The main difference between the phenomena of interference and diffraction is that:

1.	diffraction is caused by reflected waves from a source whereas interference is caused due to the refraction of waves from a source.
2.	diffraction is caused due to the interaction of waves derived from the same source, whereas interference is the bending of light from the same wavefront.
3.	diffraction is caused due to the interaction of light from the same wavefront, whereas the interference is the interaction of two waves derived from the same source.
4.	diffraction is caused due to the interaction of light from the same wavefront whereas interference is the interaction of waves from two isolated sources.

In Young's double-slit experiment, the phase difference  between light waves reaching 3rd bright fringe from the central fringe when $\lambda =5000 \stackrel{0}{A}$, is:

1. $6\pi$

2. $2\pi$

3. $4\pi$

4. zero

In an interference pattern produced by two identical slits, the intensity at the position of the central maximum is \(I.\) The intensity at the same spot when either of the slits is closed is \(I_0.\) Therefore,
1. \(I=I_0\)
2. \(I=2I_0\)
3. \(I=4I_0\)
4. \(I\) and \(I_0\) are not related to each other

The separation between successive fringes in a double-slit arrangement is x. If the whole arrangement is dipped underwater, what will be the new fringe separation? [The wavelength of light being used is 5000 $\stackrel{0}{A}$]

1. 1.5 x

2. x

3. 0.75 x

4. 2 x

If the intensities of the two interfering beams in Young's double-slit experiment be ${\mathrm{I}}_1 \mathrm{and} {\mathrm{I}}_2$, then the contrast between the maximum and minimum intensity is good when

1. ${\mathrm{l}}_1$ is much greater than ${\mathrm{I}}_2$

2. ${\mathrm{l}}_1$ is much smaller than ${\mathrm{I}}_2$

3. ${\mathrm{I}}_1={\mathrm{I}}_2$

4. Either ${\mathrm{I}}_1=0 \mathrm{or} {\mathrm{I}}_2=0$

The fringe width at a distance of 50 cm from the slits in Young's experiment for light of wavelength 6000 $\stackrel{0}{A}$ is 0.048 cm. The fringe width at the same distance for $\lambda =5000 \stackrel{0}{A}$, will be:

1. 0.04 cm

2. 0.4 cm

3. 0.14 cm

4. 0.45 cm

In an experiment, the two slits are 0.5 mm apart and the fringes are observed from a distance of 100 cm from the plane of the slits. The distance of the 11th bright fringe from the 1st bright fringe is 9.72 mm. The wavelength is: 

1. $4.86\times {10}^{-5} cm$

2. $5.72\times {10}^{-4} cm$

3. $5.87\times {10}^{-4} cm$

4. $3.25\times {10}^{-4} cm$

Two light rays having the same wavelength $\lambda$ in vacuum are in phase initially. Then the first ray travels a path $L_1$ through a medium of refractive index $n_1$, while the second ray travels a path of length $L_2$ through a medium of refractive index $n_2$. The two waves are then combined to observe interference. The phase difference between the two waves is:

1. $\frac{2\mathrm{\pi }}{\lambda }(L_1-L_1)$

2. $\frac{2\mathrm{\pi }}{\lambda }(n_1{L}_1-n_2{L}_2)$

3. $\frac{2\mathrm{\pi }}{\lambda }(n_2{L}_1-n_1{L}_2)$

4. $\frac{2\mathrm{\pi }}{\lambda }\left(\frac{L_1}{n_1}-\frac{L_2}{n_2}\right)$

The equations of two interfering waves are $y_1=b \cos \omega t and y_2=b \cos (\omega t+\varphi )$. For destructive interference, the phase difference $\varphi$ is :

1. $0°$

2. $360°$

3. $180°$

4. $720°$

When two coherent monochromatic light beams of intestines I and 4I are superimposed, what are the maximum and minimum possible intensities in the resulting beams?

1. 5l and l

2. 5l and 3l

3. 9l and l

4. 9l and 3l