What is the smallest separation between two slits that will produce a second-order minimum for 762 nm red light?
1 answer:
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Actually Welcome to the Concept of the Projectile Motion.
Since, here given that, vertical velocity= 50m/s
we know that u*sin(theta) = vertical velocity
so the time taken to reach the maximum height or the time of Ascent is equal to
T = Usin(theta) ÷ g, here g = 9.8 m/s^2
so we get as,
T = 50/9.8
T = 5.10 seconds
thus the time taken to reach max height is 5.10 seconds.
Answer:
λ = 5.85 x 10⁻⁷ m = 585 nm
f = 5.13 x 10¹⁴ Hz
Explanation:
We will use Young's Double Slit Experiment's Formula here:
where,
λ = wavelength = ?
Y = Fringe Spacing = 6.5 cm = 0.065 m
d = slit separation = 0.048 mm = 4.8 x 10⁻⁵ m
L = screen distance = 5 m
Therefore,
<u>λ = 5.85 x 10⁻⁷ m = 585 nm</u>
Now, the frequency can be given as:
where,
f = frequency = ?
c = speed of light = 3 x 10⁸ m/s
Therefore,
<u>f = 5.13 x 10¹⁴ Hz</u>