Question 155: Waves Optics

Question

Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16. The intensity of the waves are in the ratio :

Accepted Answer

Given that,
$${{{{m I}_{\max }}} \over {{{\mathop{m I}
olimits} _{min}}}} = {{16} \over 1}$$
We  know,
I
max
$$=$$ $${\left( {\sqrt {{{m I}_1}}  + \sqrt {{{m I}_2}} } ight)^2}$$
and I
min
$$ = {\left( {\sqrt {{{m I}_1}}  - \sqrt {{{m I}_2}} } ight)^2}$$
$$ 	herefore $$   $${{{{\left( {\sqrt {{{m I}_1}}  + \sqrt {{{m I}_2}} } ight)}^2}} \over {{{\left( {\sqrt {{{m I}_1}}  - \sqrt {{{m I}_2}} } ight)}^2}}} = {{16} \over 1}$$
$$ \Rightarrow $$   $${{\sqrt {{{m I}_1}}  + \sqrt {{{m I}_2}} } \over {\sqrt {{{m I}_1}}  - \sqrt {{{m I}_2}} }} = {4 \over 1}$$
$$ \Rightarrow $$   $$4\sqrt {{{m I}_1}}  - 4\sqrt {{{m I}_2}}  = \sqrt {{{m I}_1}}  + \sqrt {{{m I}_2}} $$
$$ \Rightarrow $$   $$3\sqrt {{{m I}_1}}  = 5\sqrt {{{m I}_2}} $$
$$ \Rightarrow $$   $${{\sqrt {{{m I}_1}} } \over {\sqrt {{{m I}_2}} }} = {5 \over 3}$$
$$ \Rightarrow $$   $${{{{m I}_1}} \over {{{m I}_2}}} = {{25} \over 9}$$

Answer

16 : 9

Answer

25 : 9

Answer

4 : 1

Answer

5 : 3

Waves Optics

Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16. The intensity of the waves are in the ratio :

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