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2 years ago

Compare and contrast between reflection and refraction

Physics

1 answer:

Tcecarenko [31]2 years ago

3 0

Differences:

Reflection is described as the reverting back of light or sound waves in the same medium, when it falls on plane.

Refraction means the shift in direction of the radio waves, when it enters medium with different density.

Comparisons:

Reflection and refraction both change the direction of light and both do not absorb light. All light waves can be reflected and refracted.

Both Reflection and Refraction are phenomenon related with wave motion of sound, and electromagnetic waves, thus including light.

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Two mirrors are touching so they have an angle of 35.4 degrees with one another. A light ray is incident on the first at an angl

alexandr1967 [171]

Answer:

54.6°

Explanation:

From law of reflection i=r.

So, construct the reflected ray at 55.7°degrees from the normal and let it fall on the other mirror.

Now draw the second normal at the point of incidence and again measure the angle of incidence, and draw the angle of reflection.

If you consider triangle AOB, one angle is ∠AOB=90°

and ∠OAB is 54.6°

From angle sum property third angle ie ∠ABO=180°-90°-54.6°=35.4°

So, the second incident angle will be 54.6°

Hence, the second reflected angle will be 54.6 degrees.

8 0

3 years ago

A(n) ________ microscope keeps an object in focus when the objective lens is changed.

Anvisha [2.4K]

Parfocal is the term used to describe a microscope that maintains focus when the objective lenses are replaced.

<h3>What is the name of the objective lens ?</h3>

For observing minute features within a specimen sample, a high-powered objective lens, often known as a "high dry" lens, is perfect. You can see a very detailed image of the specimen on your slide thanks to the 400x total magnification that a high-power objective lens and a 10x eyepiece provide.

The four objective lenses on your microscope are for scanning (4x), low (10x), high (40x), and oil immersion (100x).

The first-stage lens used to create a picture from electrons leaving the specimen is referred to as the "objective lens." The objective lens is the most crucial component of the imaging system since the quality of the images is determined by how well it performs (resolution, contrast, etc.,).

To learn more than objective lens , visit

brainly.com/question/17307577

#SPJ4

6 0

1 year ago

Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k

natima [27]

Answer:

part (a) $a_1\ =\ 2.9\ kg$

Part (b) $a_2\ =\ 6.25\ kg$

Explanation:

Given,

mass of the smaller disk = $M_1\ =\ 0.900\ kg$
Radius of the smaller disk = $R_1\ =\ 2.45\ cm\ =\ 0.0245\ m$

mass of the larger disk = $M_2\ =\ 1.6\ kg$
Radius of the larger disk = $R_2\ =\ 5.0\ cm\ =\ 0.05\ m$
mass of the hanging block = m = 1.60 kg

Let I be the moment of inertia of the both disk after the welding, $\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}(M_1R_1^2\ +\ M_2R_2^2)\\\Rightarrow I\ =\ 0.5\times (0.9\times 0.0245^2\ +\ 1.6\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2$

part (a)

A block of mass m is hanging on the smaller disk,

From the f.b.d. of the block,

Let 'a' be the acceleration of the block and 'T' be the tension in the string.

$mg\ -\ T\ =\ mg\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)$

Net torque on the smaller disk,

$\therefore \tau\ =\ I\alpha\\\Rightarrow TR_1\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{Ia}{R_1^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,enq (2)$

From eqn (1) and (2), we get,

$mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.027^2}\ +\ 1.60}\\\Rightarrow a\ =\ 2.91\ m/s^2$

part (b)

In this case the mass is rapped on the larger disk,

From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,

Let ' $a_2$ ' be the acceleration of the block in the second case,

From the above expression,

$\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.05^2}\ +\ 1.60}\\\Rightarrow a\ =\ 6.25\ m/s^2$

5 0

3 years ago

A small block of mass m on a horizontal frictionless surface is attached to a horizontal spring that has force constant k. The b

Rashid [163]

Answer:

$v_{max} = |d|\cdot \sqrt{\frac{k}{m} }$

Explanation:

The maximum speed of the block occurs when spring has no deformation, that is, there is no elastic potential energy, which can be remarked from appropriate application of the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot d^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

$k \cdot d^{2} = m\cdot v^{2}$

$v_{max} = |d|\cdot \sqrt{\frac{k}{m} }$

5 0

3 years ago

Please help on this one

ale4655 [162]

That thing is used as a lever.

4 0

3 years ago