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PHXII10:WAVE OPTICS

367799 The equations of displacement of two waves are given as $y_{1} = 10 \sin (3 π t + π / 3)$ , $y_{2} = 5 (\sin 3 π t + \sqrt{3} \cos 3 π t)$ , then what is the ratio of their amplitude?

1 None of these

2

1 : 1

3

2 : 1

4

1 : 2

Explanation:

The given waves are
$y_{1} = 10 \sin (3 π t + \frac{π}{3})$
$y_{2} = 10 (\frac{1}{2} \sin 3 π t + \frac{\sqrt{3}}{2} \cos 3 π t)$
$= 10 (\cos 60^{\circ} \sin 3 π t + \sin 60^{\circ} \cos 3 π t)$
$= 10 \sin (3 π t + 60^{\circ})$
$\Rightarrow \frac{a}{b} = \frac{10}{10} = 1 : 1$

PHXII10:WAVE OPTICS

367800 Two monochromatic light waves of amplitudes $3 A$ and $2 A$ interfering at a point have a phase difference of $60^{\circ}$ . The intensity at that point will be proportional to

1

5 A^{2}

2

13 A^{2}

3

7 A^{2}

4

19 A^{2}

Explanation:

Here, $A_{1} = 3 A, A_{2} = 2 A, ϕ = 60^{\circ}$
The resultant amplitude at a point is
$R = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}$
$= \sqrt{{(3 A)}^{2} + {(2 A)}^{2} + 2 \times 3 A \times 2 A \times \cos 60^{\circ}}$
$= \sqrt{9 A^{2} + 4 A^{2} + 6 A^{2}} = A \sqrt{19}$
As intensity $\propto {(A m p l i t u d e)}^{2}$
Therefore, intensity at the same point is
$I \propto 19 A^{2}$

KCET - 2011

PHXII10:WAVE OPTICS

367801 In a double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in the interference pattern

1 The intensity of maxima decreases and the minima has zero intensity

2 The intensity of maxima increases and the minima has zero intensity

3 The intensities of both the maxima and minima increases

4 The intensity of maxima decreases and that of the minima increases

Explanation:

In interference between waves of equal amplitude $a$ , the minimum intensity is zero and the maximum intensity is proportional to $4 a^{2}$ . For waves of unequal amplitudes $a$ and $2 a$ , the minimum intensity is non- zero and the maximum intensity is proportional to ${(9 a)}^{2}$ which is greater than $4 a^{2}$ .

PHXII10:WAVE OPTICS

367802 If the ratio of the intensity of two coherent sources is 4 then the visibility
$[(I_{max} - I_{min}) / (I_{max} + I_{min})]$ of the fringes is

1 4

2

4 / 5

3

3 / 5

4 9

Explanation:

We have,
$\frac{I_{max}}{I_{min}} = {(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2} = {(\frac{\sqrt{4} + \sqrt{1}}{\sqrt{4} - \sqrt{1}})}^{2} = 9$
$\frac{I_{m a x} - I_{m i n}}{I_{m a x} + I_{m i n}} = \frac{(\frac{I_{m a x}}{I_{m i n}} - 1)}{(\frac{I_{m a x}}{I_{m i n}} + 1)} = \frac{9 - 1}{9 + 1}$
$= \frac{4}{5} [\frac{I_{m a x}}{I_{m i n}} = 9]$

PHXII10:WAVE OPTICS

367799 The equations of displacement of two waves are given as $y_{1} = 10 \sin (3 π t + π / 3)$ , $y_{2} = 5 (\sin 3 π t + \sqrt{3} \cos 3 π t)$ , then what is the ratio of their amplitude?

1 None of these

2

1 : 1

3

2 : 1

4

1 : 2

Explanation:

The given waves are
$y_{1} = 10 \sin (3 π t + \frac{π}{3})$
$y_{2} = 10 (\frac{1}{2} \sin 3 π t + \frac{\sqrt{3}}{2} \cos 3 π t)$
$= 10 (\cos 60^{\circ} \sin 3 π t + \sin 60^{\circ} \cos 3 π t)$
$= 10 \sin (3 π t + 60^{\circ})$
$\Rightarrow \frac{a}{b} = \frac{10}{10} = 1 : 1$

PHXII10:WAVE OPTICS

367800 Two monochromatic light waves of amplitudes $3 A$ and $2 A$ interfering at a point have a phase difference of $60^{\circ}$ . The intensity at that point will be proportional to

