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.

If a team wins 2 sets to 1, the losing team won 1 set.

In a dynamic equilibrium, the rates of the forward and reverse reactions are equal, but the concentrations of reactants and products may not be equal.

If there are at least 2 boys, the possible combinations are (2B, 2G), (3B, 1G), (4B). Thus, the probability of having at least one girl is 3/4.

From the relationships, E is the youngest as she is the sister of B and C.

D is the mother of C and the wife of A.

A is the oldest as he is the father.

The order of ages is A > B > C > D. Therefore, B is the middle child.

The daughter is the youngest since she is younger than the son.

C is the grandmother of D as B is the daughter of C.

The mother is a doctor as stated in the question.

D is the grandson of F as F is the grandfather of B, and D is B's brother.

D is the granddaughter of F.

Since A is the mother of B and E is the sister of B, D is the cousin of E.