Red and blue light are simultaneously passed through a diffraction grating. the first-order maxima for the red light is located 20 cm from middle of the central bright fringe.

where is the first-order maxima for the blue light located? a) also at 20 cm from the middle of the central bright fringe. b) greater than 20 cm from the middle of the central bright fringe. c) less than 20 cm from the middle of the central bright fringe.

c) True.  Less than 20 cm from the middle of the central bright fringe.

Explanation:

The diffraction pattern of a grid is given by the expression

     d sin θ = m λ

Where d is the separation between the slits, λ the wavelength and m the order of spectrum diffraction

Let's apply this equation to our case

Red light.

Its wavelength is in the 700 nm range

    λ₁ = 700 10⁻⁹ m

    sin θ₁ = (m / d) lam1

For the first diffraction order m = 1

    sin θ₁ = (1 / d) 7 10⁻⁷

Blue light

Length wave range 450 nm

    Lam2 = 450 10⁻⁹ m

    sin θ₂ = (1 / d) lam₂

    sin θ₂ = (1 / d) 4.5 10⁻⁷

Let's divide these two relationships

    sin θ₂ / sin θ₁ = 4.5 10⁻⁷ / 7 10⁻⁷

     sin θ₂ / sin θ₁ = 4.5 / 7

     sin θ₂ = 0.64 sin θ₁

In diffraction measurements the distance to the screen in general is much greater than the separation of the spectral lines, so the hypotenuse will be considered constant

     sin θ = Y / h

     sin θ₂ = 0.64 sin θ₁

     y₂ / h = 0.64 y₁ / h

     y₂ = 0.64 y₁

     y₂ = 0.64 20

     y₂ = 12.8 cm

let's review the claims

a) False

b) False

c) True

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

a) the event horizon, singularity, and the chute located between the two.