1

5 A^{2}

2

13 A^{2}

3

7 A^{2}

4

19 A^{2}

Explanation:

Here, $A_{1} = 3 A, A_{2} = 2 A, ϕ = 60^{\circ}$
The resultant amplitude at a point is
$R = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}$
$= \sqrt{{(3 A)}^{2} + {(2 A)}^{2} + 2 \times 3 A \times 2 A \times \cos 60^{\circ}}$
$= \sqrt{9 A^{2} + 4 A^{2} + 6 A^{2}} = A \sqrt{19}$
As intensity $\propto {(A m p l i t u d e)}^{2}$
Therefore, intensity at the same point is
$I \propto 19 A^{2}$

KCET - 2011

PHXII10:WAVE OPTICS

367801 In a double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in the interference pattern

1 The intensity of maxima decreases and the minima has zero intensity

2 The intensity of maxima increases and the minima has zero intensity

3 The intensities of both the maxima and minima increases

4 The intensity of maxima decreases and that of the minima increases

Explanation:

In interference between waves of equal amplitude $a$ , the minimum intensity is zero and the maximum intensity is proportional to $4 a^{2}$ . For waves of unequal amplitudes $a$ and $2 a$ , the minimum intensity is non- zero and the maximum intensity is proportional to ${(9 a)}^{2}$ which is greater than $4 a^{2}$ .

PHXII10:WAVE OPTICS

367802 If the ratio of the intensity of two coherent sources is 4 then the visibility
$[(I_{max} - I_{min}) / (I_{max} + I_{min})]$ of the fringes is

1 4

2

4 / 5

3

3 / 5

4 9

Explanation:

We have,
$\frac{I_{max}}{I_{min}} = {(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2} = {(\frac{\sqrt{4} + \sqrt{1}}{\sqrt{4} - \sqrt{1}})}^{2} = 9$
$\frac{I_{m a x} - I_{m i n}}{I_{m a x} + I_{m i n}} = \frac{(\frac{I_{m a x}}{I_{m i n}} - 1)}{(\frac{I_{m a x}}{I_{m i n}} + 1)} = \frac{9 - 1}{9 + 1}$
$= \frac{4}{5} [\frac{I_{m a x}}{I_{m i n}} = 9]$

PHXII10:WAVE OPTICS

367799 The equations of displacement of two waves are given as $y_{1} = 10 \sin (3 π t + π / 3)$ , $y_{2} = 5 (\sin 3 π t + \sqrt{3} \cos 3 π t)$ , then what is the ratio of their amplitude?

1 None of these

2

1 : 1

3

2 : 1

4

1 : 2

Explanation:

The given waves are
$y_{1} = 10 \sin (3 π t + \frac{π}{3})$
$y_{2} = 10 (\frac{1}{2} \sin 3 π t + \frac{\sqrt{3}}{2} \cos 3 π t)$
$= 10 (\cos 60^{\circ} \sin 3 π t + \sin 60^{\circ} \cos 3 π t)$
$= 10 \sin (3 π t + 60^{\circ})$
$\Rightarrow \frac{a}{b} = \frac{10}{10} = 1 : 1$

PHXII10:WAVE OPTICS

367800 Two monochromatic light waves of amplitudes $3 A$ and $2 A$ interfering at a point have a phase difference of $60^{\circ}$ . The intensity at that point will be proportional to

1

5 A^{2}

2

13 A^{2}

3

7 A^{2}

4

19 A^{2}

Explanation:

Here, $A_{1} = 3 A, A_{2} = 2 A, ϕ = 60^{\circ}$
The resultant amplitude at a point is
$R = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}$
$= \sqrt{{(3 A)}^{2} + {(2 A)}^{2} + 2 \times 3 A \times 2 A \times \cos 60^{\circ}}$
$= \sqrt{9 A^{2} + 4 A^{2} + 6 A^{2}} = A \sqrt{19}$
As intensity $\propto {(A m p l i t u d e)}^{2}$
Therefore, intensity at the same point is
$I \propto 19 A^{2}$

KCET - 2011

PHXII10:WAVE OPTICS

367801 In a double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in the interference pattern

1 The intensity of maxima decreases and the minima has zero intensity

2 The intensity of maxima increases and the minima has zero intensity

3 The intensities of both the maxima and minima increases

4 The intensity of maxima decreases and that of the minima increases

Explanation:

In interference between waves of equal amplitude $a$ , the minimum intensity is zero and the maximum intensity is proportional to $4 a^{2}$ . For waves of unequal amplitudes $a$ and $2 a$ , the minimum intensity is non- zero and the maximum intensity is proportional to ${(9 a)}^{2}$ which is greater than $4 a^{2}$ .

