#7 | Solved Problems on YDSE
(Physics) > Wave Optics
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The intensity at the maximum in Young's double-slit experiment is ${\mathrm{I}}_{0}$ when the distance between two slits is d=5$\lambda$, where $\lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D= 10 d?

(1) $\frac{{\mathrm{I}}_{0}}{4}$

(2) $\frac{3}{4}{\mathrm{I}}_{0}$

(3) $\frac{{\mathrm{I}}_{0}}{2}$

(4) ${\mathrm{I}}_{0}$

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Two slits in Young's experiment have widths in the ratio 1:25. The ratio of intensity at the maxima and minima in the interference pattern Imax/Imin is

1. 9/4

2. 121/49

3. 49/121

4. 4/9

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In the Young's double-slit experiment, the intensity of light at a point on the screen (where the path difference is λ ) is K where λ being the wavelength of light used. The intensity at a point where the path difference is λ /4 will be

(1) K

(2) K/4

(3) K/2

(4) zero

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In Young’s double slit experiment. the slits are 2 mm apart and are illuminated by photons of two wavelengths , λ1= 12000Å and , λ2= 10000Å. At what minimum distance from the common central bright fringe on the screen 2m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?

(a) 8mm

(b) 6mm

(c) 4 mm

(d) 3mm

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In a double slit experiment, the distance between the slits is 3 mm and the slits are 2 m away from the screen one due to light with wavelength 480 nm, and the other due to light with wavelength 600 nm. What is the separation on the screen between the fifth order bright fringes of the two interference patterns ?

1. $4×{10}^{-4}$

2. $9×{10}^{-4}$

3. $3×{10}^{-4}$

4. $5×{10}^{-4}$

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