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3 years ago

In single-slit diffraction, what happens to the dark fringes if the slit gets wider?.

Physics

1 answer:

Reptile [31]3 years ago

5 0

The spacing between the fringes does not depend on the width of the slit.

The light intensity produced by the slits increases with increased width of the slit. The screen thus develops brighter fringes as the slit width increases.

On the screen, coloured fringes will be created if the white light replaces monochromatic light.

<h3>What is diffraction?</h3>

Diffraction is the breaking up of an electromagnetic wave as it passes a geometric structure (e.g. a slit), followed by reconstruction of the wave by interference. Diffraction occurs for all types of waves, including sound waves.

Learn more about diffraction:

brainly.com/question/16749356

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Two capacitors A and B are connected in

fredd [130]

Answer:

$Capacitor A = 0.24\mu F\\Capacitor B = 0.16\mu F$

Explanation:

I'll upload my work shortly as an attachment, but here is my process in words:

In our first situation we have two capacitors in parallel, which means the charge distribution on both of them is the same. With that, we can find a ratio between the values of Capacitors A and B.
In our second situation , we add a capacitor parallel to A (I called it C). Because A and C are in parallel, we know that they must have the same potential difference; which should come to be 10V since 90V of the total 100V is on B. Also, the equivalent charge distribution across A and C must be equal to that of the charge distribution at B, because A&C are in series with B. So I added the charges on A&C and set that equal to the charge on B.
Next, I used the ratio from the first situation to substitute Capacitor A out of the equation. This allows us to solve for B's capacitance. (Note: You could have also substituted B for A and solved for A first if you wanted to.)
Finally, I used B's capacitance to plug back into the ratio from the first situation to find A's capacitance. And they wanted the answer in micro-Farads, so I went ahead and converted each answer to micro.

5 0

4 years ago

An antelope moving with constant acceleration covers the distance 68.0 m between two points in time 7.50 s. Its speed as it pass

Darya [45]

Answer:

A)The speed of the antelope at the first point is 2.43 m/s.

B) The acceleration of the antelope is 1.77 m/s²

Explanation:

The equations of the position and velocity of the antelope is given by the following expressions:

x = x0 + v0 · t + 1/2 · a ·t²

v = v0 + a · t

Where:

x = position of the antelope at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

A) Let´s place the center of the frame of reference at the first point. The equation of position at t = 7.50 s will be:

x = x0 + v0 · t + 1/2 · a ·t²

68.0 m = 0 m + v0 · 7.50 s + 1/2 · a · (7.50 s)²

We also know that at the second point the velocity is 15.7 m/s. Then at t = 7.50 the velocity will be 15.7 m/s.

v = v0 + a · t

15.7 m/s = v0 + a · 7.50 s

We can solve this equation for "a" and replace it in the equation of height to obtain "v0". Then:

a = (15.7 m/s - v0) / 7.50 s

Replacing it in the equation for position:

68.0 m = 0 m + v0 · 7.50 s + 1/2 · a · (7.50 s)²

68.0 m = v0 · 7.50 s + 1/2 · (15.7 m/s - v0) / 7.50 s · (7.50 s)²

68.0 m = v0 · 7.50 s + 7.85 m/s · 7.50 s - 3.75 s · v0

68.0 m - 7.85 m/s · 7.50 s = 3.75 s · v0

(68.0 m - 7.85 m/s · 7.50 s) / 3.75 s = v0

v0 = 2.43 m/s

The speed of the antelope at the first point is 2.43 m/s.

B) The acceleration of the antelope will be:

a = (15.7 m/s - v0) / 7.50 s

a = (15.7 m/s - 2.43 m/s) / 7.50 s

a = 1.77 m/s²

The acceleration of the antelope is 1.77 m/s²

5 0

3 years ago

Air at 80 kPa and 127 °C enters an adiabatic diffuser steadily at a rate of 6000 kg/h and leaves at 100 kPa. The velocity of the

Vinil7 [7]

Answer:

a) The exit temperature is 430 K

b) The inlet and exit areas are 0.0096 m² and 0.051 m²

Explanation:

a) Given:

T₁ = 127°C = 400 K

At 400 K, h₁ = 400.98 kJ/kg (ideal gas properties table)

The energy equation is:

$q-w=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2} }{2} +delta-p$

For a diffuser, w = Δp = 0

The diffuser is adiabatic, q = 0

Replacing:

$0-0=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2} }{2} +0$

Where

V₁ = 250 m/s

V₂ = 40 m/s

Replacing:

$0-0=h_{2} -400x10^{3} +\frac{40^{2}-250^{2} }{2} +0\\h_{2} =431430 J/kg=431.43kJ/kg$

Using tables, at 431.43 kJ/kg the temperature is 430 K

b) The inlet area is:

$m=\frac{P_{1} }{RT_{1} } A_{1} V_{1} \\\frac{6000}{3600} =\frac{80}{0.287*400} A_{1} *250\\A_{1} =0.0096m^{2}$

The exit area is:

$m=\frac{P_{2} }{RT_{2} } A_{2} V_{2} \\\frac{6000}{3600} =\frac{100}{0.287*430} A_{2} *40\\A_{2} =0.051m^{2}$

6 0

3 years ago

A car driving down the road speeds up to pass bus. While passing changed velocity from 10 m/s to 35 m/s in 5 seconds. What is th

VMariaS [17]

Answer:

The ans is 3

Explanation:

First write the given values and write the formula of acceleration than do velocity - final velocity by time taken is 10-35%5 so, therefore acceleration of car is 3

5 0

3 years ago

Which type of rock can only form below earths surface?

Inessa [10]

Answer: Igneous it forms because of magma but magma is under the earths surface so its Igneous

Explanation:

7 0

2 years ago

Read 2 more answers