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JEE Mains Previous Paper 1 (Held On: 10 Apr 2019 Shift 2)

Option 2 : 9 : 1

**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 I_{1}, I_{2} are intensity of light coming from two slits.

As, I ∝ W

From given,

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

I_{1} = 4I_{0} and I_{2 }= I_{0}

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

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

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

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

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

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

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

Therefore, the ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be 9 : 1.JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

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