A screen is at a distance of 2 m from narrow slits
that are illuminated with light of 589 nm. The 10th
minimum lies at 0.005 m on either side of the central
maximum, then the distance between the slits will be
Answer this please , with perfect explanation
Answers
Answer:
Answer:
Answer: L=2m,
Answer: L=2m,d=3mm,A=
Answer: L=2m,d=3mm,A= 4
Answer: L=2m,d=3mm,A= 49π
Answer: L=2m,d=3mm,A= 49π
Answer: L=2m,d=3mm,A= 49π ×10
Answer: L=2m,d=3mm,A= 49π ×10 −6
Answer: L=2m,d=3mm,A= 49π ×10 −6 m
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL=
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 4
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm .
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm .
Given:
Distance D
minimum 10th lies at
wavelength of light
To Find:
distance between the slits (d)=?
Solution:
The position of 10th minimum from the center
The above formula can be written as
The distance between the slits is .