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WAVE OPTICS

283257 A beam of light consisting of two wavelengths $6500 \AA$ and $5200 \AA$ , is used to obtain interference fringes in Young's double slit experiment. Suppose the $m^{th}$ bright fringe due to $6500 \AA$ coincides with $n^{th}$ bright fringe due to $5200 \AA$ at a minimum distance from the central maximum. Then :

1

m = 10, n = 8

2

m = 8, n = 10

3

m = 5, n = 4

4

m = 4, n = 5

Explanation:

: Given, $λ_{1} = 6500 \AA λ_{2} = 5200 \AA$ For $m^{th}$ bright fringe due to light,
$x_{m} = \frac{nD λ_{1}}{d}$
For nth bright fringe due to light,
$x_{n} = \frac{nD λ_{2}}{d}$
$∵$ Fringe are coincides
$x_{m} = x_{n} \Rightarrow \frac{mD λ_{1}}{d} = \frac{nd λ_{2}}{d}$
$m λ_{1} = n λ_{2}$
$\frac{m}{n} = \frac{λ_{2}}{λ_{1}} = \frac{5200}{6500} = \frac{52}{65} = \frac{4}{5}$
$m = 4, n = 5$

Karnataka CET-2001

WAVE OPTICS

283260 Unpolarized light falls on two polarizing sheets placed one on top of other. If the intensity of transmitted light is one fourth of the incident light, the angle between them is

1

35^{\circ}

2

40^{\circ}

3

45^{\circ}

4

50^{\circ}

Explanation:

: After the polarizing, intensity of light $= \frac{I_{0}}{\sqrt{2}}$
$I_{0}^{'} = \frac{I_{0}}{2} \cos^{2} θ$
According to question,
$I_{0}^{'} = \frac{I_{0}}{4}$
$\frac{I_{0}}{4} = \frac{I_{0}}{2} \cos^{2} θ$
$\frac{1}{2} = \cos^{2} θ$
$\frac{1}{\sqrt{2}} = \cos θ$
$\cos 45^{\circ} = \cos θ$ So, $θ = 45^{\circ}$

J and K-CET-2013

WAVE OPTICS

283263 If two mono-chromatic and phase related beams of light having intensities I and 4I superimposed on each other, then the possible maximum and minimum intensities in the resultant beam obtained are

1

5 I

and

3 I

2

9 I

and

3 I

3

9 I

and I

4

5 I

and I

Explanation:

: Given,
$I_{1} = I$
$I_{2} = 4 I$
Maximum intensity, $I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
$= (\sqrt{I} + \sqrt{4 I})^{2}$
$= 9 I$
Minimum intensity $(I_{min}) = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$= (\sqrt{I} - \sqrt{4 I})^{2}$
$= I$

COMEDK 2015

WAVE OPTICS

283268 If the sun rays are incidenting at $60^{\circ}$ angle and intensity is $I$ . if the sun rays are made incident at $30^{\circ}$ angle, then what will be the intensity?

1

\frac{I}{\sqrt{3}}

2

\sqrt{3} I

3

3 I

4

\frac{I}{3}

Explanation:

: Given, $I_{1} = I, θ_{1} = 60^{\circ}, θ_{2} = 30^{\circ}$ , Intensity of sunrays that making angle $θ$ is given by,
$∵ I = I_{0} \cos θ$
$\frac{I_{1}}{I_{2}} = \frac{I_{0} \cos 60^{\circ}}{I_{0} \cos 30^{\circ}} = \frac{1 / 2}{\sqrt{3} / 2}$
$\frac{I_{1}}{I_{2}} = \frac{1}{\sqrt{3}}$
$I_{2} = \sqrt{3} I_{1}$
$I_{2} = \sqrt{3} I$

JIPMER-2014

WAVE OPTICS

283257 A beam of light consisting of two wavelengths $6500 \AA$ and $5200 \AA$ , is used to obtain interference fringes in Young's double slit experiment. Suppose the $m^{th}$ bright fringe due to $6500 \AA$ coincides with $n^{th}$ bright fringe due to $5200 \AA$ at a minimum distance from the central maximum. Then :

1

m = 10, n = 8

2

m = 8, n = 10

3

m = 5, n = 4

4

m = 4, n = 5

Explanation:

: Given, $λ_{1} = 6500 \AA λ_{2} = 5200 \AA$ For $m^{th}$ bright fringe due to light,
$x_{m} = \frac{nD λ_{1}}{d}$
For nth bright fringe due to light,
$x_{n} = \frac{nD λ_{2}}{d}$
$∵$ Fringe are coincides
$x_{m} = x_{n} \Rightarrow \frac{mD λ_{1}}{d} = \frac{nd λ_{2}}{d}$
$m λ_{1} = n λ_{2}$
$\frac{m}{n} = \frac{λ_{2}}{λ_{1}} = \frac{5200}{6500} = \frac{52}{65} = \frac{4}{5}$
$m = 4, n = 5$

