- 1)
If double slit apparatus is immersed in a liquid of refractive index, I-l the wavelength of light reduces to \(\lambda\) and fringe width also reduces to \(\beta^{\prime}=\frac{\beta}{\mu} \text { . }\)
The given figure shows a double-slit experiment in which coherent monochromatic light of wavelength A from a distant source is incident upon the two slits, each of width w(w >> \(\lambda\)) and the interference pattern is viewed on a distant screen. A thin piece of glass of thickness t and refractive index n is placed between one of the slit and the screen, perpendicular to the light path.

(i) In Young's double slit interference pattern, the fringe width
(a) can be changed only by changing the wavelength of incident light |
(b) can be changed only by changing the separation between the two slits |
(c) can be changed either by changing the wavelength or by changing the separation between two sources |
(d) is a universal constant and hence cannot be changed |
(ii) If the width w of one of the slits is increased to 2w, the become the amplitude due to slit
(a) 1.5a |
(b) a/2 |
(c) 2a |
(d) no change |
(iii) In YDSE, let A and B be two slits. Films of thicknesses tA and tB and refractive indices mA and mB are placed in front of A and B, respectively. If \(\mu_{\mathrm{A}} t_{A}=\mu_{B} t_{B}\) then the central maxima will
(a) not shift |
(b) shift towards A |
(c) shift towards B |
(d) shift towards A if tB = tA and shift towards B if tB < tA |
(iv) In Young's double slit experiment, a third slit is made in between the double slits. Then
(a) fringes of unequal width are formed. |
(b) contrast between bright and dark fringes is reduced |
(c) intensity of fringes totally disappears |
(d) only bright light is observed on the screen |
(v) In Young's double slit experiment, if one of the slits is covered with a microscope cover slip, then
(a) fringe pattern disappears |
(b) the screen just gets illuminated |
(c) in the fringe pattern, the brightness of the bright fringes will decreases and the dark fringes will become more dark |
(d) bright fringes will be more bright and dark fringes will become more dark. |
- 2)
Distance between two successive bright or dark fringes is called fringe width.
\(\beta=Y_{n+1}-Y_{n}=\frac{(n+1) \lambda D}{d}-\frac{n \lambda D}{d}=\frac{\lambda D}{d}\)
Fringe width is independent of the order of the maxima. If whole apparatus is immersed in liquid of refractive index \(\mu\) then \(\beta=\frac{\lambda D}{\mu d}\) (fringe width decreases). Angular fringe width (\(\theta\)) is the angular separation between two consecutive maxima or minima \(\theta=\frac{\beta}{D}=\frac{\lambda}{d}\)
In the arrangement shown in figure, slit S3 and S4 are having a variable separation Z. Point 0 on the screen is at the common perpendicular bisector of S1S2 and S3S4.

(i) The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young's double-slit experiment, is
(a) infinite |
(b) five |
(c) three |
(d) zero |
(ii) In Young's double - slit experiment if yellow light is replaced by blue light, the interference fringes become
(a) wider |
(b) brighter |
(c) narrower |
(d) darker |
(iii) In Young's double slit experiment, if the separation between the slits is halved and the distance between the slits and the screen is doubled, then the fringe width compared to the unchanged one will be
(a) Unchanged |
(b) Halved |
(c) Doubled |
(d) Quadrupled |
(iv) When the complete Young's double slit experiment is immersed in water, the fringes
(a) remain unaltered |
(b) become wider |
(c) become narrower |
(d) disappear |
(v) In a two slit experiment with white light, a white fringe is observed on a screen kept behind the slits. When the screen is moved away by 0.05 m, this white fringe
(a) does not move at all |
(b) gets displaced from its earlier position |
(c) becomes coloured |
(d) disappears |
- 3)
When light from a monochromatic source is incident on a single narrow slit, it gets diffracted and a pattern of alternate bright and dark fringes is obtained on screen, called "Diffraction Pattern" of single slit. In diffraction pattern of single slit, it is found that
(I) Central bright fringe is ·of maximum intensity and the intensity of any secondary bright fringe decreases with increase in its order.
(II) Central bright fringe is twice as wide as any other secondary bright or dark fringe .