PHXII10:WAVE OPTICS

367802 If the ratio of the intensity of two coherent sources is 4 then the visibility
$[(I_{max} - I_{min}) / (I_{max} + I_{min})]$ of the fringes is

1 4

2

4 / 5

3

3 / 5

4 9

Explanation:

We have,
$\frac{I_{max}}{I_{min}} = {(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2} = {(\frac{\sqrt{4} + \sqrt{1}}{\sqrt{4} - \sqrt{1}})}^{2} = 9$
$\frac{I_{m a x} - I_{m i n}}{I_{m a x} + I_{m i n}} = \frac{(\frac{I_{m a x}}{I_{m i n}} - 1)}{(\frac{I_{m a x}}{I_{m i n}} + 1)} = \frac{9 - 1}{9 + 1}$
$= \frac{4}{5} [\frac{I_{m a x}}{I_{m i n}} = 9]$

PHXII10:WAVE OPTICS

367799 The equations of displacement of two waves are given as $y_{1} = 10 \sin (3 π t + π / 3)$ , $y_{2} = 5 (\sin 3 π t + \sqrt{3} \cos 3 π t)$ , then what is the ratio of their amplitude?

1 None of these

2

1 : 1

3

2 : 1

4

1 : 2

Explanation:

The given waves are
$y_{1} = 10 \sin (3 π t + \frac{π}{3})$
$y_{2} = 10 (\frac{1}{2} \sin 3 π t + \frac{\sqrt{3}}{2} \cos 3 π t)$
$= 10 (\cos 60^{\circ} \sin 3 π t + \sin 60^{\circ} \cos 3 π t)$
$= 10 \sin (3 π t + 60^{\circ})$
$\Rightarrow \frac{a}{b} = \frac{10}{10} = 1 : 1$

PHXII10:WAVE OPTICS

367800 Two monochromatic light waves of amplitudes $3 A$ and $2 A$ interfering at a point have a phase difference of $60^{\circ}$ . The intensity at that point will be proportional to

1

5 A^{2}

2

13 A^{2}

3

7 A^{2}

4

19 A^{2}

Explanation:

Here, $A_{1} = 3 A, A_{2} = 2 A, ϕ = 60^{\circ}$
The resultant amplitude at a point is
$R = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}$
$= \sqrt{{(3 A)}^{2} + {(2 A)}^{2} + 2 \times 3 A \times 2 A \times \cos 60^{\circ}}$
$= \sqrt{9 A^{2} + 4 A^{2} + 6 A^{2}} = A \sqrt{19}$
As intensity $\propto {(A m p l i t u d e)}^{2}$
Therefore, intensity at the same point is
$I \propto 19 A^{2}$

KCET - 2011

PHXII10:WAVE OPTICS

367801 In a double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in the interference pattern

1 The intensity of maxima decreases and the minima has zero intensity

2 The intensity of maxima increases and the minima has zero intensity

3 The intensities of both the maxima and minima increases

4 The intensity of maxima decreases and that of the minima increases

Explanation:

In interference between waves of equal amplitude $a$ , the minimum intensity is zero and the maximum intensity is proportional to $4 a^{2}$ . For waves of unequal amplitudes $a$ and $2 a$ , the minimum intensity is non- zero and the maximum intensity is proportional to ${(9 a)}^{2}$ which is greater than $4 a^{2}$ .

PHXII10:WAVE OPTICS

367802 If the ratio of the intensity of two coherent sources is 4 then the visibility
$[(I_{max} - I_{min}) / (I_{max} + I_{min})]$ of the fringes is

1 4

2

4 / 5

3

3 / 5

4 9

Explanation:

We have,
$\frac{I_{max}}{I_{min}} = {(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2} = {(\frac{\sqrt{4} + \sqrt{1}}{\sqrt{4} - \sqrt{1}})}^{2} = 9$
$\frac{I_{m a x} - I_{m i n}}{I_{m a x} + I_{m i n}} = \frac{(\frac{I_{m a x}}{I_{m i n}} - 1)}{(\frac{I_{m a x}}{I_{m i n}} + 1)} = \frac{9 - 1}{9 + 1}$
$= \frac{4}{5} [\frac{I_{m a x}}{I_{m i n}} = 9]$

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