In Young's experiment, the ratio of the intensities at the maxima and minima in the interference pattern is 36:16. what is the ratio of the widths of the two beams?

Options

A. 15
B. 20
C. 25
D. 30

Answer

C

Explanation

The intensity of a wave is directly proportional to `("amplitude")^(2)`. As the wave is coming out of the slit,
`Iprop("width of the slit")(w)`.
`(I_(max))/((I_("min"))=((A_(1)+A_(2))^(2))/((A_(1)+A_(2))^(2)))=(36)/(16)`
where `A_(2) and A_(2)` are the amplitudes of the two waves.
`therefore (A_(1)+A_(2))/(A_(1)-A_(2))=(6)/(4)=(3)/(2)`
By componendo, and dividendo
`(A_(1)+A_(2)+A_(1)-A_(2))/(A_(1)+A_(2)-A_(1)+A_(2))=(3+2)/(3-2)`
`therefore (A_(1))/(A_(2))=(5)/(1)`
`therefore (I_(1))/(I_(2))=((A_(1))/(A_(2)))^(2)=25`
and `(I_(1))/(I_(2))=(w_(1))/(w_(2))=25`