$\sin C = \tan r = \tan i (i = r)$
$C = \sin^{- 1} (\tan i)$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364961 The critical angle of light going from medium $A$ to medium $B$ is $θ$ . The speed of light in medium $A$ is $v$ . The speed of light in medium $B$ is:

1

\frac{v}{\sin θ}

2

v \sin θ

3

v \cot θ

4

v \tan θ

Explanation:

$\sin θ = \frac{1}{μ} = \frac{C_{A}}{C_{B}} \Rightarrow C_{B} = \frac{V}{\sin θ}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364958 Critical angle of incidence for a pair of optical media is $45^{\circ}$ . The refractive indices of first and second media are in the ratio

1

1 : 2

2

1 : \sqrt{2}

3

\sqrt{2} : 1

4

2 : 1

Explanation:

$\sin c = \frac{n_{2}}{n_{1}} \Rightarrow \frac{n_{2}}{n_{1}} = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$
$\Rightarrow \frac{n_{1}}{n_{2}} = \sqrt{2}$

JEE - 2024

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364959 Assertion :
The images formed by total internal reflections are much brighter than those formed by mirrors or lenses.
Reason :
There is no loss of intensity in total internal refraction.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Total internal reflection (TIR) retains $100 %$ of incident light without intensity loss, creating
$\Rightarrow$ Brighter images are formed by TIR (compared to reflection from mirrors or refraction through lenses, which involve absorption and so their images are less intense). So correct option is (1).

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364960 A ray of light is incident at an angle $i$ from denser to rarer medium. The reflected and the refracted rays are mutually perpendicular. The angle of reflection and the angle of refraction are respectively $r$ and $r^{'}$ , then the critical angle will be
supporting img

1

\sin^{- 1} (\tan r^{'})

2

\sin^{- 1} (\sin r)

3

\tan^{- 1} (\sin i)

4

\sin^{- 1} (\tan i)

Explanation:

Let $μ_{1} & μ_{2}$ are the refractive indices of denser and rarer mediums respectively.
$μ_{1} \sin i = μ_{2} \sin r^{'} (1)$
The critical angle is
$μ_{1} \sin C = μ_{2} \sin 90^{\circ}$
$\Rightarrow \frac{μ_{1}}{μ_{2}} = \frac{1}{\sin C} (2)$
From (1) & (2)
$\frac{1}{\sin C} = \frac{\sin r^{'}}{\sin r} = \frac{\sin (90 - r)}{\sin r} = \frac{\cos r}{\sin r}$
$\sin C = \tan r = \tan i (i = r)$
$C = \sin^{- 1} (\tan i)$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364961 The critical angle of light going from medium $A$ to medium $B$ is $θ$ . The speed of light in medium $A$ is $v$ . The speed of light in medium $B$ is:

1

\frac{v}{\sin θ}

2

v \sin θ

3

v \cot θ

4

v \tan θ

Explanation:

$\sin θ = \frac{1}{μ} = \frac{C_{A}}{C_{B}} \Rightarrow C_{B} = \frac{V}{\sin θ}$