Following figures shows sources `S_1` and `S_2` that emits light of wavelength `lambda` in all directions. The sources are exactly in phase and are separated by a distance equal to `1.5lambda`. If we start at the indicated start point and travel along path 1 and 2, the interference produce a maxima all along A. (a) Path 1B. (b) Path 2C. (c) Any pathD. (d) None of these

Following figures shows sources `S_1` and `S_2` that emits light of wa

1.	Following figures shows sources `S_1` and `S_2` that emits light of wavelength `lambda` in all directions. The sources are exactly in phase and are separated by a distance equal to `1.5lambda`. If we start at the indicated start point and travel along path 1 and 2, the interference produce a maxima all along A. (a) Path 1B. (b) Path 2C. (c) Any pathD. (d) None of these
Answer» Correct Answer - A At any point along the path 1, path difference between the wave is 0. Hence maxima is obtained all along the path 1. At any point along the path 2, path difference is `1.5lambda` which is odd multiple of `lambda/2`, so minima is obtained all along the path 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?

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment