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PHXII10:WAVE OPTICS

367811 If an interference pattern has maximum and minimum intensities in $36 : 1$ ratio then what will be the ratio of amplitudes?

1

5 : 7

2

7 : 4

3

4 : 7

4

7 : 5

Explanation:

$\frac{I_{max}}{I_{min}} = {[\frac{r + 1}{r - 1}]}^{2}$ ${(\frac{\frac{a_{1}}{a_{2}} + 1}{\frac{a_{1}}{a_{2}} - 1})}^{2} = 36$
$7 a_{2} = 5 a_{1}$ .

PHXII10:WAVE OPTICS

367812 The interference pattern is obtained with two coherent light sources of intensity ratio $n$ . In the interference pattern, the ratio $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$ will be

1

\frac{\sqrt{n}}{n + 1}

2

\frac{2 \sqrt{n}}{n + 1}

3

\frac{\sqrt{n}}{{(n + 1)}^{2}}

4

\frac{2 \sqrt{n}}{{(n + 1)}^{2}}

Explanation:

Given that $\frac{I_{1}}{I_{2}} = n$
Let $I_{2} = I_{0} \Rightarrow I_{1} = n I_{0}$
$I_{min} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = {(\sqrt{I_{0}} + \sqrt{n} \sqrt{I_{0}})}^{2}$
$I_{max} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2} = {(\sqrt{I_{0}} - \sqrt{n} \sqrt{I_{0}})}^{2}$
$\frac{I_{max} - I_{min}}{I_{max} + I_{min}} = \frac{{(\sqrt{I_{0}} + \sqrt{n} \sqrt{I_{0}})}^{2} - {(\sqrt{I_{0}} - \sqrt{n} \sqrt{I_{0}})}^{2}}{{(\sqrt{I_{0}} + \sqrt{n} \sqrt{I_{0}})}^{2} + {(\sqrt{I_{0}} - \sqrt{n} \sqrt{I_{0}})}^{2}}$
$= \frac{2 \sqrt{n}}{1 + n}$

NEET - 2016

PHXII10:WAVE OPTICS

367813 Assertion :
The maximum intensity in interference pattern is four times the intensity due to each slit.
Reason :
Intensity is directly proportional to square of amplitude.

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:

$I \propto a^{2}$ ,
$I_{max} \propto (a + a)^{2} \propto (4 a^{2})$
Intensity in double-slit interference is proportional to the square of amplitude.
$I_{min} \propto {(a_{1} - a_{2})}^{2}$
Destructive interference results in zero amplitude and minimum intensity, while constructive interference yields a fourfold increase in intensity. So correct option is (2).

PHXII10:WAVE OPTICS

367814 Two light beams of intensities in the ratio of 9 : 4 are allowed to interface. The ratio of the intensity of maxima and minima will be:

1

2 : 3

2

16 : 81

3

25 : 169

4

25 : 1

Explanation:

$\sqrt{\frac{I_{1}}{I_{2}}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$
$\sqrt{I_{2}} = \frac{2}{3} \sqrt{I_{1}}$
$\Rightarrow \frac{I_{max}}{I_{min}} = {(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2}$
$= {[\frac{\sqrt{I_{1}} + \frac{2}{3} \sqrt{I_{1}}}{\sqrt{I_{1}} - \frac{2}{3} \sqrt{I_{1}}}]}^{2}$
$= \frac{\sqrt{I_{1}}}{\sqrt{I_{1}}} {(\frac{5}{1})}^{2} = 25$

JEE - 2022

PHXII10:WAVE OPTICS

367815 Two coherent sources have intensity ratio of $100 : 1$ , and are used for obtaining the phenomenon of interference. Then the ratio of maximum and minimum intensity will be-

1

100 : 1

2

121 : 81

3

1 : 1

4

5 : 1

Explanation:

Given $I_{1} : I_{2} = 100 : 1$
$\frac{\sqrt{I_{1}}}{\sqrt{I_{2}}} = 10 : 1$
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = (11)^{2} = 121$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2} = (9)^{2} = 81$
$\frac{I_{max}}{I_{min}} = \frac{121}{81}$ .

PHXII10:WAVE OPTICS

367811 If an interference pattern has maximum and minimum intensities in $36 : 1$ ratio then what will be the ratio of amplitudes?

1

5 : 7

2

7 : 4

3

4 : 7

4

7 : 5

Explanation:

$\frac{I_{max}}{I_{min}} = {[\frac{r + 1}{r - 1}]}^{2}$ ${(\frac{\frac{a_{1}}{a_{2}} + 1}{\frac{a_{1}}{a_{2}} - 1})}^{2} = 36$
$7 a_{2} = 5 a_{1}$ .

PHXII10:WAVE OPTICS

367812 The interference pattern is obtained with two coherent light sources of intensity ratio $n$ . In the interference pattern, the ratio $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$ will be

1

\frac{\sqrt{n}}{n + 1}

2

\frac{2 \sqrt{n}}{n + 1}

3

\frac{\sqrt{n}}{{(n + 1)}^{2}}

4

\frac{2 \sqrt{n}}{{(n + 1)}^{2}}

Explanation:

Given that $\frac{I_{1}}{I_{2}} = n$
Let $I_{2} = I_{0} \Rightarrow I_{1} = n I_{0}$
$I_{min} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = {(\sqrt{I_{0}} + \sqrt{n} \sqrt{I_{0}})}^{2}$
$I_{max} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2} = {(\sqrt{I_{0}} - \sqrt{n} \sqrt{I_{0}})}^{2}$
$\frac{I_{max} - I_{min}}{I_{max} + I_{min}} = \frac{{(\sqrt{I_{0}} + \sqrt{n} \sqrt{I_{0}})}^{2} - {(\sqrt{I_{0}} - \sqrt{n} \sqrt{I_{0}})}^{2}}{{(\sqrt{I_{0}} + \sqrt{n} \sqrt{I_{0}})}^{2} + {(\sqrt{I_{0}} - \sqrt{n} \sqrt{I_{0}})}^{2}}$
$= \frac{2 \sqrt{n}}{1 + n}$

NEET - 2016

PHXII10:WAVE OPTICS

367813 Assertion :
The maximum intensity in interference pattern is four times the intensity due to each slit.
Reason :
Intensity is directly proportional to square of amplitude.

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:

$I \propto a^{2}$ ,
$I_{max} \propto (a + a)^{2} \propto (4 a^{2})$
Intensity in double-slit interference is proportional to the square of amplitude.
$I_{min} \propto {(a_{1} - a_{2})}^{2}$
Destructive interference results in zero amplitude and minimum intensity, while constructive interference yields a fourfold increase in intensity. So correct option is (2).

PHXII10:WAVE OPTICS

367814 Two light beams of intensities in the ratio of 9 : 4 are allowed to interface. The ratio of the intensity of maxima and minima will be:

1

2 : 3

2

16 : 81

3

25 : 169

4

25 : 1

Explanation:

$\sqrt{\frac{I_{1}}{I_{2}}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$
$\sqrt{I_{2}} = \frac{2}{3} \sqrt{I_{1}}$
$\Rightarrow \frac{I_{max}}{I_{min}} = {(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2}$
$= {[\frac{\sqrt{I_{1}} + \frac{2}{3} \sqrt{I_{1}}}{\sqrt{I_{1}} - \frac{2}{3} \sqrt{I_{1}}}]}^{2}$
$= \frac{\sqrt{I_{1}}}{\sqrt{I_{1}}} {(\frac{5}{1})}^{2} = 25$

JEE - 2022

PHXII10:WAVE OPTICS

367815 Two coherent sources have intensity ratio of $100 : 1$ , and are used for obtaining the phenomenon of interference. Then the ratio of maximum and minimum intensity will be-

1

100 : 1

2

121 : 81

3

1 : 1

4

5 : 1

Explanation:

Given $I_{1} : I_{2} = 100 : 1$
$\frac{\sqrt{I_{1}}}{\sqrt{I_{2}}} = 10 : 1$
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = (11)^{2} = 121$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2} = (9)^{2} = 81$
$\frac{I_{max}}{I_{min}} = \frac{121}{81}$ .

PHXII10:WAVE OPTICS

367811 If an interference pattern has maximum and minimum intensities in $36 : 1$ ratio then what will be the ratio of amplitudes?

1

5 : 7

2

7 : 4

3

4 : 7

4

7 : 5

Explanation:

$\frac{I_{max}}{I_{min}} = {[\frac{r + 1}{r - 1}]}^{2}$ ${(\frac{\frac{a_{1}}{a_{2}} + 1}{\frac{a_{1}}{a_{2}} - 1})}^{2} = 36$
$7 a_{2} = 5 a_{1}$ .

PHXII10:WAVE OPTICS

367812 The interference pattern is obtained with two coherent light sources of intensity ratio $n$ . In the interference pattern, the ratio $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$ will be

1

\frac{\sqrt{n}}{n + 1}

2

\frac{2 \sqrt{n}}{n + 1}

3

\frac{\sqrt{n}}{{(n + 1)}^{2}}

4

\frac{2 \sqrt{n}}{{(n + 1)}^{2}}

Explanation:

Given that $\frac{I_{1}}{I_{2}} = n$
Let $I_{2} = I_{0} \Rightarrow I_{1} = n I_{0}$
$I_{min} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = {(\sqrt{I_{0}} + \sqrt{n} \sqrt{I_{0}})}^{2}$
$I_{max} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2} = {(\sqrt{I_{0}} - \sqrt{n} \sqrt{I_{0}})}^{2}$
$\frac{I_{max} - I_{min}}{I_{max} + I_{min}} = \frac{{(\sqrt{I_{0}} + \sqrt{n} \sqrt{I_{0}})}^{2} - {(\sqrt{I_{0}} - \sqrt{n} \sqrt{I_{0}})}^{2}}{{(\sqrt{I_{0}} + \sqrt{n} \sqrt{I_{0}})}^{2} + {(\sqrt{I_{0}} - \sqrt{n} \sqrt{I_{0}})}^{2}}$
$= \frac{2 \sqrt{n}}{1 + n}$

NEET - 2016

PHXII10:WAVE OPTICS

367813 Assertion :
The maximum intensity in interference pattern is four times the intensity due to each slit.
Reason :
Intensity is directly proportional to square of amplitude.

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:

$I \propto a^{2}$ ,
$I_{max} \propto (a + a)^{2} \propto (4 a^{2})$
Intensity in double-slit interference is proportional to the square of amplitude.
$I_{min} \propto {(a_{1} - a_{2})}^{2}$
Destructive interference results in zero amplitude and minimum intensity, while constructive interference yields a fourfold increase in intensity. So correct option is (2).

PHXII10:WAVE OPTICS

367814 Two light beams of intensities in the ratio of 9 : 4 are allowed to interface. The ratio of the intensity of maxima and minima will be:

1

2 : 3

2

16 : 81

3

25 : 169

4

25 : 1

Explanation:

$\sqrt{\frac{I_{1}}{I_{2}}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$
$\sqrt{I_{2}} = \frac{2}{3} \sqrt{I_{1}}$
$\Rightarrow \frac{I_{max}}{I_{min}} = {(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2}$
$= {[\frac{\sqrt{I_{1}} + \frac{2}{3} \sqrt{I_{1}}}{\sqrt{I_{1}} - \frac{2}{3} \sqrt{I_{1}}}]}^{2}$
$= \frac{\sqrt{I_{1}}}{\sqrt{I_{1}}} {(\frac{5}{1})}^{2} = 25$

JEE - 2022

PHXII10:WAVE OPTICS

367815 Two coherent sources have intensity ratio of $100 : 1$ , and are used for obtaining the phenomenon of interference. Then the ratio of maximum and minimum intensity will be-

1

100 : 1

2

121 : 81

3

1 : 1

4

5 : 1

Explanation:

Given $I_{1} : I_{2} = 100 : 1$
$\frac{\sqrt{I_{1}}}{\sqrt{I_{2}}} = 10 : 1$
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = (11)^{2} = 121$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2} = (9)^{2} = 81$
$\frac{I_{max}}{I_{min}} = \frac{121}{81}$ .

PHXII10:WAVE OPTICS

367811 If an interference pattern has maximum and minimum intensities in $36 : 1$ ratio then what will be the ratio of amplitudes?

1

5 : 7

2

7 : 4

3

4 : 7

4

7 : 5

Explanation:

$\frac{I_{max}}{I_{min}} = {[\frac{r + 1}{r - 1}]}^{2}$ ${(\frac{\frac{a_{1}}{a_{2}} + 1}{\frac{a_{1}}{a_{2}} - 1})}^{2} = 36$
$7 a_{2} = 5 a_{1}$ .

PHXII10:WAVE OPTICS

367812 The interference pattern is obtained with two coherent light sources of intensity ratio $n$ . In the interference pattern, the ratio $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$ will be

1

\frac{\sqrt{n}}{n + 1}

2

\frac{2 \sqrt{n}}{n + 1}

3

\frac{\sqrt{n}}{{(n + 1)}^{2}}

4

\frac{2 \sqrt{n}}{{(n + 1)}^{2}}

Explanation:

Given that $\frac{I_{1}}{I_{2}} = n$
Let $I_{2} = I_{0} \Rightarrow I_{1} = n I_{0}$
$I_{min} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = {(\sqrt{I_{0}} + \sqrt{n} \sqrt{I_{0}})}^{2}$
$I_{max} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2} = {(\sqrt{I_{0}} - \sqrt{n} \sqrt{I_{0}})}^{2}$
$\frac{I_{max} - I_{min}}{I_{max} + I_{min}} = \frac{{(\sqrt{I_{0}} + \sqrt{n} \sqrt{I_{0}})}^{2} - {(\sqrt{I_{0}} - \sqrt{n} \sqrt{I_{0}})}^{2}}{{(\sqrt{I_{0}} + \sqrt{n} \sqrt{I_{0}})}^{2} + {(\sqrt{I_{0}} - \sqrt{n} \sqrt{I_{0}})}^{2}}$
$= \frac{2 \sqrt{n}}{1 + n}$

NEET - 2016

PHXII10:WAVE OPTICS

367813 Assertion :
The maximum intensity in interference pattern is four times the intensity due to each slit.
Reason :
Intensity is directly proportional to square of amplitude.

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:

$I \propto a^{2}$ ,
$I_{max} \propto (a + a)^{2} \propto (4 a^{2})$
Intensity in double-slit interference is proportional to the square of amplitude.
$I_{min} \propto {(a_{1} - a_{2})}^{2}$
Destructive interference results in zero amplitude and minimum intensity, while constructive interference yields a fourfold increase in intensity. So correct option is (2).

PHXII10:WAVE OPTICS

367814 Two light beams of intensities in the ratio of 9 : 4 are allowed to interface. The ratio of the intensity of maxima and minima will be:

1

2 : 3

2

16 : 81

3

25 : 169

4

25 : 1

Explanation:

$\sqrt{\frac{I_{1}}{I_{2}}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$
$\sqrt{I_{2}} = \frac{2}{3} \sqrt{I_{1}}$
$\Rightarrow \frac{I_{max}}{I_{min}} = {(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2}$
$= {[\frac{\sqrt{I_{1}} + \frac{2}{3} \sqrt{I_{1}}}{\sqrt{I_{1}} - \frac{2}{3} \sqrt{I_{1}}}]}^{2}$
$= \frac{\sqrt{I_{1}}}{\sqrt{I_{1}}} {(\frac{5}{1})}^{2} = 25$

JEE - 2022

PHXII10:WAVE OPTICS

367815 Two coherent sources have intensity ratio of $100 : 1$ , and are used for obtaining the phenomenon of interference. Then the ratio of maximum and minimum intensity will be-

1

100 : 1

2

121 : 81

3

1 : 1

4

5 : 1

Explanation:

Given $I_{1} : I_{2} = 100 : 1$
$\frac{\sqrt{I_{1}}}{\sqrt{I_{2}}} = 10 : 1$
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = (11)^{2} = 121$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2} = (9)^{2} = 81$
$\frac{I_{max}}{I_{min}} = \frac{121}{81}$ .

