a young's double slit arrangement produces interferences fringes which are 0.004 rad apapt for wave lenght 589nm. For what wavelenght the angular seperation would be 10% greater?
To find the wavelength for which the angular separation would be 10% greater, we need to start by understanding the Young's double-slit experiment and how it produces interference fringes.
In a Young's double-slit arrangement, a coherent light source (usually a laser) is directed at a barrier with two small slits. When the light passes through these slits, it diffracts and creates waves that overlap and interfere with each other.
The interference pattern is observed on a screen placed behind the slits and consists of alternating bright and dark fringes. The separation between adjacent fringes, called the angular separation or angular fringe width, can be calculated using the formula:
θ = λ / d
where θ is the angular separation, λ is the wavelength of light, and d is the distance between the slits.
We are given that the angular separation is 0.004 rad for a wavelength of 589 nm (or 589 x 10^-9 meters). Let's call this initial wavelength λ1 and the corresponding angular separation θ1.
θ1 = 0.004 rad
λ1 = 589 x 10^-9 meters
To find the wavelength for which the angular separation would be 10% greater, we need to calculate the new angular separation θ2.
θ2 = 1.10 θ1
θ2 = 1.10 * 0.004 rad
θ2 = 0.0044 rad
Now we can rearrange the formula θ = λ / d to solve for the wavelength λ2, which corresponds to θ2.
λ2 = θ2 * d
λ2 = 0.0044 rad * d
To calculate the new wavelength, we also need to determine the value of d. Unfortunately, the distance between the slits is not provided in the question. Without this information, it is not possible to determine the exact wavelength for the 10% greater angular separation.
Therefore, to find the new wavelength, we would need to know the distance between the slits (d). With this value, we could substitute it into the equation to find the corresponding wavelength (λ2) for the 10% greater angular separation.