Question 145: Waves Optics

Question

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 :

Accepted Answer

$${I_1} = 4{I_0}$$
$${I_2} = {I_0}$$
$${I_{\max }} = {\left( {\sqrt {{I_0}}  + \sqrt {{I_2}} } ight)^2}$$
$$ = {\left( {2\sqrt {{I_0}}  + \sqrt {{I_0}} } ight)^2} = 9{I_0}$$
$${I_{\min }} = {\left( {\sqrt {{I_1}}  - \sqrt {{I_2}} } ight)^2}$$
$$ = {\left( {2\sqrt {{I_0}}  - \sqrt {{I_0}} } ight)^2} = {I_0}$$
$$ 	herefore $$ $${{{{\mathop{m I}
olimits} _{\max }}} \over {{I_{\min }}}} = {{9{I_0}} \over {{I_0}}} = {9 \over 1}$$

Answer

25 : 9

Answer

4 : 1

Answer

$${\left( {\sqrt 3  + 1} \right)^4}:16$$

Answer

9 : 1

Waves Optics

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 :

Choose the correct answer: