All the given pairs can undergo double displacement reactions.

In a double-elimination tournament, the minimum number of matches needed is 2n - 2, where n is the number of teams. For 8 teams, it is 2(8) - 2 = 14.

Fringe separation is directly proportional to the distance from the slits to the screen; increasing this distance increases fringe separation.

Fringe separation (β) = λD/d = (500 x 10^-9 m)(1 m)/(0.1 x 10^-3 m) = 0.5 mm.

Fringe width β = λD/d = (600 x 10^-9 m)(1 m)/(0.2 x 10^-3 m) = 0.003 m = 0.6 mm.

Distance between fringes = β = λD/d. For first and second bright fringes, the distance is 2β.

Using the formula for fringe separation, y = (λD)/d, we find y = (500 x 10^-9 m * 2 m) / (0.5 x 10^-3 m) = 2 cm.

Fringe width (β) is given by β = λD/d, where d is the distance between the slits. If d is doubled, β is halved.

Fringe width (β) is inversely proportional to the distance between the slits (d). If d is doubled, β is halved.

Fringe width (β) is given by β = λD/d, where d is the distance between the slits. If d is doubled, β becomes β/2, hence the fringe width halves.

The fringe separation is inversely proportional to the distance between the slits; thus, if the distance is doubled, the fringe separation halves.

The fringe separation is inversely proportional to the distance between the slits; halving the distance doubles the fringe separation.

Increasing the distance between the slits (d) causes the fringe width (β) to decrease, making the fringes narrower.

Increasing the distance between the slits decreases the fringe width, which can lead to more visible fringes within a given distance on the screen.

Increasing the slit separation d decreases the fringe width, causing the fringes to move closer together.

The distance between the fringes decreases as fringe width β = λD/d; increasing d reduces β.

Increasing the distance to the screen increases the fringe width, as fringe width is proportional to the distance from the slits.

Fringe separation is directly proportional to the distance from the slits to the screen (D), so increasing D increases the fringe separation.

Increasing the distance to the screen increases the fringe width, as fringe width is proportional to the distance from the slits.

At the first minimum, the intensity is 0 due to destructive interference.

Using polychromatic light results in overlapping patterns of different wavelengths, causing the overall pattern to become broader.

Fringe separation is directly proportional to the distance from the slits to the screen (D), hence it increases.

Moving the screen further away increases the fringe spacing, as fringe width is directly proportional to the distance from the slits.

Moving the screen further away increases the distance (D) in the fringe width formula, causing the fringes to become wider.

Moving the screen further away increases the fringe separation, as fringe width β = λD/d; increasing D increases β.

Fringe width β = λD/d = (600 x 10^-9 m)(1 m)/(0.5 x 10^-3 m) = 0.12 mm.

Fringe width (β) is given by β = λD/d. If λ is halved, β will also be halved.

The distance between the fringes increases with an increase in wavelength, as fringe separation is directly proportional to wavelength.

Increasing the distance between the slits decreases the fringe width, making the pattern narrower.

Increasing the distance between the slits increases the fringe width because the angle of diffraction increases.