The relation between the fringe width for the red light and yellow light is: (all other things being the same.)

$1. {\beta }_{red}<{\beta }_{yellow}$
$2. {\beta }_{red}>{\beta }_{yellow}$
$3. {\beta }_{red}={\beta }_{yellow}$
$4. {\beta }_{red}=2{\beta }_{yellow}$

Fringe width in a particular YDSE is measured to be $\beta$. What will be the fringe width if the wavelength of the light is doubled, the separation between the slits is halved and the separation between the screen and slits is tripled?

1. 10 times

2. 11 times

3. Same

4. 12 times

Two coherent sources are 0.3 mm apart. They are 1 m away from the screen. The second dark fringe is at a distance of 0.3 cm from the center. The distance of the fourth bright fringe from the centre is:

1.  0.6 cm

2.  0.8 cm

3.  1.2 cm

4.  0.12 cm

In Young's double-slit experiment the light emitted from the source has $\lambda$ = 6.5 × 10^–7 m and the distance between the two slits is 1 mm. The distance between the screen and slits is 1 metre. Distance between third dark and fifth bright fringe will be:

1. 3.2 mm 

2. 1.63 mm 

3. 0.585 mm 

4. 2.31 mm

In Young's double-slit experiment, the separation d between the slits is 2 mm, the wavelength $\lambda$ of the light used is 5896 Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular width to 0.21° (with same $\lambda$ and D) the separation between the slits needs to be changed to:

1. 1.8 mm

2. 1.9 mm

3. 2.1 mm

4. 1.7 mm

In two separate set-ups of the Young's double slit experiment, fringes of equal width are observed when lights of wavelengths in the ratio 1 : 2 are used. If the ratio of the slit separation in the two cases is 2 : 1, the ratio of the distances between the plane of the slits and the screen in the two set-ups is: 

1. 4 : 1

2. 1 : 1

3. 1 : 4

4. 2 : 1

In Young's double-slit experiment, the slit separation is doubled. This results in: