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 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) $\sqrt{3\lambda D}$

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

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

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