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Korvikt [17]
3 years ago
14

A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the

width of the central bright spot on the screen change if the slit width is doubled? A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if the slit width is doubled? It will be cut in half. It will double. It will become eight times as large. It will be cut to one-quarter its original size. It will become four times as large.
Physics
1 answer:
Novay_Z [31]3 years ago
4 0

Answer:

It will be cut in half

Explanation:

The diffraction of a slit is given by the formula

a sin θ = m where

a = width of the slit,

λ = wavelength and

m = integer that determines the order of diffraction.

Next we divide both sides by a, we have

sin θ = m λ / a

Also, recall that

a’ = 2 a

Then we substitute in the previous equation

2asin θ' = m λ, if divide by 2a, we have

sin θ' = (m λ / 2a).

Now again, from the first equation, we said that sin θ = m λ / a, so we substitute

sin θ ’= sin θ / 2

Then we use trigonometry to find the width, we say

tan θ = y / L

Since the angle is small, we then have

tan θ = sin θ / cos θ

tan θ = sin θ, this then means that

sin θ = y / L

we will then substitute

y’ / L = y/L 1/2

y' = y / 2

this means that when the slit width is doubled the pattern width will then be halved

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If you walk 30 meters forwards, and then turn around and walk 25 meters backwards, what is the distance that you walked? What di
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Given :

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I need an answer for this plzz!!<br>number 2 <br>anybody can help ??
Anuta_ua [19.1K]
2.1) (i) W = mg downwards
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s
a \:  =  \:  \frac{dv}{dt}  \:  =  \:  \frac{25 \:  \frac{m}{s} }{50 \: s} \:  =  \: 0.5 \:  \frac{m}{ {s}^{2} }
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2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)
2ax \:  =  \:  {v}^{2}  \:  -  \:  {u}^{2}
x \:  =  \:  \frac{ {v}^{2}  \:   -  \:  {u}^{2} }{2a}  \:  =  \:   \frac{{(25 \:  \frac{m}{s})}^{2}  \:  -  \:  {(0 \:  \frac{m}{s} )}^{2} }{2 \:  \times  \: 0.5 \:  \frac{m}{ {s}^{2} } } \:  =  \: 625 \: m
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Fe = f + ma
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Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.
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What kind of law of motion A car still moves for a short period even after the brakes
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Explanation:

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