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?​

sorry bro don't know the answer

Explanation:

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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

C) The fringe width becomes four times.

Explanation:

Fringe width β = Dλ/d

where D is the distance between slits and screen  

and d is the distance between the slits.

Thus, From qn D’ = 2D and d’ = d/2

When D is doubled and d is reduced to half, then  the fringe width becomes

β’ = λ2D/ (d/2)

= 4λD/d

= 4β

Thus, when the double slit experiment the distance between the slits and the screen is doubled and the separation between the slits is reduced to half, the fringe width becomes four times.

Explanation:

answer in attachment.

Hope it helps duck!!

Explanation:

hard question

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In a young's double slit experiment the distance between the two identical slits is  equal.

The angular position of the first dark fringe is 0.00546

Explanation:

Given that,

Distance between two sources = 0.1 mm

Distance of the screen from the sources = 20 cm

Wave length

The formula of position is given by

For first dark fringe

n = 1

The angular position of the first dark fringe is

Put the value into the formula

Hence, The angular position of the first dark fringe is 0.00546

X DJs

Explanation:

Explanation:

In young's double slit experiment, the distance between the slits is gradually increased. The width of the fringes system:

Options:

(a) Increase

(b) Decrease

(c) Remains the same

(d) Halved

In young's double slit experiment, the distance between the slits is gradually increased. The width of the fringes system increases.