A 550-nm light is incident on a double slit interference arrangements and fringes are viewed on screen. A thin flake of mica (n = 1.58) is used to cover one of slits. As a result the central point on the viewing screen is now at the location of seventh bright fringe (m = 7) when mica was not there. What is the thickness of the mica?
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To find the thickness of the mica, we can use the formula for the fringe shift caused by a thin film. The formula is given by:
Δy = λ * D / (2 * d * (n - 1))
Where:
Δy is the fringe shift,
λ is the wavelength of light,
D is the distance between the double slits and the screen,
d is the thickness of the mica,
n is the refractive index of the mica.
In this case, the fringe shift is given by the location of the central point on the screen when the mica is present (7th bright fringe), minus the location when the mica is not present (0th bright fringe). So, Δy = 7 * λ.
Rearranging the formula, we get:
d = (λ * D) / (2 * (n - 1) * Δy)
Substituting the given values:
λ = 550 nm = 550 x 10^-9 m
D is not given in the question.
n = 1.58
Δy = 7 * λ
We can't calculate the thickness of the mica without knowing the value of D.