Answer
64.8k+ views
Hint:YDSE (Young’s double slit experiment) shows both energy and matter show both wave and particle characteristics. Apply the relation between the maximum intensities and minimum intensities of the two-sources used in Young’s double slit experiment. From the equation we can ratio between the intensities of two interfering sources.
Formula used:
$\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = {\dfrac{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} + 1} \right)}}{{{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} - 1} \right)}^2}}}^2}$
Where ${I_1}$ and ${I_{_2}}$ are the two intensities.
Complete step by step solution:
Two coherent sources of lights placed at a small distance apart are used in Young’s double slit experiment. Usually only magnitudes greater than wavelength of light is used. Young’s double slit experiment helps in understanding the wave theory of light. Commonly used coherent sources in the modern-day experiments is Laser. Young’s double slit experiment firmly establishes that light behaves as a particle and wave.
Now lets us use the formula
$\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = {\dfrac{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} + 1} \right)}}{{{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} - 1} \right)}^2}}}^2}$
Now in the question they have mentioned that the amplitude of intensity variation of two sources is found to be 5% of the average intensity.
Let us assume the average intensity $I$ to be 100. Then ${I_{\max }}$ will be $5\% $ more than the average intensity $I$. Then ${I_{\max }} = 105$ units and ${I_{\min }}$ will be $5\% $ less than the average intensity $I$. ${I_{\min }} = 95$ units.
$\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = {\dfrac{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} + 1} \right)}}{{{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} - 1} \right)}^2}}}^2}$
$\dfrac{{105}}{{95}} = {\dfrac{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} + 1} \right)}}{{{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} - 1} \right)}^2}}}^2}$
We will the value of $\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} $is $40$
Hence$\dfrac{{{I_1}}}{{{I_2}}} = \left( {\dfrac{{{{\left( {40} \right)}^2}}}{{{{\left( 1 \right)}^2}}}} \right)$
$\dfrac{{{I_1}}}{{{I_2}}} = 1600$
Hence option (C) is the right option.
Note: Later they conducted Young’s double slit experiment using electrons and the pattern generated a similar result as light. It behaves both as a particle and wave. Young’s double slit experiment shows both these characteristics prominently.
Formula used:
$\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = {\dfrac{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} + 1} \right)}}{{{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} - 1} \right)}^2}}}^2}$
Where ${I_1}$ and ${I_{_2}}$ are the two intensities.
Complete step by step solution:
Two coherent sources of lights placed at a small distance apart are used in Young’s double slit experiment. Usually only magnitudes greater than wavelength of light is used. Young’s double slit experiment helps in understanding the wave theory of light. Commonly used coherent sources in the modern-day experiments is Laser. Young’s double slit experiment firmly establishes that light behaves as a particle and wave.
Now lets us use the formula
$\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = {\dfrac{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} + 1} \right)}}{{{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} - 1} \right)}^2}}}^2}$
Now in the question they have mentioned that the amplitude of intensity variation of two sources is found to be 5% of the average intensity.
Let us assume the average intensity $I$ to be 100. Then ${I_{\max }}$ will be $5\% $ more than the average intensity $I$. Then ${I_{\max }} = 105$ units and ${I_{\min }}$ will be $5\% $ less than the average intensity $I$. ${I_{\min }} = 95$ units.
$\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = {\dfrac{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} + 1} \right)}}{{{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} - 1} \right)}^2}}}^2}$
$\dfrac{{105}}{{95}} = {\dfrac{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} + 1} \right)}}{{{{\left( {\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} - 1} \right)}^2}}}^2}$
We will the value of $\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} $is $40$
Hence$\dfrac{{{I_1}}}{{{I_2}}} = \left( {\dfrac{{{{\left( {40} \right)}^2}}}{{{{\left( 1 \right)}^2}}}} \right)$
$\dfrac{{{I_1}}}{{{I_2}}} = 1600$
Hence option (C) is the right option.
Note: Later they conducted Young’s double slit experiment using electrons and the pattern generated a similar result as light. It behaves both as a particle and wave. Young’s double slit experiment shows both these characteristics prominently.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)