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Ray Optics

282207 A beam of light is incident from air on the surface of a liquid. The angle of incidence is $θ$ and the angle of refraction is $α$ . If the critical angle for liquid when surrounded by air is $θ_{c}$ then $\sin θ_{c}$ is

1

\frac{\sin (α)}{\sin (θ)}

2

\sin (α) \times \sin (θ)

3

\frac{\sin (θ)}{\sin (α)}

4

\frac{\sin (α)}{\cos (θ)}

Explanation:

A: Refractive index, $μ = \frac{\sin i}{\sin r}$
$μ = \frac{\sin θ}{\sin α}$
We know that, critical angle $(θ_{c}) = \sin^{- 1} (\frac{1}{μ})$
$\sin θ_{c} = \frac{\sin α}{\sin θ}$

AP EAMCET-07.07.2022

Ray Optics

282208 Light travels in two media $M_{1}$ and $M_{2}$ with speeds $1.5 \times 10^{8} {ms}^{- 1}$ and $2.0 \times 10^{8} {ms}^{- 1}$ respectively. The critical angle between them is:

1

\tan^{- 1} (\frac{3}{\sqrt{7}})

2

\tan^{- 1} (\frac{2}{3})

3

\cos^{- 1} (\frac{3}{4})

4

\sin^{- 1} (\frac{2}{3})

Explanation:

A: Given that, speed in medium $M_{1}$ and $M_{2}$
$\begin{array}{r} v_{m_{1}} = 1.5 \times 10^{8} m / s \\ v_{m_{2}} = 2.0 \times 10^{8} m / s \end{array}$
Critical angle,
$\begin{aligned} \sin θ_{c} & = \frac{μ_{2}}{μ_{1}} = \frac{v_{m_{1}}}{V_{m_{2}}} \\ \sin θ_{c} & = \frac{1.5 \times 10^{8}}{2.0 \times 10^{8}} \\ \sin θ_{c} & = \frac{3}{4} \\ θ_{c} & = \sin^{- 1} (\frac{3}{4}) \\ θ_{c} & = \tan^{- 1} (\frac{3}{\sqrt{7}}) \end{aligned}$

JEE Main-26.07.2022

Ray Optics

282209 Consider a glass slab which is silvered at one side and the other side is transparent. Given the refractive index of the glass slab to be 1.5 . If a ray of light is incident at an angle of $45^{\circ}$ on the transparent side, the deviation of the ray of light from its initial path, when it comes out of the slab is

1

120^{\circ}

2

45^{\circ}

3

90^{\circ}

4

180^{\circ}

Explanation:

B: Given, $μ = 1.5$ , and $i = 45^{\circ}$
From figure it is clear that -

$\begin{array}{r} i = e \\ e = 45^{\circ} \end{array}$

Karnataka CET-2022

Ray Optics

282210 A ray is incident from a medium of refractive index 2 into a medium of refractive index 1 . The critical angle is

1

30^{\circ}

2

60^{\circ}

3

45^{\circ}

4

90^{\circ}

Explanation:

A: Given,
Refractive index $(μ_{2}) = 2$ and $μ_{1} = 1$
$_{1} μ_{2} = \frac{μ_{2}}{μ_{1}} = \frac{2}{1} = 2$
Critical angle $(θ_{c}) = \sin^{- 1} (\frac{1}{_{1} μ_{2}})$
$θ_{c} = \sin^{- 1} (\frac{1}{2}) = 30^{\circ}$

AP EAMCET-04.07.2022

Ray Optics

282207 A beam of light is incident from air on the surface of a liquid. The angle of incidence is $θ$ and the angle of refraction is $α$ . If the critical angle for liquid when surrounded by air is $θ_{c}$ then $\sin θ_{c}$ is

1

\frac{\sin (α)}{\sin (θ)}

2

\sin (α) \times \sin (θ)

3

\frac{\sin (θ)}{\sin (α)}

4

\frac{\sin (α)}{\cos (θ)}

Explanation:

A: Refractive index, $μ = \frac{\sin i}{\sin r}$
$μ = \frac{\sin θ}{\sin α}$
We know that, critical angle $(θ_{c}) = \sin^{- 1} (\frac{1}{μ})$
$\sin θ_{c} = \frac{\sin α}{\sin θ}$

AP EAMCET-07.07.2022

Ray Optics

282208 Light travels in two media $M_{1}$ and $M_{2}$ with speeds $1.5 \times 10^{8} {ms}^{- 1}$ and $2.0 \times 10^{8} {ms}^{- 1}$ respectively. The critical angle between them is:

1

\tan^{- 1} (\frac{3}{\sqrt{7}})

