# In a Young’s double slit experiment, the ratio of the slit’s width is 4 : 1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be:

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JEE Mains Previous Paper 1 (Held On: 10 Apr 2019 Shift 2)
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1. 25 : 9
2. 9 : 1
3. 4 : 1
4. $${\left( {\sqrt 3 + 1} \right)^4}:16$$

Option 2 : 9 : 1

## Detailed Solution

Concept:

Maximum intensity:

Maximum intensity is the square of the sum of both the amplitude and minimum intensity is the square of the difference between both the amplitude.

$${{\rm{I}}_{{\rm{max}}}} = {\left( {\sqrt {{{\rm{I}}_1}} + \sqrt {{{\rm{I}}_2}} } \right)^2}$$

Minimum intensity:

Minimum intensity is the square of the difference of both the amplitude and minimum intensity is the square of the difference between both the amplitude.

$${{\rm{I}}_{{\rm{min}}}} = {\left( {\sqrt {{{\rm{I}}_1}} - \sqrt {{{\rm{I}}_2}} } \right)^2}$$

As we know, width is proportional to the intensity and intensity is directly proportional to the square of amplitude.

Calculation:

Given, Ratio of the slits width is 4 : 1

We know that, width is proportional to the intensity and intensity is directly proportional to the square of amplitude.

Maximum intensity is the square of the sum of both the amplitude and minimum intensity is the square of the difference between both the amplitude.

i.e., $$\frac{{{{\rm{W}}_1}}}{{{{\rm{W}}_2}}} = \frac{{{{\rm{I}}_1}}}{{{{\rm{I}}_2}}} = \frac{4}{1}$$

Let I1, I2 are intensity of light coming from two slits.

As, I ∝ W

From given,

$$\therefore \frac{{{{\rm{I}}_1}}}{{{{\rm{I}}_2}}} = \frac{4}{1}$$

I1 = 4I0 and I2 = I0

$${{\rm{I}}_{{\rm{max}}}} = {\left( {\sqrt {{{\rm{I}}_1}} + \sqrt {{{\rm{I}}_2}} } \right)^2}$$

$${{\rm{I}}_{{\rm{min}}}} = {\left( {\sqrt {{{\rm{I}}_1}} - \sqrt {{{\rm{I}}_2}} } \right)^2}$$

$$\frac{{{{\rm{I}}_{{\rm{max}}}}}}{{{{\rm{I}}_{{\rm{min}}}}}} = \frac{{{{\left( {\sqrt {{{\rm{I}}_1}} + \sqrt {{{\rm{I}}_2}} } \right)}^2}}}{{{{\left( {\sqrt {{{\rm{I}}_1}} - \sqrt {{{\rm{I}}_2}} } \right)}^2}}}$$

$$\frac{{{{\rm{I}}_{{\rm{max}}}}}}{{{{\rm{I}}_{{\rm{min}}}}}} = {\left( {\frac{{\sqrt {\frac{{{{\rm{I}}_1}}}{{{{\rm{I}}_2}}}} + 1}}{{\sqrt {\frac{{{{\rm{I}}_1}}}{{{{\rm{I}}_2}}}} - 1}}} \right)^2}$$

$$\frac{{{{\rm{I}}_{{\rm{max}}}}}}{{{{\rm{I}}_{{\rm{min}}}}}} = {\left( {\frac{{\sqrt 4 + 1}}{{\sqrt 4 - 1}}} \right)^2}$$

$$\frac{{{{\rm{I}}_{{\rm{max}}}}}}{{{{\rm{I}}_{{\rm{min}}}}}} = {\left( {\frac{{2 + 1}}{{2 - 1}}} \right)^2}$$

$$\frac{{{{\rm{I}}_{{\rm{max}}}}}}{{{{\rm{I}}_{{\rm{min}}}}}} = \frac{9}{1}$$

Therefore, the ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be 9 : 1.
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JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)
8425
90 Questions 360 Marks 180 Mins
The text passage contains a physics question related to the Young’s double slit experiment, followed by multiple-choice answers and a detailed solution explaining the calculation of the ratio of the intensity of maxima to minima.
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Given a text passage, determine:
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    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

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# In a Young’s double slit experiment, the ratio of the slit’s width is 4 : 1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be:

This question was previously asked in
JEE Mains Previous Paper 1 (Held On: 10 Apr 2019 Shift 2)
View all JEE Main Papers >
1. 25 : 9
2. 9 : 1
3. 4 : 1
4. $${\left( {\sqrt 3 + 1} \right)^4}:16$$

Option 2 : 9 : 1

## Detailed Solution

Concept:

Maximum intensity:

Maximum intensity is the square of the sum of both the amplitude and minimum intensity is the square of the difference between both the amplitude.

$${{\rm{I}}_{{\rm{max}}}} = {\left( {\sqrt {{{\rm{I}}_1}} + \sqrt {{{\rm{I}}_2}} } \right)^2}$$

Minimum intensity:

Minimum intensity is the square of the difference of both the amplitude and minimum intensity is the square of the difference between both the amplitude.

$${{\rm{I}}_{{\rm{min}}}} = {\left( {\sqrt {{{\rm{I}}_1}} - \sqrt {{{\rm{I}}_2}} } \right)^2}$$

As we know, width is proportional to the intensity and intensity is directly proportional to the square of amplitude.

Calculation:

Given, Ratio of the slits width is 4 : 1

We know that, width is proportional to the intensity and intensity is directly proportional to the square of amplitude.

Maximum intensity is the square of the sum of both the amplitude and minimum intensity is the square of the difference between both the amplitude.

i.e., $$\frac{{{{\rm{W}}_1}}}{{{{\rm{W}}_2}}} = \frac{{{{\rm{I}}_1}}}{{{{\rm{I}}_2}}} = \frac{4}{1}$$

Let I1, I2 are intensity of light coming from two slits.

As, I ∝ W

From given,

$$\therefore \frac{{{{\rm{I}}_1}}}{{{{\rm{I}}_2}}} = \frac{4}{1}$$

I1 = 4I0 and I2 = I0

$${{\rm{I}}_{{\rm{max}}}} = {\left( {\sqrt {{{\rm{I}}_1}} + \sqrt {{{\rm{I}}_2}} } \right)^2}$$

$${{\rm{I}}_{{\rm{min}}}} = {\left( {\sqrt {{{\rm{I}}_1}} - \sqrt {{{\rm{I}}_2}} } \right)^2}$$

$$\frac{{{{\rm{I}}_{{\rm{max}}}}}}{{{{\rm{I}}_{{\rm{min}}}}}} = \frac{{{{\left( {\sqrt {{{\rm{I}}_1}} + \sqrt {{{\rm{I}}_2}} } \right)}^2}}}{{{{\left( {\sqrt {{{\rm{I}}_1}} - \sqrt {{{\rm{I}}_2}} } \right)}^2}}}$$

$$\frac{{{{\rm{I}}_{{\rm{max}}}}}}{{{{\rm{I}}_{{\rm{min}}}}}} = {\left( {\frac{{\sqrt {\frac{{{{\rm{I}}_1}}}{{{{\rm{I}}_2}}}} + 1}}{{\sqrt {\frac{{{{\rm{I}}_1}}}{{{{\rm{I}}_2}}}} - 1}}} \right)^2}$$

$$\frac{{{{\rm{I}}_{{\rm{max}}}}}}{{{{\rm{I}}_{{\rm{min}}}}}} = {\left( {\frac{{\sqrt 4 + 1}}{{\sqrt 4 - 1}}} \right)^2}$$

$$\frac{{{{\rm{I}}_{{\rm{max}}}}}}{{{{\rm{I}}_{{\rm{min}}}}}} = {\left( {\frac{{2 + 1}}{{2 - 1}}} \right)^2}$$

$$\frac{{{{\rm{I}}_{{\rm{max}}}}}}{{{{\rm{I}}_{{\rm{min}}}}}} = \frac{9}{1}$$

Therefore, the ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be 9 : 1.
Free
JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)
8425
90 Questions 360 Marks 180 Mins
user
The text passage contains a physics question related to the Young’s double slit experiment, followed by multiple-choice answers and a detailed solution explaining the calculation of the ratio of the intensity of maxima to minima.

TYPE: SINGLE_QUESTION_WITH_ANSWER