Light from two lasers is incident on an opaque barrier with a single slit of width 4.0 x 10-5 m. One laser emits light of wavelength 4.0 x 10-7 m and the other is 6.0 x 10-7 m. A screen to view the light intensity pattern is 2.0 m behind the barrier. What is the distance from the center of the pattern to the nearest completely dark spot (dark for both colors)?

y = 6 10⁻² m

Explanation:

This is a diffraction exercise which is described by the expression

          a sin θ = m λ

we can use trigonometry to find the distance from the inside of the screen to the dark point (y)

          tan θ = y / L

angles are very small in diffraction experiments, so we can approximate

          tan θ = sin θ /cos θ = sin θ

          sin θ = y / L

substituting

         a (y / L) = m λ

Let's find the points where the intensity becomes zero

         y = m L λ / a

λ₁ = 4.0 10⁻⁷ m

m = 1

          y = 1 2.0 4.0 10⁻⁷/ 4.0 10⁻⁵

          y = 2 10⁻² m

m = 2

          y = 4 10⁻² m

λ₂ = 6.0 10⁻⁷ m

m = 1

          y = 1 2.0 6.0 10⁻⁷ / 4.0 10⁻⁵

          y = 3 10⁻² m

m = 2

          y = 6 10⁻² m

if we want a point where the two colors are dark, we set the two expressions equal

          y₁ = y₂

          m₁ L  λ₁ / a = m₂ L  λ₂ / a

          m₁/m₂ =  λ₂/λ₁

          m₁ / m₂ = 6 10⁻⁷ / 4 10⁻⁷

          m₁ / m₂ = 1.5

since the quantities m must be integers, the smallest relation that the relation fulfills is

            m₁ = 3

            m₂ = 2

the distance for this destructive interference is

           y = 3 2 4.0 10⁻⁷ / 4.0 10⁻⁵

           y = 6 10⁻² m

this is the first point where the minimum of the two wavelengths coincide

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j