monochromatic light of wavelength 500 nm is incident normally on a diffraction grating. if the third order maximum is 32. how many total number of maximuima can be seen

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

talk back to him about the sport and show him your expertise or leave the room