Physics, asked by niharika1411, 9 months ago

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

Answered by anubhabkumar2020
2

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 .

Answered by mad210201
0

Given:

Distance D=2\ m

minimum 10th lies at

Y_{10}=0.005 \ m

wavelength of light=589\  nm=589\times10^{-9} \ m

To Find:

distance between the slits (d)=?

Solution:

The position of 10th minimum from the center

Y_{10}= \dfrac{(9+\dfrac{1}{2})\lambda D}{d}

The above formula can be written as

d=\dfrac{19\lambda D}{2Y_{10} }\\\\d= \dfrac{19\times589\times10^{-9}\times2}{2\times0.005}\\=2.23\times10^{-3} \ m\\=2.23\  mm.

The distance between the slits is 2.23 \ mm.

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