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PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364928 A beam of light of converging towards a point $I$ on a screen. A plane glass plate whose thickness in the direction of the beam = t, Refractive index = $μ$ , is introduced in the path of the beam. The convergence point is shifted by

1

t (1 + \frac{1}{μ}) n e a r e r

2

t (1 - \frac{1}{μ}) n e a r e r

3

t (1 - \frac{1}{μ}) a w a y

4

t (1 + \frac{1}{μ}) a w a y

Explanation:

Normal
supporting img
shift $Δ x = (1 - \frac{1}{μ}) t$ and shift takes place in direction of ray.

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364929 A diver at a depth of $12 m$ in water $(μ = 4 / 3)$ sees the sky in a cone of semi-vertical angle

1

90^{\circ}

2

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

3

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

4

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

Explanation:

Semi vertical angle $= C = \sin^{- 1} (\frac{1}{μ}) = \sin^{- 1} (\frac{3}{4})$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364930 ‘Mirage’ is a phenomenon due to

1 Refraction of light

2 Reflection of the light

3 Diffraction of light

4 Total internal reflection of light

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364931 A point source of light $S$ is placed at the bottom of a vessel containing a liquid of refractive index $5 / 3$ . A person is viewing the source from above the surface. There is an opaque disc $D$ of radius $1 c m$ floating on the surface of the liquid. The centre of the disc lies vertically above the source. The liquid from the vessel is gradually drained out through a tap. The maximum height of the liquid for which the source cannot be seen at all from above is
supporting img

1

1.50 c m

2

1.64 c m

3

1.33 c m

4

1.86 c m

Explanation:

Suppose the maximum height of the liquid is $h$ for which the source is not visible.
Hence radius of the disc
supporting img
We have,
$\sin i_{c} = \frac{r}{\sqrt{r^{2} + h^{2}}}$
$\Rightarrow \frac{1}{μ} = \frac{r}{\sqrt{r^{2} + h^{2}}}$
$\begin{aligned} \Rightarrow \frac{1}{μ^{2}} = \frac{r^{2}}{r^{2} + h^{2}} \\ \Rightarrow r = \frac{h}{\sqrt{μ^{2} - 1}} \end{aligned}$
$\begin{aligned} \Rightarrow 1 = \frac{h}{{(\frac{5}{3})}^{2} - 1} \\ \Rightarrow h = 1.33 c m \end{aligned}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364928 A beam of light of converging towards a point $I$ on a screen. A plane glass plate whose thickness in the direction of the beam = t, Refractive index = $μ$ , is introduced in the path of the beam. The convergence point is shifted by

1

t (1 + \frac{1}{μ}) n e a r e r

2

t (1 - \frac{1}{μ}) n e a r e r

3

t (1 - \frac{1}{μ}) a w a y

4

t (1 + \frac{1}{μ}) a w a y

Explanation:

Normal
supporting img
shift $Δ x = (1 - \frac{1}{μ}) t$ and shift takes place in direction of ray.

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364929 A diver at a depth of $12 m$ in water $(μ = 4 / 3)$ sees the sky in a cone of semi-vertical angle

1

90^{\circ}

2

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

3

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

4

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

Explanation:

Semi vertical angle $= C = \sin^{- 1} (\frac{1}{μ}) = \sin^{- 1} (\frac{3}{4})$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364930 ‘Mirage’ is a phenomenon due to

1 Refraction of light

2 Reflection of the light

3 Diffraction of light

4 Total internal reflection of light

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364931 A point source of light $S$ is placed at the bottom of a vessel containing a liquid of refractive index $5 / 3$ . A person is viewing the source from above the surface. There is an opaque disc $D$ of radius $1 c m$ floating on the surface of the liquid. The centre of the disc lies vertically above the source. The liquid from the vessel is gradually drained out through a tap. The maximum height of the liquid for which the source cannot be seen at all from above is
supporting img

1

1.50 c m

2

1.64 c m

3

1.33 c m

4

1.86 c m

Explanation:

Suppose the maximum height of the liquid is $h$ for which the source is not visible.
Hence radius of the disc
supporting img
We have,
$\sin i_{c} = \frac{r}{\sqrt{r^{2} + h^{2}}}$
$\Rightarrow \frac{1}{μ} = \frac{r}{\sqrt{r^{2} + h^{2}}}$
$\begin{aligned} \Rightarrow \frac{1}{μ^{2}} = \frac{r^{2}}{r^{2} + h^{2}} \\ \Rightarrow r = \frac{h}{\sqrt{μ^{2} - 1}} \end{aligned}$
$\begin{aligned} \Rightarrow 1 = \frac{h}{{(\frac{5}{3})}^{2} - 1} \\ \Rightarrow h = 1.33 c m \end{aligned}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364928 A beam of light of converging towards a point $I$ on a screen. A plane glass plate whose thickness in the direction of the beam = t, Refractive index = $μ$ , is introduced in the path of the beam. The convergence point is shifted by

1

t (1 + \frac{1}{μ}) n e a r e r

2

t (1 - \frac{1}{μ}) n e a r e r

3

t (1 - \frac{1}{μ}) a w a y

4

t (1 + \frac{1}{μ}) a w a y

Explanation:

Normal
supporting img
shift $Δ x = (1 - \frac{1}{μ}) t$ and shift takes place in direction of ray.

