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

In a Young's double-slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case:

1. there shall be alternate interference patterns of red and blue

2. there shall be an interference pattern for red distinct from that for blue

3. there shall be no interference fringes

4. there shall be an interference pattern for red mixing with one for blue

Two non-coherent sources emit a light beam of intensities I and 4I. The maximum and minimum intensities in the resulting beam are: 

1. 9l and l

2. 9l and 3l

3. 5l and l

4. there is no interference, the intensity is 5I everywhere

In Young's double-slit experiment, the distance of the nth dark fringe from the centre is:

1. $n\left(\frac{\lambda D}{2d}\right)$

2. $n\left(\frac{2d}{\lambda D}\right)$

3. $(2n-1)\frac{\lambda D}{2d}$

4. $(2n-1)\frac{4d}{\lambda D}$

If the path difference between the interfering waves is, $n\lambda$ then the fringes obtained on the screen will be:

1. Dark

2. Bright

3. colored

4. White

The fringe width in a Young's double-slit experiment can be increased. If we decrease:

1. Width of the slits

2. Separation of the slits

3. The wavelength of the light used

4. Distance between slits and screen

In a YDSE, the central bright fringe can be identified.

1. as it has a greater intensity than the other bright fringes

2. as it is wider than the other bright fringes

3. as it is narrower than the other bright fringes

4. by using white light instead of single-wavelength light