A beam of electron is used in a YDSE experiment. The slit width is \(d\). When the velocity of the electron is increased, then,

1.	No interference is observed
2.	Fringe width increases
3.	Fringe width decreases
4.	Fringe width remains the same

If the separation between screen and source is increased by 2% what would be the effect on the intensity?

(1) Increases by 4%

(2) Increases by 2%

(3) Decreases by 2%

(4) Decreases by 4%

In Young's double-slit experiment, the intensity at a point is (1/4) of the maximum intensity. The angular position of this point is:

(1) sin^-1(λ/d)

(2) sin^-1(λ/2d)

(3) sin^-1(λ/3d)

(4) sin^-1(λ/4d)

In a YDSE bi-chromatic light of wavelengths, 400 nm and 560 nm are used. The distance between the slits is 0.1 mm and the distance between the plane of the slits and the screen is 1 m. The minimum distance between two successive regions of complete darkness is: 

(1) 4 mm

(2) 5.6 mm

(3) 14 mm

(4) 28 mm

In a single slit diffraction of light of wavelength λ by a slit of width e, the size of the central maximum on a screen at a distance b is

(1) $2b\lambda +e$

(2) $\frac{2b\lambda }{e}$

(3) $\frac{2b\lambda }{e}+e$

(4) $\frac{2b\lambda }{e}-e$

Two coherent sources separated by distance \(d\) are radiating in a phase having wavelength \(\lambda.\) A detector moves in a big circle around the two sources in the plane of the two sources. The angular position of \(n=4\) interference maxima is given as:

            
1. \(\text{sin}^{-1}\left(\frac{n\lambda}{d}\right )\)
2. \(\text{cos}^{-1}\left(\frac{4\lambda}{d}\right)\)
3. \(\text{tan}^{-1}\left(\frac{d}{4\lambda}\right)\)
4. \(\text{cos}^{-1}\left(\frac{\lambda}{4d}\right)\)

A beam with wavelength λ falls on a stack of partially reflecting planes with separation d. The angle θ that the beam should make with the planes so that the beams reflected from successive planes may interfere constructively is (where n =1, 2, ……)

(1) ${\sin }^{-1}\left(\frac{{\displaystyle n\lambda }}{{\displaystyle d}}\right)$

(2) ${\tan }^{-1}\left(\frac{{\displaystyle n\lambda }}{{\displaystyle d}}\right)$

(3) ${\sin }^{-1}\left(\frac{{\displaystyle n\lambda }}{{\displaystyle 2d}}\right)$

(4) ${\cos }^{-1}\left(\frac{{\displaystyle n\lambda }}{{\displaystyle 2d}}\right)$

Two point sources X and Y emit waves of same frequency and speed but Y lags in phase behind X by 2πl radian. If there is a maximum in direction D the distance XO using n as an integer is given by

(1) $\frac{\lambda }{2}(n-l)$

(2) $\lambda (n+l)$

(3) $\frac{\lambda }{2}(n+l)$

(4) $\lambda (n-l)$

Two ideal slits S₁ and S₂ are at a distance d apart and illuminated by the light of wavelength λ passing through an ideal source slit S placed on the line through S₂ as shown. The distance between the planes of slits and the source slit is D. A screen is held at a distance D from the plane of the slits. The minimum value of d for which there is darkness at O is:

(1) $\sqrt{\frac{3\lambda D}{2}}$

(2) $\sqrt{\lambda D}$

(3) $\sqrt{\frac{\lambda D}{2}}$

(4) $\sqrt{3\lambda D}$

A monochromatic beam of light falls on the YDSE apparatus at some angle (say θ) as shown in the figure. A thin sheet of glass is inserted in front of the lower slit S₂. The central bright fringe (path difference = 0) will be obtained:

(1) At O

(2) Above O

(3) Below O

(4) Anywhere depending on angle θ, the thickness of plate t and refractive index of glass μ