(i) A single slit of width 0.1 mm is illuminated by a parallel beam of light of wavelength 6000 A and diffraction bands are observed on a screen 0.5 m from the slit. The distance of the third dark band from the central bright band is
(a) 3 mm |
(b) 1.5 mm |
(c) 9 mm |
(d) 4.5 mm |
(ii) In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 2 m away from the lens. If wavelength of light used is 5000 \(\lambda\) then the distance between the first minimum on either side the central maximum is
(a) 10-1 m |
(b) 10-2 m |
(c) 2 x 10-2 m |
(d) 2 x 10-1 m |
(iii) Light of wavelength 600 nm is incident normally on a slit of width 0.2 mm. The angular width of central maxima in the diffraction pattern is (measured from minimum to minimum)
(a) 6 x 10-3 rad |
(b) 4 x 10-3 rad |
(c) 2.4 x 10-3 rad |
(d) 4.5 x 10-3 rad |
(iv) A diffraction pattern is obtained by using a beam of red light. What will happen, if the red light is replaced by the blue light?
(a) bands disappear |
(b) bands become broader and farther apart |
(c) no change will take place |
(d) diffraction bands become narrower and crowded together. |
(v) To observe diffraction, the size of the obstacle
(a) should be A/2, where A is the wavelength. |
(b) should be of the order of wavelength. |
(c) has no relation to wavelength. |
(d) should be much larger than the wavelength. |
- 4)
In Young's double slit experiment, the width of the central bright fringe is equal to the distance between the first dark fringes on the two sides of the central bright fringe.
In given figure below a screen is placed normal to the line joining the two point coherent source SI and S2' The interference pattern consists of concentric circles.

(i) The optical path difference at P is
\((a) d\left[1+\frac{y^{2}}{2 D}\right]\) |
\((b) d\left[1+\frac{2 D}{y^{2}}\right]\) |
\((c) d\left[1-\frac{y^{2}}{2 D^{2}}\right]\) |
\((d) d\left[2 D-\frac{1}{y^{2}}\right]\) |
(ii) Find the radius of the nth bright fringe.
\((a) D \sqrt{1\left(1-\frac{n \lambda}{d}\right)}\) |
\((b) D \sqrt{2\left(1-\frac{n \lambda}{d}\right)}\) |
\((c) 2 D \sqrt{2\left(1-\frac{n \lambda}{d}\right)}\) |
\((d) D \sqrt{2\left(1-\frac{n \lambda}{2 d}\right)}\) |
(iii) If d = 0.5 mm, \(\lambda\)= 5000 \(\dot A \)and D = 100 em, find the value of n for the closest second bright fringe
(a) 888 |
(b) 830 |
(c) 914 |
(d) 998 |
(iv) The coherence of two light sources means that the light waves emitted have
(a) same frequency |
(b) same intensity |
(c) constant phase difference |
(d) same velocity. |
(v) The phenomenon of interference is shown by
(a) longitudinal mechanical waves only |
(b) transverse mechanical waves only |
(c) electromagnetic waves only |
(d) all of these |
- 5)
A narrow tube is bent in the form of a circle of radius R, as shown in figure. Two small holes S and D are made in the tube at the positions at right angle to each other. A source placed at S generates a wave of intensity Io which is equally divided into two parts: one part travels along the longer path, while the other travels along the shorter path. Both the waves meet at point D where a detector is placed.

(i) If a maxima is formed at a detector, then the magnitude of wavelength \(\lambda\) of the wave produced is given by
\((a) \pi R\) |
\((b) \frac{\pi R}{2}\) |
\((c) \frac{\pi R}{4}\) |
(d) all of these |
(ii) If the intensity ratio of two coherent sources used in Young's double slit experiment is 49 : 1, then the ratio between the maximum and minimum intensities in the interference pattern is
(a) 1: 9 |
(b) 9: 16 |
(c) 25: 16 |
(d) 16: 9 |
(iii) The maximum intensity produced at D is given by
(a) 4I0 |
(b) 2I0 |
(c) I0 |
(d) 3I0 |
(iv) In a Young's double slit experiment, the intensity at a point where the path difference is \(\lambda\)/6 (\(\lambda\) - wavelength of the light) is I. If I0 denotes the maximum intensity, then I/I0 is equal to
\((a) \frac{1}{2}\) |
\((b) \frac{\sqrt{3}}{2}\) |
\((c) \frac{1}{\sqrt{2}}\) |
\((d) \frac{3}{4}\) |
(v) Two identical light waves, propagating in the same direction, have a phase difference d. After they superpose the intensity of the resulting wave will be proportional to
\((a) \cos \delta\) |
\((b) \cos (\delta / 2)\) |
\((c) \cos ^{2}(\delta / 2)\) |
\((d) \cos ^{2} \delta\) |