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364929 A diver at a depth of $12 m$ in water $(μ = 4 / 3)$ sees the sky in a cone of semi-vertical angle

1

90^{\circ}

2

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

3

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

4

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

Explanation:

Semi vertical angle $= C = \sin^{- 1} (\frac{1}{μ}) = \sin^{- 1} (\frac{3}{4})$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364930 ‘Mirage’ is a phenomenon due to

1 Refraction of light

2 Reflection of the light

3 Diffraction of light

4 Total internal reflection of light

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364931 A point source of light $S$ is placed at the bottom of a vessel containing a liquid of refractive index $5 / 3$ . A person is viewing the source from above the surface. There is an opaque disc $D$ of radius $1 c m$ floating on the surface of the liquid. The centre of the disc lies vertically above the source. The liquid from the vessel is gradually drained out through a tap. The maximum height of the liquid for which the source cannot be seen at all from above is
supporting img

1

1.50 c m

2

1.64 c m

3

1.33 c m

4

1.86 c m

Explanation:

Suppose the maximum height of the liquid is $h$ for which the source is not visible.
Hence radius of the disc
supporting img
We have,
$\sin i_{c} = \frac{r}{\sqrt{r^{2} + h^{2}}}$
$\Rightarrow \frac{1}{μ} = \frac{r}{\sqrt{r^{2} + h^{2}}}$
$\begin{aligned} \Rightarrow \frac{1}{μ^{2}} = \frac{r^{2}}{r^{2} + h^{2}} \\ \Rightarrow r = \frac{h}{\sqrt{μ^{2} - 1}} \end{aligned}$
$\begin{aligned} \Rightarrow 1 = \frac{h}{{(\frac{5}{3})}^{2} - 1} \\ \Rightarrow h = 1.33 c m \end{aligned}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364928 A beam of light of converging towards a point $I$ on a screen. A plane glass plate whose thickness in the direction of the beam = t, Refractive index = $μ$ , is introduced in the path of the beam. The convergence point is shifted by

1

t (1 + \frac{1}{μ}) n e a r e r

2

t (1 - \frac{1}{μ}) n e a r e r

3

t (1 - \frac{1}{μ}) a w a y

4

t (1 + \frac{1}{μ}) a w a y

Explanation:

Normal
supporting img
shift $Δ x = (1 - \frac{1}{μ}) t$ and shift takes place in direction of ray.

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364929 A diver at a depth of $12 m$ in water $(μ = 4 / 3)$ sees the sky in a cone of semi-vertical angle

1

90^{\circ}

2

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

3

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

4

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

Explanation:

Semi vertical angle $= C = \sin^{- 1} (\frac{1}{μ}) = \sin^{- 1} (\frac{3}{4})$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364930 ‘Mirage’ is a phenomenon due to

1 Refraction of light

2 Reflection of the light

3 Diffraction of light

4 Total internal reflection of light

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364931 A point source of light $S$ is placed at the bottom of a vessel containing a liquid of refractive index $5 / 3$ . A person is viewing the source from above the surface. There is an opaque disc $D$ of radius $1 c m$ floating on the surface of the liquid. The centre of the disc lies vertically above the source. The liquid from the vessel is gradually drained out through a tap. The maximum height of the liquid for which the source cannot be seen at all from above is
supporting img

1

1.50 c m

2

1.64 c m

3

1.33 c m

4

1.86 c m

Explanation:

Suppose the maximum height of the liquid is $h$ for which the source is not visible.
Hence radius of the disc
supporting img
We have,
$\sin i_{c} = \frac{r}{\sqrt{r^{2} + h^{2}}}$
$\Rightarrow \frac{1}{μ} = \frac{r}{\sqrt{r^{2} + h^{2}}}$
$\begin{aligned} \Rightarrow \frac{1}{μ^{2}} = \frac{r^{2}}{r^{2} + h^{2}} \\ \Rightarrow r = \frac{h}{\sqrt{μ^{2} - 1}} \end{aligned}$
$\begin{aligned} \Rightarrow 1 = \frac{h}{{(\frac{5}{3})}^{2} - 1} \\ \Rightarrow h = 1.33 c m \end{aligned}$

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