36. In a Young's double slit experiment the distance between the slits is 1.2 mm
and the screen is at 0.75 m from the slits. If the distance of the 5th bright fringe
from the central fringe on the screen is 1.8 mm. Calculate the wavelength of
the light used. What will be the distance of the 5th dark fringe from the centre of
the screen?
Answers
Answer:
ANSWER
d=0.28mm=0.28\times { 10 }^{ -3 }md=0.28mm=0.28×10
−3
m
D=1.4mD=1.4m
{ x }_{ n }=1.35cm=1.35\times { 10 }^{ -2 }mx
n
=1.35cm=1.35×10
−2
m
n=5n=5
For n^{th}n
th
dark fringe,
{ x }_{ n }=\cfrac { (2n-1) }{ 2 } \cfrac { \lambda D }{ d }x
n
=
2
(2n−1)
d
λD
1.35\times { 10 }^{ -2 }=\cfrac { \left[ \left( 2\times 5 \right) -1 \right] }{ 2 } \times \cfrac { \lambda \times 1.4 }{ 0.28\times { 10 }^{ -3 } }1.35×10
−2
=
2
[(2×5)−1]
×
0.28×10
−3
λ×1.4
\lambda =\cfrac { 1.35\times { 10 }^{ -2 }\times 0.28\times { 10 }^{ -3 }\times 2 }{ 1.4\times \left[ \left( 2\times 5 \right) -1 \right] } =0.06\times { 10 }^{ -5 }=6000Aλ=
1.4×[(2×5)−1]
1.35×10
−2
×0.28×10
−3
×2
=0.06×10
−5
=6000A
When D=1.4-0.4m=1mD=1.4−0.4m=1m
\beta =\cfrac { \lambda D }{ d } =\cfrac { 6000\times { 10 }^{ -10 }\times 1 }{ 0.28\times { 10 }^{ -3 } } =21428.5714\times { 10 }^{ -7 }=0.21cm=2.1429mmβ=
d
λD
=
0.28×10
−3
6000×10
−10
×1
=21428.5714×10
−7
=0.21cm=2.1429mm