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xenn [34]
3 years ago
9

Light from a laser strikes a diffraction grating that has 5500 lines per centimeter. The central and first-order maxima are sepa

rated by 0.455 m on a wall, 1.62 m from the grating, calculate the wavelength of the laser light.

Physics
1 answer:
WINSTONCH [101]3 years ago
3 0

.Answer:

491.4 nm

Explanation:

The distance between central and first maxima is,

y=0.455m

And the distance between screen abnd grating is,

L=1.62 m

Now the angle can be find as,

tan\theta=\frac{y}{L} \\\theta=tan^{-1}(\frac{0.455}{1.62})  \\\theta=15.68^{\circ}

Now the grating distance is,

d=\frac{1}{5500} cm\\d=1.82\times 10^{-6}m

Now with m=1 condition will become,

\lambda=dsin\theta

So,

\lambda=1.82\times 10^{-6}m\times sin(15.68^{\circ})\\\lambda=1.82\times 10^{-6}m\times 0.270\\\lambda=491.4\times 10^{-9}m\\\lambda=491.4 nm

Therefore the wavelength of laser light is 491.4 nm.

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4. I drop a pufferfish of mass 5 kg from a height of 5.5 m onto an upright spring of total length 0.5 m and spring constant 3000
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Answer:

a)  0.28 m or 28 cm is the minimum  height above ground the fish reaches.

b)  at the height of 0.484 m height , the pufferfish will eventually come to rest.

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Gravitational potential energy  = 23.72J

Spring potential energy   = 0.384 J

Explanation:

Given that :

Mass of the pufferfish m =5kg

initial height of the fish h =5.5m

length of the spring l =0.5m

Spring constant K =3000N/m

a)

Assuming no energy loss to friction, what is the minimum height above the ground that the pufferfish reaches?

Lets assume that the minimum height the fish reaches is = x meters

Now by using the conservation of energy; we realize that :

Initial total energy = final total energy

Gravitational potential energy =

Gravitational potential energy' + Spring potential energy (kinetic energy is zero in both cases)

mgh = mgx + \frac{1}{2}K(l-x)^2

Replacing our given values into the above equation; we have :

(5)(9.8)(5.5) = (5)(9.5)(x) + \frac{1}{2}(3000)(0.5-x)^2

269.5 = 47.5 x + 1500(0.5 -x )²

269.5 = 47.5 x + 1500(0.25 - x²)

269.5 = 47.5 x + 375 - 1500 x²

269.5 - 375 = 47.5 x - 1500 x²

-105.5 = 47.5 x - 1500 x²

-105.5 + 1500 x² - 47.5 x = 0

1500 x² - 47.5 x - 105.5 = 0

By using quadratic equation and taking the positive value;

x = 0.28 m or 28 cm is the minimum height above ground the fish reaches.

b)

At the equilibrium position the weight of fish will be equal to the force applied by the spring thus

mg = kx

substituting  our given values ; we have:

(5)(9.8) = 3000x

x = 61.22

x = 0.016m  : so this is the compression in the spring

Now; to determine the height  the pufferfish gets to before  it eventually come to rest; we have

(0.5-0.016) m = 0.484m

therefore, at the height of 0.484 m height , the pufferfish will eventually come to rest.

c)

There exists  two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy  = mgh' = (5)(9.8)(0.484)

= 23.72J

and spring potential energy  

=\frac{1}{2}Kx^2\\ = \frac{1}{2}(3000)(0.016)^2\\= 0.384J

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