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3 years ago

What is reflection? Reflection is light that has struck a surface and has __________. .

2 answers:

8 0

The correct answer should be bounced off since the light ray hit the surface and reflected towards a new location. It therefore bounced off.

katovenus [111]3 years ago

5 0

Correct answer choice is :

C) Bounced off

Explanation:

Reflection is the difference in the path of a wavefront at an interface within two various media so that the wavefront returns into the tool from which it began. Current models involve the reflection of light, sound and water waves. Reflection is when light bounces off an object. If the cover is soft and shiny, like glass, water or bright metal, the light will reflect at the same point as it hit the cover.

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Light of wavelength 500 nm is incident perpendicularly from air on a film 10-4cm thick and of refractive index 1.375. Part of th

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Answer

given,

wavelength (λ)= 500 n m

thickness of film= 10⁻⁴ cm

refractive index = μ = 1.375

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3 years ago

A uniform marble rolls down a symmetrical bowl, start- ing from rest at the top of the left side. The top of each side is a dist

Paha777 [63]

Answer:

Part a)

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

Explanation:

As we know by energy conservation the total energy at the bottom of the bowl is given as

$\frac{1}{2} mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that on the left side the ball is rolling due to which it is having rotational and transnational both kinetic energy

now on the right side of the bowl there is no friction

so its rotational kinetic energy will not change and remains the same

so it will have

$\frac{1}{2}mv^2 = mgh'$

now we know that

$I = \frac{2}{5}mr^2$

$\omega = \frac{v}{r}$

so we have

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh$

$\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh$

$\frac{7}{10}mv^2 = mgh$

$\frac{1}{2}mv^2 = \frac{10}{14}mgh$

so the height on the smooth side is given as

$h' = \frac{10}{14} h$

Part b)

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Part c)

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