Karnataka CET-2001

WAVE OPTICS

283260 Unpolarized light falls on two polarizing sheets placed one on top of other. If the intensity of transmitted light is one fourth of the incident light, the angle between them is

1

35^{\circ}

2

40^{\circ}

3

45^{\circ}

4

50^{\circ}

Explanation:

: After the polarizing, intensity of light $= \frac{I_{0}}{\sqrt{2}}$
$I_{0}^{'} = \frac{I_{0}}{2} \cos^{2} θ$
According to question,
$I_{0}^{'} = \frac{I_{0}}{4}$
$\frac{I_{0}}{4} = \frac{I_{0}}{2} \cos^{2} θ$
$\frac{1}{2} = \cos^{2} θ$
$\frac{1}{\sqrt{2}} = \cos θ$
$\cos 45^{\circ} = \cos θ$ So, $θ = 45^{\circ}$

J and K-CET-2013

WAVE OPTICS

283263 If two mono-chromatic and phase related beams of light having intensities I and 4I superimposed on each other, then the possible maximum and minimum intensities in the resultant beam obtained are

1

5 I

and

3 I

2

9 I

and

3 I

3

9 I

and I

4

5 I

and I

Explanation:

: Given,
$I_{1} = I$
$I_{2} = 4 I$
Maximum intensity, $I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
$= (\sqrt{I} + \sqrt{4 I})^{2}$
$= 9 I$
Minimum intensity $(I_{min}) = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$= (\sqrt{I} - \sqrt{4 I})^{2}$
$= I$

COMEDK 2015

WAVE OPTICS

283268 If the sun rays are incidenting at $60^{\circ}$ angle and intensity is $I$ . if the sun rays are made incident at $30^{\circ}$ angle, then what will be the intensity?

1

\frac{I}{\sqrt{3}}

2

\sqrt{3} I

3

3 I

4

\frac{I}{3}

Explanation:

: Given, $I_{1} = I, θ_{1} = 60^{\circ}, θ_{2} = 30^{\circ}$ , Intensity of sunrays that making angle $θ$ is given by,
$∵ I = I_{0} \cos θ$
$\frac{I_{1}}{I_{2}} = \frac{I_{0} \cos 60^{\circ}}{I_{0} \cos 30^{\circ}} = \frac{1 / 2}{\sqrt{3} / 2}$
$\frac{I_{1}}{I_{2}} = \frac{1}{\sqrt{3}}$
$I_{2} = \sqrt{3} I_{1}$
$I_{2} = \sqrt{3} I$

JIPMER-2014

WAVE OPTICS

283257 A beam of light consisting of two wavelengths $6500 \AA$ and $5200 \AA$ , is used to obtain interference fringes in Young's double slit experiment. Suppose the $m^{th}$ bright fringe due to $6500 \AA$ coincides with $n^{th}$ bright fringe due to $5200 \AA$ at a minimum distance from the central maximum. Then :

1

m = 10, n = 8

2

m = 8, n = 10

3

m = 5, n = 4

4

m = 4, n = 5

Explanation:

: Given, $λ_{1} = 6500 \AA λ_{2} = 5200 \AA$ For $m^{th}$ bright fringe due to light,
$x_{m} = \frac{nD λ_{1}}{d}$
For nth bright fringe due to light,
$x_{n} = \frac{nD λ_{2}}{d}$
$∵$ Fringe are coincides
$x_{m} = x_{n} \Rightarrow \frac{mD λ_{1}}{d} = \frac{nd λ_{2}}{d}$
$m λ_{1} = n λ_{2}$
$\frac{m}{n} = \frac{λ_{2}}{λ_{1}} = \frac{5200}{6500} = \frac{52}{65} = \frac{4}{5}$
$m = 4, n = 5$

Karnataka CET-2001

WAVE OPTICS

283260 Unpolarized light falls on two polarizing sheets placed one on top of other. If the intensity of transmitted light is one fourth of the incident light, the angle between them is

1

35^{\circ}

2

40^{\circ}

3

45^{\circ}

4

50^{\circ}

Explanation:

: After the polarizing, intensity of light $= \frac{I_{0}}{\sqrt{2}}$
$I_{0}^{'} = \frac{I_{0}}{2} \cos^{2} θ$
According to question,
$I_{0}^{'} = \frac{I_{0}}{4}$
$\frac{I_{0}}{4} = \frac{I_{0}}{2} \cos^{2} θ$
$\frac{1}{2} = \cos^{2} θ$
$\frac{1}{\sqrt{2}} = \cos θ$
$\cos 45^{\circ} = \cos θ$ So, $θ = 45^{\circ}$

J and K-CET-2013

WAVE OPTICS

283263 If two mono-chromatic and phase related beams of light having intensities I and 4I superimposed on each other, then the possible maximum and minimum intensities in the resultant beam obtained are

1

5 I

and

3 I

2

9 I

and

3 I

3

9 I

and I

4

5 I

and I

Explanation:

: Given,
$I_{1} = I$
$I_{2} = 4 I$
Maximum intensity, $I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
$= (\sqrt{I} + \sqrt{4 I})^{2}$
$= 9 I$
Minimum intensity $(I_{min}) = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$= (\sqrt{I} - \sqrt{4 I})^{2}$
$= I$

COMEDK 2015

WAVE OPTICS

283268 If the sun rays are incidenting at $60^{\circ}$ angle and intensity is $I$ . if the sun rays are made incident at $30^{\circ}$ angle, then what will be the intensity?

1

\frac{I}{\sqrt{3}}

2

\sqrt{3} I

3

3 I

4

\frac{I}{3}

Explanation:

: Given, $I_{1} = I, θ_{1} = 60^{\circ}, θ_{2} = 30^{\circ}$ , Intensity of sunrays that making angle $θ$ is given by,
$∵ I = I_{0} \cos θ$
$\frac{I_{1}}{I_{2}} = \frac{I_{0} \cos 60^{\circ}}{I_{0} \cos 30^{\circ}} = \frac{1 / 2}{\sqrt{3} / 2}$
$\frac{I_{1}}{I_{2}} = \frac{1}{\sqrt{3}}$
$I_{2} = \sqrt{3} I_{1}$
$I_{2} = \sqrt{3} I$

JIPMER-2014

WAVE OPTICS

283257 A beam of light consisting of two wavelengths $6500 \AA$ and $5200 \AA$ , is used to obtain interference fringes in Young's double slit experiment. Suppose the $m^{th}$ bright fringe due to $6500 \AA$ coincides with $n^{th}$ bright fringe due to $5200 \AA$ at a minimum distance from the central maximum. Then :

1

m = 10, n = 8

2

m = 8, n = 10

3

m = 5, n = 4

4

m = 4, n = 5

Explanation:

: Given, $λ_{1} = 6500 \AA λ_{2} = 5200 \AA$ For $m^{th}$ bright fringe due to light,
$x_{m} = \frac{nD λ_{1}}{d}$
For nth bright fringe due to light,
$x_{n} = \frac{nD λ_{2}}{d}$
$∵$ Fringe are coincides
$x_{m} = x_{n} \Rightarrow \frac{mD λ_{1}}{d} = \frac{nd λ_{2}}{d}$
$m λ_{1} = n λ_{2}$
$\frac{m}{n} = \frac{λ_{2}}{λ_{1}} = \frac{5200}{6500} = \frac{52}{65} = \frac{4}{5}$
$m = 4, n = 5$

Karnataka CET-2001

WAVE OPTICS

283260 Unpolarized light falls on two polarizing sheets placed one on top of other. If the intensity of transmitted light is one fourth of the incident light, the angle between them is

1

35^{\circ}

2

40^{\circ}

3

45^{\circ}

4

50^{\circ}

Explanation:

: After the polarizing, intensity of light $= \frac{I_{0}}{\sqrt{2}}$
$I_{0}^{'} = \frac{I_{0}}{2} \cos^{2} θ$
According to question,
$I_{0}^{'} = \frac{I_{0}}{4}$
$\frac{I_{0}}{4} = \frac{I_{0}}{2} \cos^{2} θ$
$\frac{1}{2} = \cos^{2} θ$
$\frac{1}{\sqrt{2}} = \cos θ$
$\cos 45^{\circ} = \cos θ$ So, $θ = 45^{\circ}$

J and K-CET-2013

WAVE OPTICS

283263 If two mono-chromatic and phase related beams of light having intensities I and 4I superimposed on each other, then the possible maximum and minimum intensities in the resultant beam obtained are

1

5 I

and

3 I

2

9 I

and

3 I

3

9 I

and I

4

5 I

and I

Explanation:

: Given,
$I_{1} = I$
$I_{2} = 4 I$
Maximum intensity, $I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
$= (\sqrt{I} + \sqrt{4 I})^{2}$
$= 9 I$
Minimum intensity $(I_{min}) = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$= (\sqrt{I} - \sqrt{4 I})^{2}$
$= I$

COMEDK 2015

WAVE OPTICS

283268 If the sun rays are incidenting at $60^{\circ}$ angle and intensity is $I$ . if the sun rays are made incident at $30^{\circ}$ angle, then what will be the intensity?

1

\frac{I}{\sqrt{3}}

2

\sqrt{3} I

3

3 I

4

\frac{I}{3}

Explanation:

: Given, $I_{1} = I, θ_{1} = 60^{\circ}, θ_{2} = 30^{\circ}$ , Intensity of sunrays that making angle $θ$ is given by,
$∵ I = I_{0} \cos θ$
$\frac{I_{1}}{I_{2}} = \frac{I_{0} \cos 60^{\circ}}{I_{0} \cos 30^{\circ}} = \frac{1 / 2}{\sqrt{3} / 2}$
$\frac{I_{1}}{I_{2}} = \frac{1}{\sqrt{3}}$
$I_{2} = \sqrt{3} I_{1}$
$I_{2} = \sqrt{3} I$

JIPMER-2014

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