2

\tan^{- 1} (\frac{2}{3})

3

\cos^{- 1} (\frac{3}{4})

4

\sin^{- 1} (\frac{2}{3})

Explanation:

A: Given that, speed in medium $M_{1}$ and $M_{2}$
$\begin{array}{r} v_{m_{1}} = 1.5 \times 10^{8} m / s \\ v_{m_{2}} = 2.0 \times 10^{8} m / s \end{array}$
Critical angle,
$\begin{aligned} \sin θ_{c} & = \frac{μ_{2}}{μ_{1}} = \frac{v_{m_{1}}}{V_{m_{2}}} \\ \sin θ_{c} & = \frac{1.5 \times 10^{8}}{2.0 \times 10^{8}} \\ \sin θ_{c} & = \frac{3}{4} \\ θ_{c} & = \sin^{- 1} (\frac{3}{4}) \\ θ_{c} & = \tan^{- 1} (\frac{3}{\sqrt{7}}) \end{aligned}$

JEE Main-26.07.2022

Ray Optics

282209 Consider a glass slab which is silvered at one side and the other side is transparent. Given the refractive index of the glass slab to be 1.5 . If a ray of light is incident at an angle of $45^{\circ}$ on the transparent side, the deviation of the ray of light from its initial path, when it comes out of the slab is

1

120^{\circ}

2

45^{\circ}

3

90^{\circ}

4

180^{\circ}

Explanation:

B: Given, $μ = 1.5$ , and $i = 45^{\circ}$
From figure it is clear that -

$\begin{array}{r} i = e \\ e = 45^{\circ} \end{array}$

Karnataka CET-2022

Ray Optics

282210 A ray is incident from a medium of refractive index 2 into a medium of refractive index 1 . The critical angle is

1

30^{\circ}

2

60^{\circ}

3

45^{\circ}

4

90^{\circ}

Explanation:

A: Given,
Refractive index $(μ_{2}) = 2$ and $μ_{1} = 1$
$_{1} μ_{2} = \frac{μ_{2}}{μ_{1}} = \frac{2}{1} = 2$
Critical angle $(θ_{c}) = \sin^{- 1} (\frac{1}{_{1} μ_{2}})$
$θ_{c} = \sin^{- 1} (\frac{1}{2}) = 30^{\circ}$

AP EAMCET-04.07.2022

Ray Optics

282207 A beam of light is incident from air on the surface of a liquid. The angle of incidence is $θ$ and the angle of refraction is $α$ . If the critical angle for liquid when surrounded by air is $θ_{c}$ then $\sin θ_{c}$ is

1

\frac{\sin (α)}{\sin (θ)}

2

\sin (α) \times \sin (θ)

3

\frac{\sin (θ)}{\sin (α)}

4

\frac{\sin (α)}{\cos (θ)}

Explanation:

A: Refractive index, $μ = \frac{\sin i}{\sin r}$
$μ = \frac{\sin θ}{\sin α}$
We know that, critical angle $(θ_{c}) = \sin^{- 1} (\frac{1}{μ})$
$\sin θ_{c} = \frac{\sin α}{\sin θ}$

AP EAMCET-07.07.2022

Ray Optics

282208 Light travels in two media $M_{1}$ and $M_{2}$ with speeds $1.5 \times 10^{8} {ms}^{- 1}$ and $2.0 \times 10^{8} {ms}^{- 1}$ respectively. The critical angle between them is:

1

\tan^{- 1} (\frac{3}{\sqrt{7}})

2

\tan^{- 1} (\frac{2}{3})

3

\cos^{- 1} (\frac{3}{4})

4

\sin^{- 1} (\frac{2}{3})

Explanation:

A: Given that, speed in medium $M_{1}$ and $M_{2}$
$\begin{array}{r} v_{m_{1}} = 1.5 \times 10^{8} m / s \\ v_{m_{2}} = 2.0 \times 10^{8} m / s \end{array}$
Critical angle,
$\begin{aligned} \sin θ_{c} & = \frac{μ_{2}}{μ_{1}} = \frac{v_{m_{1}}}{V_{m_{2}}} \\ \sin θ_{c} & = \frac{1.5 \times 10^{8}}{2.0 \times 10^{8}} \\ \sin θ_{c} & = \frac{3}{4} \\ θ_{c} & = \sin^{- 1} (\frac{3}{4}) \\ θ_{c} & = \tan^{- 1} (\frac{3}{\sqrt{7}}) \end{aligned}$

JEE Main-26.07.2022

Ray Optics

282209 Consider a glass slab which is silvered at one side and the other side is transparent. Given the refractive index of the glass slab to be 1.5 . If a ray of light is incident at an angle of $45^{\circ}$ on the transparent side, the deviation of the ray of light from its initial path, when it comes out of the slab is

