Maxima & Minima in YDSE
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Q. Young's double-slit experiment is carried with two thin sheets of thickness 10.4 μm each and refractive index μ1=1.52 and μ2=1.40 covering the slits S1 and S2 respectively. If white light of range 400 nm to 780 nm is used, then which wavelength will form maxima at point O, the center of the screen?
- 416 nm only
- 624 nm only
- 416 nm and 624 nm only
- None of these
Q.
In Young's double slit experiment, white light is used. The separation between the slits is b. The screen is at a distance d (d>> b) from the slits. Some wavelengths are missing exactly in front of one slit. These wavelengths are
[IIT 1984; AIIMS 1995]
λ=b2d
λ=2b2d
λ=b23d
λ=2b23d
Q. Two slits separated by a distance of 1 mm are illuminated with red light of wavelength 6.5×10−7 m. The interference fringes are observed on a screen placed 1 m from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to
- 0.65 mm
- 1.625 mm
- 3.25 mm
- 0.975 mm
Q. Consider two coherent monochromatic sources S1 and S2, each of wavelength λ and separated by a distance d. The ratio of the intensity of S1 and S2 is 4. A detector moves on the line perpendicular to S1S2 as shown in the figure. If the resultant intensity at point P is equal to 94 times intensity of S1, then the distance of P from S1 is,
(Given : d>0 and n is a positive integer)
(Given : d>0 and n is a positive integer)
- d2−n2λ22nλ
- d2+n2λ22nλ
- nλd√d2−n2λ2
- 2nλd√d2−n2λ2
Q. In YDSE apparatus shown in figure wavelength of light used is λ. The screen is moved away from the source with a constant speed v. Initial distance between screen and plane of slits was D and at point P on screen, nth order bright fringe is observed.
As the screen moves away value of n will
As the screen moves away value of n will
- increase
- decrease
- remain constant
- first increase then decrease
Q. Consider the arrangement shown in the figure (17-E7). By some mechanism, the separation between the slits S3 and S4 can be changed. The intensity is measured at the point P, which is at the common perpendicular bisector of S1S2 and S2S4. When , the intensity measured at P is I. Find the intensity when z is equal to
Figure
Figure