text: # 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
explanation: The document contains a physics question related to a Young’s double slit experiment, specifically asking about the ratio of the intensity of maxima to minima. The question is followed by multiple-choice options and a detailed solution that leads to the correct answer, which is 9 : 1.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # 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
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a physics question related to a Young’s double slit experiment, specifically asking about the ratio of the intensity of maxima to minima. The question is followed by multiple-choice options and a detailed solution that leads to the correct answer, which is 9 : 1.
  
  Label: question-and-answer