Ratio waves originating from sources `S_(1)` and `S_(2)` having zero phase difference and common wavelength `lambda` will show completely destructive interference at a point `P` is `S_(1)P-S_(2)P` isA. `5lambda`B. `3lambda//4`C. `2lambda`D. `11lambda//2`

Ratio waves originating from sources `S_(1)` and `S_(2)` having zero p

1.	Ratio waves originating from sources `S_(1)` and `S_(2)` having zero phase difference and common wavelength `lambda` will show completely destructive interference at a point `P` is `S_(1)P-S_(2)P` isA. `5lambda`B. `3lambda//4`C. `2lambda`D. `11lambda//2`
Answer» Answer: For destructive interference: Path difference =`S_(1)P-S_(2)P=(2n-1)lambda//2` `n=1,S_(1)P-S_(2)P=(2xx1-1)lambda//2=lambda//2` `n=2,S_(1)P-S_(2)P=(2xx2-1)lambda//2=3lambda//2` `n=3,S_(1)P-S_(2)P=(2xx3-1)lambda//2=5lambda//2` `n=4,S_(1)P-S_(2)P=(2xx4-1)lambda//2=7lambda//2` `n=5,S_(1)P-S_(2)P=(2xx5-1)lambda//2=9lambda//2` `n=6,S_(1)P-S_(2)P=(2xx6-1)lambda//2=11lambda//2` So,destructive pattern is possible only for path difference=`11lambda//2`.

Discussion

No Comment Found

Related InterviewSolutions

Angular width of central maxima in the Fraunhofer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength `6000Å`. When the slit is illuminated by light of another wavelength, the angular width decreases by 30%. The wavelength of this light will beA. `3500 Å`B. `4200 Å`C. `4700 Å`D. `6000 Å`
Find the half angular width of the central bright maximum in the fraunhofer diffraction pattern of a slit of width `12xx10^(-5)cm` when the slit is illuminated by monochromatic light of wavelength 6000 Å.
Find the half angular width of the central bright maximum in the fraunhofer diffraction pattern of a slit of width `12xx10^(-5)cm` when the slit is illuminated by monochromatic light of wavelength 6000 Å.A. `40^(@)`B. `45^(@)`C. `30^(@)`D. `60^(@)`
In a single slit diffraction, the width of slits is `0.5cm`, focal length of lens is 40 cm and of first dark fringe is
What will be the angular width of central maxima in Fraunhofer diffraction when light of wavelength `6000Å` is used and slit width is `12xx10^-5cm`?A. (a) `2rad`B. (b) `3rad`C. (c) `1rad`D. (d) `8rad`
A light wave is incident normally over a slit of width `24xx10^-5cm`. The angular position of second dark fringe from the central maxima is `30^@`. What is the wavelength of light?A. `6000 Å`B. `5000 Å`C. `3000 Å`D. `1500 Å`
Light of wavelength `6328 Å` is incident normally on a slit of width 0.2 mm. Angular width of the central maximum on the screen will be :A. `0.9^(@)`B. `0.18^(@)`C. `0.54^(@)`D. `0.36^(@)`
Light of wavelength `6328 Å` is incident normally on a slit of width 0.2 mm. Angular width of the central maximum on the screen will be :A. `0.36^(@)`B. `0.18^(@)`C. `0.72^(@)`D. `0.09^(@)`
Light wavelength 6000 Å is incident normally on a slit of width `24xx10^(-5)` cm. find out the angular position of seconds minimum from central maximum?A. `12 xx 10^(-5)cm`B. `18xx10^(-5)cm`C. `24xx10^(-5)cm`D. `36xx10^(-5)cm`
Light wavelength 6000 Å is incident normally on a slit of width `24xx10^(-5)` cm. find out the angular position of seconds minimum from central maximum?

Ratio waves originating from sources `S_(1)` and `S_(2)` having zero phase difference and common wavelength `lambda` will show completely destructive interference at a point `P` is `S_(1)P-S_(2)P` isA. `5lambda`B. `3lambda//4`C. `2lambda`D. `11lambda//2`

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment