In a single-slit diffraction pattern on a flat screen, the central bright fringe in 1.0 cm wide when the slit width is 4.70 10-5 m. When the slit is replaced by a second slit, the wavelength of the light and the distance to the screen remaining unchanged, the central bright fringe broadens to a width of 2.4 cm. What is the width of the second slit? It may be assumed that è is so small that sin è = tan è
To solve this problem, we can use the concept of diffraction and apply the formula for the angular width of the central bright fringe in a single-slit diffraction pattern.
First, let's calculate the angular width, θ, for the first slit using the formula:
θ = λ / (a * sin θ)
Where:
θ is the angular width of the central bright fringe
λ is the wavelength of light
a is the width of the slit
Given:
θ = 1.0 cm (converted to meters, so θ = 0.01 m)
λ is constant throughout the problem
a = 4.70 * 10^(-5) m
Rearrange the formula to solve for λ:
λ = θ * a / sin θ
Substitute the given values:
λ = (0.01 m) * (4.70 * 10^(-5) m) / sin θ
Now, we know that λ is the same for both slits (since the wavelength of light remains unchanged). Therefore, we can use this value to calculate the angular width, θ, for the second slit.
Using the same formula:
θ = λ / (a_2 * sin θ)
Where:
θ is the angular width of the central bright fringe for the second slit
λ is the previously calculated wavelength
a_2 is the width of the second slit (what we need to find)
Given:
θ = 2.4 cm (converted to meters, so θ = 0.024 m)
Rearrange the formula to solve for a_2:
a_2 = λ / (θ * sin θ)
Substitute the known values:
a_2 = λ / (0.024 m * sin θ)
Now, substitute the previously calculated value for λ:
a_2 = [(0.01 m) * (4.70 * 10^(-5) m) / sin θ] / (0.024 m * sin θ)
Simplifying the expression:
a_2 = (0.01 m * 4.70 * 10^(-5) m) / 0.024 m
a_2 ≈ 1.96 * 10^(-6) m
Therefore, the width of the second slit is approximately 1.96 * 10^(-6) meters.