Question 121: Wave Optics

Question

In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

Accepted Answer

Given, amplitude $$\propto$$ width of slit
$$\Rightarrow$$ A
2
= 3A
1
We know that,
$${{{I_{\max }}} \over {{I_{\min }}}} = {{{{(\sqrt {{I_1}}  + \sqrt {{I_2}} )}^2}} \over {{{(\sqrt {{I_1}}  - \sqrt {{I_2}} )}^2}}}$$
$$\because$$ Intensity, I $$\propto$$ A
2
$$	herefore$$ $${{{I_{\max }}} \over {{I_{\min }}}} = {{{{({A_1} + {A_2})}^2}} \over {{{(|{A_1} - {A_2}|)}^2}}}$$
$$ = {\left( {{{{A_1} + 3{A_1}} \over {|{A_1} - 3{A_1}|)}}} ight)^2} = {\left( {{{4{A_1}} \over {2{A_1}}}} ight)^2} = {4 \over 1}$$
$$	herefore$$ $${I_{\max }}:{I_{\min }} = 4:1$$

Answer

4 : 1

Answer

2 : 1

Answer

1 : 4

Answer

3 : 1

Wave Optics

In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

Choose the correct answer: