Question

In youngs double slit experiment how slit width is related to intensity?

Answer 1

In young's double slit experiment

Intensity of light due to slit is directly proportional to width of slit

Answer 2

I hope you know that intensity (I)(I) of light at any point on the screen due to interference in the Young's Double Slit experiment can be given as

A2=I=a21+a22+2a1a2cosϕ
A2=I=a12+a22+2a1a2cos⁡ϕ
where a1,a2a1,a2 are the amplitudes of the light waves with constant phase difference of ϕϕ, AA is the amplitude of the resultant displaement at the point on the screen. For simplicity, we can assume that intensity of light to be equal to square of the amplitude as given above.

Thus,
Imax=a21+a22+2a1a2(1)=(a1+a2)2
Imax=a12+a22+2a1a2(1)=(a1+a2)2
Imin=a21+a22+2a1a2(−1)=(a1−a2)2
Imin=a12+a22+2a1a2(−1)=(a1−a2)2
Therefore, ImaxImin=(a1+a2)2(a1−a2)2=259ImaxImin=(a1+a2)2(a1−a2)2=259
Thus, a1+a2=5,a1−a2=3a1+a2=5,a1−a2=3

a1+(a1−3)=5=2a1−3a1+(a1−3)=5=2a1−3
Thus, a1=8/2=4,a2=1a1=8/2=4,a2=1

The intensity of light due to a slit (source of light) is directly proportional to width of the slit. Therefore, if w1w1 and w2w2 are widths of the tow slits S1S1 and S2S2; I1I1 and I2I2 are intensities of light due to the respective slits on the screen, then

w1w2=I1I2=a21a22=4212=16