Answer:
The total number of maxima that can be seen is 11
Explanation:
Given the data in the question
wavelength λ = 500 nm = 5 × 10⁻⁷ m
if the third order maximum is 32
i.e m = 3 and θ = 32°
Now, we know that condition for diffraction maximum is as follows;
d × sinθ = m × λ
so we substitute in our given values
d × sin( 32° ) = 3 × 5 × 10⁻⁷ m
d × sin( 32° ) = 1.5 × 10⁻⁶ m
d = [ 1.5 × 10⁻⁶ m ] / sin( 32° )
d = 2.83 × 10⁻⁶ m
Now, maxima n when θ = 90° will be;
sin( 90° ) = nλ / d
1 = nλ / d
d = nλ
n = d / λ
we substitute
n = [ 2.83 × 10⁻⁶ m ] / [ 5 × 10⁻⁷ m ]
n = 5.66
so 5 is the max value
hence, total maxima value is;
⇒ 2n + 1 = 2( 5 ) + 1 = 10 + 1 = 11
Therefore, total number of maxima that can be seen is 11
