A physics student uses a spring loaded launcher to fire a 58 gram projectile horizontally. The projectile leaves the launcher at
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Answer:
43.16°
Explanation:
λ = Wavelength = 1.4×10⁻¹⁰ m
θ₁ = 20°
n can be any integer
d = distance between the two slits
Since for the first bright fringe, n₁ = 1
n₂ = 2 for second order line
The relation between the distance of the slits and the angle through which it is passed is:
dsinθ=nλ
As d and λ are constant
∴ Angle by which the second order line appear is 43.16°