PHXII10:WAVE OPTICS

367811 If an interference pattern has maximum and minimum intensities in $36 : 1$ ratio then what will be the ratio of amplitudes?

1

5 : 7

2

7 : 4

3

4 : 7

4

7 : 5

Explanation:

$\frac{I_{max}}{I_{min}} = {[\frac{r + 1}{r - 1}]}^{2}$ ${(\frac{\frac{a_{1}}{a_{2}} + 1}{\frac{a_{1}}{a_{2}} - 1})}^{2} = 36$
$7 a_{2} = 5 a_{1}$ .

PHXII10:WAVE OPTICS

367812 The interference pattern is obtained with two coherent light sources of intensity ratio $n$ . In the interference pattern, the ratio $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$ will be

1

\frac{\sqrt{n}}{n + 1}

2

\frac{2 \sqrt{n}}{n + 1}

3

\frac{\sqrt{n}}{{(n + 1)}^{2}}

4

\frac{2 \sqrt{n}}{{(n + 1)}^{2}}

Explanation:

Given that $\frac{I_{1}}{I_{2}} = n$
Let $I_{2} = I_{0} \Rightarrow I_{1} = n I_{0}$
$I_{min} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = {(\sqrt{I_{0}} + \sqrt{n} \sqrt{I_{0}})}^{2}$
$I_{max} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2} = {(\sqrt{I_{0}} - \sqrt{n} \sqrt{I_{0}})}^{2}$
$\frac{I_{max} - I_{min}}{I_{max} + I_{min}} = \frac{{(\sqrt{I_{0}} + \sqrt{n} \sqrt{I_{0}})}^{2} - {(\sqrt{I_{0}} - \sqrt{n} \sqrt{I_{0}})}^{2}}{{(\sqrt{I_{0}} + \sqrt{n} \sqrt{I_{0}})}^{2} + {(\sqrt{I_{0}} - \sqrt{n} \sqrt{I_{0}})}^{2}}$
$= \frac{2 \sqrt{n}}{1 + n}$

NEET - 2016

PHXII10:WAVE OPTICS

367813 Assertion :
The maximum intensity in interference pattern is four times the intensity due to each slit.
Reason :
Intensity is directly proportional to square of amplitude.

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:

$I \propto a^{2}$ ,
$I_{max} \propto (a + a)^{2} \propto (4 a^{2})$
Intensity in double-slit interference is proportional to the square of amplitude.
$I_{min} \propto {(a_{1} - a_{2})}^{2}$
Destructive interference results in zero amplitude and minimum intensity, while constructive interference yields a fourfold increase in intensity. So correct option is (2).

PHXII10:WAVE OPTICS

367814 Two light beams of intensities in the ratio of 9 : 4 are allowed to interface. The ratio of the intensity of maxima and minima will be:

1

2 : 3

2

16 : 81

3

25 : 169

4

25 : 1

Explanation:

$\sqrt{\frac{I_{1}}{I_{2}}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$
$\sqrt{I_{2}} = \frac{2}{3} \sqrt{I_{1}}$
$\Rightarrow \frac{I_{max}}{I_{min}} = {(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2}$
$= {[\frac{\sqrt{I_{1}} + \frac{2}{3} \sqrt{I_{1}}}{\sqrt{I_{1}} - \frac{2}{3} \sqrt{I_{1}}}]}^{2}$
$= \frac{\sqrt{I_{1}}}{\sqrt{I_{1}}} {(\frac{5}{1})}^{2} = 25$

JEE - 2022

PHXII10:WAVE OPTICS

367815 Two coherent sources have intensity ratio of $100 : 1$ , and are used for obtaining the phenomenon of interference. Then the ratio of maximum and minimum intensity will be-

1

100 : 1

2

121 : 81

3

1 : 1

4

5 : 1

Explanation:

Given $I_{1} : I_{2} = 100 : 1$
$\frac{\sqrt{I_{1}}}{\sqrt{I_{2}}} = 10 : 1$
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = (11)^{2} = 121$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2} = (9)^{2} = 81$
$\frac{I_{max}}{I_{min}} = \frac{121}{81}$ .