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3 years ago

What could you do to make the beam of light from the laser refract less as it exits the prism?

Physics

2 answers:

Sergeeva-Olga [200]3 years ago

6 0

<span>Increase the wavelength of the light

</span>

Bumek [7]3 years ago

3 0

Answer:

We want to make the beam to refract less as it exits the prism, so we want to increase the internal reflection inside the prism.

We could do this by approaching the critical angle of incidence when the beam wants to leave the prism, this is:

By using snell law:

n1*sen(a1) = n2*sen(a2)

Here n1 is the material of the prism and n2 is air, so we can suppose that n1 >n2

And a1 and a2 are the angles of incidence and refraction.

so we have that:

sen(a1) = (n2/n1)sen(a2)

Now, the max value of the left side is smaller than 1, so if we take sen(a2) = 1 (this means that the refracted beam is tangential to the wall of the prism, so there is no refracted beam after this angle) then we have:

Sen(a1) = n2/n1

a1 = asin(n2/n1) is the critical angle, and for angles bigger than that there is total internal reflection.

So approaching this angle is a good way to reduce the "amount" of the beam that is refracted when it exits the prism.

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The forces acting on a sailboat are 390n north and 180n east. if the boat (including crew) has a mass of 270 kg, what are the ma

patriot [66]

F=ma so a=F/m

ax=180/270=0.67m/s^2
ay=390/270=1.44m/s^2

Magnitude = sqrt((0.67^2)+(1.44^2))=1.59m/s^2

Direction- Tan(x)=0.67/1.44=0.47 Tan^-1(x)=25 degrees

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3 years ago

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A 900 kg car pushes a 2400 kg truck that has a dead battery. when the driver steps on the accelerator, the drive wheels of the c

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b)magnitude of the force on the car is 500n

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3 years ago

A point on a wheel of radius 40 cm that is rotating at a constant 5.0 revolutions per second is located 0.20 m from the axis of

Margarita [4]

Answer:

$197.2 m/s^2$

Explanation:

The centripetal acceleration of a point moving by circular motion is given by:

$a=\omega^2 r$

where

$\omega$ is the angular velocity

r is the distance from the axis of rotation

The point on the wheel makes 5.0 revolutions per second, so the frequency is

$f=\frac{5}{1}=5 Hz$

and the angular velocity is

$\omega=2\pi f = 2\pi (5)=31.4 rad/s$

While the distance of the point from the axis of rotation is

$r=0.20 m$

Substituting, we find the acceleration:

$a=(31.4)^2(0.20)=197.2 m/s^2$

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3 years ago

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A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.

fredd [130]

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

$F=\dfrac{mv^2}{r}-mg$

$F=m(\dfrac{v^2}{r}-g)$

$F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)$

F = 18 N

(b) The net tension in the string when the ball is at the bottom of the circle is given by :

$F=\dfrac{mv^2}{r}+mg$

$F=m(\dfrac{v^2}{r}+g)$

$F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)$

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

$T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}$

$T=\dfrac{m}{r}(g(r-h)+v^2)$

Since, $v^2=u^2-2gh$

$T=\dfrac{m}{r}(u^2-3gh+gr)$

Hence, this is the required solution.

6 0

3 years ago

A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water

elena-14-01-66 [18.8K]

Answer:

The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Explanation:

Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ( $\dot Q$ ), measured in watts, that is, joules per second, by the following formula:

$\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}$ (1)

Where:

$m$ - Mass of the sphere, measured in kilograms.

$c$ - Specific heat of the material, measured in joules per kilogram-degree Celsius.

$T_{o}$ , $T_{f}$ - Initial and final temperatures of the sphere, measured in degrees Celsius.

$\Delta t$ - Time, measured in seconds.

In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:

$\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}$ (2)

Where:

$m_{I}$ , $m_{X}$ - Masses of the iron and unknown spheres, measured in kilograms.

$\Delta t_{I}$ , $\Delta t_{X}$ - Times of the iron and unknown spheres, measured in seconds.

$c_{I}$ , $c_{X}$ - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.

$c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}$

If we know that $\Delta t_{I} = 6.35\,s$ , $\Delta t_{X} = 4.59\,s$ , $m_{I} = 0.515\,kg$ , $m_{X} = 1.263\,kg$ and $c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}$ , then the specific heat of the unknown material is:

$c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)$

$c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}$

Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

3 0

3 years ago