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kykrilka [37]
3 years ago
12

Parallel light rays with a wavelength of 563 nm fall on a single slit. On a screen 3.30 m away, the distance between the first d

ark fringes on either side of the central maximum is 4.70 mm . Part A What is the width of the slit
Physics
1 answer:
MrMuchimi3 years ago
3 0

Answer:

The width of the slit is 0.4 mm (0.00040 m).

Explanation:

From the Young's interference expression, we have;

(λ ÷ d) = (Δy ÷ D)

where λ is the wavelength of the light, D is the distance of the slit to the screen, d is the width of slit and Δy is the fringe separation.

Thus,

d = (Dλ) ÷ Δy

D = 3.30 m, Δy = 4.7 mm (0.0047 m) and λ = 563 nm (563 ×10^{-9} m)

d = (3.30 × 563 ×10^{-9} ) ÷ (0.0047)

  = 1.8579 × 10^{-6} ÷ 0.0047

  = 0.0003951 m

d = 0.00040 m

The width of the slit is 0.4 mm (0.00040 m).

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umka21 [38]

Answer:

(d) Negative.

Explanation:

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2 years ago
Accelerates uniformly at 2.0 ms2 for 10.0s. Calculate its final velocity​
Crazy boy [7]

Answer:

The distance is

=

7

m

Explanation:

Apply the equation of motion

s

(

t

)

=

u

t

+

1

2

a

t

2

The initial velocity is

u

=

0

m

s

−

1

The acceleration is

a

=

2

m

s

−

2

Therefore, when

t

=

3

s

, we get

s

(

3

)

=

0

+

1

2

⋅

2

⋅

3

2

=

9

m

and when

t

=

4

s

s

(

4

)

=

0

+

1

2

⋅

2

⋅

4

2

=

16

m

Therefore,

The distance travelled in the fourth second is

d

=

s

(

4

)

−

s

(

3

)

=

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−

9

=

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m

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VMariaS [17]

Answer:

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