a yellow light of 500 mm emitted by a single source passes through two narrow slits 1 mm apart how far apart are two adjacent bright fringes when interference is observed on a screen 10 m away
Answers
Answer:
I always taught my students to use the fundamental idea and its corresponding formula.
Two bright fringes - this means constructive interference.
Adjacent tells you that the light orders differ by 1, so lets choose something simple- choose central maximum (n=0) and the next maximum (n= 1). (you could choose n=7 and n=8 - it makes no differenence).
We get constructive interference when the wave from one slit travels a whole ( or integer number of) wavelength further. The formula nlambda = dsin (theta) describes this situation.
the n lambda is the integer number of extra wavelengths - we ar choosing n=1.
the d sin(theta) is the extra distance travelled.
We know lambda (500 E-9) and n (1), d is the distance between the slits ( 1 E-3) spo calculate sin(theta) and hence theta- this will be a small angle.
Now calculate the opposite side of the triangle, which is the fringe separation
Tan (theta)= fringe separation/distance to screen
For the small angles involved, you will find that Sin and Tan of theta have the same values. However, it is only a few button presses ona calculator and it is worth understanding what you are calculating rather than just ‘ using a formula’.
We all use formulas but if you understand what the formula is about/doing then you are less likely to use the worng formula/ a formula in inappropriate circumstances.
Explanation:
hope it helps you please mark it as brainliest
Answer:
5mm
Explanation:
Ym = D (mλ/d)
Ym+1 = D [(m+1)λ/d]
→ Two adjacent bright fringe = Ym+1 - Ym = Dλ/d
d = 1mm = 10-3 m , D = 10 m , λ = 500 * 10-9 m
Ym+1 - Ym = Dλ/d
→ Ym+1 - Ym = (10*500*10-9)/0.001 = 0.005 m
= 5 mm