The 2nd maximum of a double slit diffraction pattern makes an angle 20 degrees when a wavelength of 600 nm is used. What is the angle for the 5th maximum when the slit separation d is increased to 1.5d keeping the wavelength the same. g

θ = 34.77°

Explanation:

From diffraction equation:

where,

m = order of diffraction

λ = wavelength of light used

d = slit separation

θ = angle

Therefore, for initial case:

m = 2

λ = 600 nm = 6 x 10⁻⁷ m

d = slit seperation = ?

θ = angle 20°

Therefore,

Now, for the second case:

m = 5

λ = 600 nm = 6 x 10⁻⁷ m

d = slit seperation = (1.5)(3.5 x 10⁻⁶ m) = 5.26 x 10⁻⁶ m

θ = angle = ?

Therefore,

θ = 34.77°

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg