A ruby laser, ? = 690 nm , shines on double-slits separated by 4.40 mm . An inte

Question

A ruby laser, ? = 690 nm, shines on double-slits separated by 4.40 mm. An interference pattern is observed on a screen at a distance R from the slits. The point C on the screen is at the center of the principal maximum of the interference pattern. The point P is the point on the principal maximum at which the intensity of light is half that of the intensity at C. What is the value of the angle ??

A ruby laser, ? = 690 nm, shines on double-slits separated by 4.40 mm. An interference pattern is observed on a screen at a distance R from the slits. The point C on the screen is at the center of the principal maximum of the interference pattern. The point P is the point on the principal maximum at which the intensity of light is half that of the intensity at C. What is the value of the angle ??

Explanation / Answer

I = Imax cos ^2(pi *d* sin theta / lamda)

here,

at C ,

I =Imax/2,

so,

Imax/2 = I max *cos ^2(pi *d* sin theta / lamda)

so,

cos ^2(pi *d* sin theta / lamda)=1/2,

putting values,

cos ^2(pi *4.4*10^-3* sin theta / 690*10^-9) = 1/2

(pi *4.4*10^-3* sin theta / 690*10^-9)= pi/4

either,

(pi *4.4*10^-3* sin theta / 690*10^-9)=pi/4

(4.4*10^-3* sin theta / 690*10^-9)=1/4

solving it ,

theta =3.92*10^-5 rad= 2.235*110^-3 degrees

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