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Verdich [7]
2 years ago
13

A laser with wavelength d/8 is shining light on a double slit with slit separation 0.500 mm. This results in an interference pat

tern on a screen a distance L away from the slits. We wish to shine a second laser, with a different wavelength, through the same slits. What is the wavelength λ of the second laser that would place its second maximum at the same location as the fourth minimum of the first laser, if d = 0.500 mm?
Physics
1 answer:
ipn [44]2 years ago
7 0

Answer:

0.000109375 m

Explanation:

d = Distance between grating = 0.5 mm

m = Order

\lambda_1=\dfrac{d}{8}

Minima relation

m-\dfrac{1}{2}

For fourth order minima

dsin\theta=(4-\dfrac{1}{2})\lambda_1

For second maxima

dsin\theta=2\lambda_2

From the two equations we get

(4-\dfrac{1}{2})\lambda_1=2\lambda_2\\\Rightarrow \lambda_2=\dfrac{(4-\dfrac{1}{2})\lambda_1}{2}\\\Rightarrow \lambda_2=\dfrac{(4-\dfrac{1}{2})\dfrac{0.5\times 10^{-3}}{8}}{2}\\\Rightarrow \lambda_2=0.000109375\ m

The wavelength is 0.000109375 m

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I pull the throttle in my racing plane at a = 12.0 m/s2. I was originally flying at v = 100. m/s. Where am I when t = 2.0s, t =
Helen [10]
Summary:
a= 12.0 m/(s^2)
v= 100m/s
t1= 2.0s => s1=?
t2=5.0s => s2=?
t3=10.0s => s3=?
——————
Solution:
• when t1=2.0 s, I have gone:
S1= v*t1 + 1/2*a*(t1^2)
=100.0 *2 + 1/2*12.0*(2.0^2)
=224 (m)

• when t2=5.0s, I have gone
S2=v*t2+ 1/2*a*(t2^2)
= 100*5.0+ 1/2*12.0*(5.0^2)
=650 (m)

•when t3= 10.0s, I have gone:
S3=v*t3+ 1/2*a*(t3^2)
=100*10.0+ 1/2*12*(10.0^2)
=1600 (m)
7 0
2 years ago
It takes 3 minutes to make toast in a 1500 watt toaster. Calculate how much work is done by the toaster.
denis23 [38]
Hello there.

<span>It takes 3 minutes to make toast in a 1500 watt toaster. Calculate how much work is done by the toaster. 

</span>270,000 J
6 0
2 years ago
Read 2 more answers
A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an en
nataly862011 [7]

Answer:

The energy of this particle in the ground state is E₁=1.5 eV.

Explanation:

The energy E_{n} of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:

                                                    E_{n}=\frac{n^{2}h^{2}}{8mL^{2}}

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

                                                    E_{1}=\frac{h^{2}}{8mL^{2}}            

                                                    E_{2}=\frac{h^{2}}{2mL^{2}}

So we can rewrite the energy in the ground state as:

                                                   E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})

                                                      E_{1}=\frac{1}{4} E_{2}

                                                   E_{1}=\frac{1}{4} ( 6.0\ eV)

Finally

                                                    E_{1}=1.5\ eV

                                                   

                                                   

6 0
2 years ago
Two convex thin lenses with focal lengths 10.0 cm and 20.0 cm are aligned on a common axis, running left to right, the 10-cm len
love history [14]

Answer:

(c) +6.67

Explanation:

f1 = 10 cm

f2 = 20 cm

u = Object distance = 15 cm

Distance between lenses = 20 cm

For first lens image distance

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{15}\\\Rightarrow \frac{1}{v}=\frac{1}{30}\\\Rightarrow v=30\ cm

Distance from second lens is 10 cm to the right

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{-10}\\\Rightarrow \frac{1}{v}=\frac{3}{20}\\\Rightarrow v=6.67\ cm

The final image will appear as +6.67 cm

3 0
2 years ago
A fly has a mass of 1 gram at rest. how fast would it have to be traveling to have the mass of a large suv, which is about 3000
Zigmanuir [339]

We solve this using special relativity. Special relativity actually places the relativistic mass to be the rest mass factored by a constant "gamma". The gamma is equal to 1/sqrt (1 - (v/c)^2). <span>

We want a ratio of 3000000 to 1, or 3 million to 1. 

</span>

<span>Therefore:
3E6 = 1/sqrt (1 - (v/c)^2) 
1 - (v/c)^2 = (0.000000333)^2 
0.99999999999999 = (v/c)^2 
0.99999999999999 = v/c 
<span>v= 99.999999999999% of the speed of light ~ speed of light
<span>v = 3 x 10^8 m/s</span></span></span>

8 0
2 years ago
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