In the Young's double slit experiment, if the phase difference between the two waves interfering at a point is ϕ, the intensity at that point can be expressed by the expression-

(where A and B depend upon the amplitudes of the two waves)

(1) $I=\sqrt{A^2+B^2{\cos }^2\varphi }$

(2) $I=\frac{A}{B}\cos \varphi$

(3) $I=A+B\cos \frac{\varphi }{2}$

(4) $I=A+B\cos \varphi$

 

When one of the slits of Young’s experiment is covered with a transparent sheet of thickness 4.8 mm, the central fringe shifts to a position originally occupied by the 30^th bright fringe. What should be the thickness of the sheet if the central fringe has to shift to the position occupied by 20^th bright fringe 

(1) 3.8 mm

(2) 1.6 mm

(3) 7.6 mm

(4) 3.2 mm

In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength λ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is 

(1) 2λ

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

(3) $\frac{\lambda }{3}$

(4) λ

In the figure is shown Young’s double-slit experiment, \(Q\) is the position of the first bright fringe on the right side of \(O.\) \(P\) is the \(11\)^th bright fringe on the other side, as measured from \(Q.\) If the wavelength of the light used is \(6000 \times10^{-10}\) m, then \(S_1B\) will be equal to:

   
1. \(6\times10^{-6}\) m
2. \(6.6\times10^{-6}\) m
3. \(3.1\times10^{-6}\) m
4. \(3.1\times10^{-7}\) m

In Young’s double-slit experiment, the two slits act as coherent sources of equal amplitude A and wavelength λ. In another experiment with the same set up, the two slits are of equal amplitude A and wavelength λ but are incoherent. The ratio of the intensity of light at the mid-point of the screen in the first case to that in the second case is:

(1) 1 : 2

(2) 2 : 1

(3) 4 : 1

(4) 1 : 1

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

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) $\lambda (n-l)$

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

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