1

120^{\circ}

2

45^{\circ}

3

90^{\circ}

4

180^{\circ}

Explanation:

B: Given, $μ = 1.5$ , and $i = 45^{\circ}$
From figure it is clear that -

$\begin{array}{r} i = e \\ e = 45^{\circ} \end{array}$

Karnataka CET-2022

Ray Optics

282210 A ray is incident from a medium of refractive index 2 into a medium of refractive index 1 . The critical angle is

1

30^{\circ}

2

60^{\circ}

3

45^{\circ}

4

90^{\circ}

Explanation:

A: Given,
Refractive index $(μ_{2}) = 2$ and $μ_{1} = 1$
$_{1} μ_{2} = \frac{μ_{2}}{μ_{1}} = \frac{2}{1} = 2$
Critical angle $(θ_{c}) = \sin^{- 1} (\frac{1}{_{1} μ_{2}})$
$θ_{c} = \sin^{- 1} (\frac{1}{2}) = 30^{\circ}$

AP EAMCET-04.07.2022

Ray Optics

282207 A beam of light is incident from air on the surface of a liquid. The angle of incidence is $θ$ and the angle of refraction is $α$ . If the critical angle for liquid when surrounded by air is $θ_{c}$ then $\sin θ_{c}$ is

1

\frac{\sin (α)}{\sin (θ)}

2

\sin (α) \times \sin (θ)

3

\frac{\sin (θ)}{\sin (α)}

4

\frac{\sin (α)}{\cos (θ)}

Explanation:

A: Refractive index, $μ = \frac{\sin i}{\sin r}$
$μ = \frac{\sin θ}{\sin α}$
We know that, critical angle $(θ_{c}) = \sin^{- 1} (\frac{1}{μ})$
$\sin θ_{c} = \frac{\sin α}{\sin θ}$

AP EAMCET-07.07.2022

Ray Optics

282208 Light travels in two media $M_{1}$ and $M_{2}$ with speeds $1.5 \times 10^{8} {ms}^{- 1}$ and $2.0 \times 10^{8} {ms}^{- 1}$ respectively. The critical angle between them is:

1

\tan^{- 1} (\frac{3}{\sqrt{7}})

2

\tan^{- 1} (\frac{2}{3})

3

\cos^{- 1} (\frac{3}{4})

4

\sin^{- 1} (\frac{2}{3})

Explanation:

A: Given that, speed in medium $M_{1}$ and $M_{2}$
$\begin{array}{r} v_{m_{1}} = 1.5 \times 10^{8} m / s \\ v_{m_{2}} = 2.0 \times 10^{8} m / s \end{array}$
Critical angle,
$\begin{aligned} \sin θ_{c} & = \frac{μ_{2}}{μ_{1}} = \frac{v_{m_{1}}}{V_{m_{2}}} \\ \sin θ_{c} & = \frac{1.5 \times 10^{8}}{2.0 \times 10^{8}} \\ \sin θ_{c} & = \frac{3}{4} \\ θ_{c} & = \sin^{- 1} (\frac{3}{4}) \\ θ_{c} & = \tan^{- 1} (\frac{3}{\sqrt{7}}) \end{aligned}$

JEE Main-26.07.2022

Ray Optics

282209 Consider a glass slab which is silvered at one side and the other side is transparent. Given the refractive index of the glass slab to be 1.5 . If a ray of light is incident at an angle of $45^{\circ}$ on the transparent side, the deviation of the ray of light from its initial path, when it comes out of the slab is

1

120^{\circ}

2

45^{\circ}

3

90^{\circ}

4

180^{\circ}

Explanation:

B: Given, $μ = 1.5$ , and $i = 45^{\circ}$
From figure it is clear that -

$\begin{array}{r} i = e \\ e = 45^{\circ} \end{array}$

Karnataka CET-2022

Ray Optics

282210 A ray is incident from a medium of refractive index 2 into a medium of refractive index 1 . The critical angle is

1

30^{\circ}

2

60^{\circ}

3

45^{\circ}

4

90^{\circ}

Explanation:

A: Given,
Refractive index $(μ_{2}) = 2$ and $μ_{1} = 1$
$_{1} μ_{2} = \frac{μ_{2}}{μ_{1}} = \frac{2}{1} = 2$
Critical angle $(θ_{c}) = \sin^{- 1} (\frac{1}{_{1} μ_{2}})$
$θ_{c} = \sin^{- 1} (\frac{1}{2}) = 30^{\circ}$

AP EAMCET-04.07.2022

Question Bank Series

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