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padilas [110]
3 years ago
5

A ray is incident on a film of thickness t and the index of refraction n=1.6 at an angle θ = 35 degrees. Find the angle of refra

ction in the film: ________degrees. If the reflected ray comes out at a distance___________ 2.5 mm. What is the thickness of the film?
Physics
1 answer:
lidiya [134]3 years ago
4 0

Answer

Given,

refractive index of film, n = 1.6

refractive index of air, n' = 1

angle of incidence, i = 35°

angle of refraction, r = ?

Using Snell's law

n' sin i = n sin r

1 x sin 35° = 1.6 x sin  r

 r = 21°

Angle of refraction is equal to 21°.

Now,

distance at which refractive angle comes out

d = 2.5 mm

α be the angle with horizontal surface and incident ray.

α = 90°-21° = 69°

t be the thickness of the film.

So,

tan \alpha = \dfrac{t}{d/2}

tan 61^0 = \dfrac{t}{2.5/2}

t = 2.26 mm

Hence, the thickness of the film is equal to 2.26 mm.

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Answer:

4.7\mu m

Explanation:

We are given that

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Surface charge density=\sigma=29nC/m^2=29\times 10^{-5} C/m^2

Using 1 nC/cm^2=10^{-5} C/m^2

We know that

C=\frac{\epsilon_0A}{d}=\frac{Q}{V}=\frac{\sigma A}{V}

d=\frac{\epsilon_0V}{\sigma}

Where

\epsilon_0=8.85\times 10^{-12}C^2/Nm^2

Using the formula

d=\frac{8.85\times 10^{-12}\times 154}{29\times 10^{-5}}

d=4.7\times 10^{-6} m=4.7\mu m

Using 1\mu m=10^{-6} m

Hence, the spacing between the plates=4.7\mu m

3 0
3 years ago
Look at this picture of a frog.
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2 years ago
Canola oil is less dense than water, so it floats on water, but its index of refraction is 1.47, higher than that of water. When
kupik [55]

Answer:

therefore critical angle c= 69.79°

Explanation:

Canola oil is less dense than water, so it floats over water.

Given n_{canola}= 1.47

which is higher than that of water

refractive index of water n_{water}=1.33

to calculate critical angle of light going from the oil into water

we know that

sinc= \frac{n_{water}}{n_{canola}}

now putting values we get

sinc= \frac{1.33}{1.47}

c= sin^{-1}(\frac{1.33}{1.47} )

c=69.79°

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8 0
3 years ago
A weight suspended from a spring is seen to bob up and down over a distance of 20 cm triply each second. What is the period? Wha
Kisachek [45]

Answer:

1) Hence, the period is 0.33 s.

2) The amplitude is 10 cm.

Explanation:

1) The period is given by:

T = \frac{1}{f}

Where:

f: is the frequency = 3 bob up and down each second = 3 s⁻¹ = 3 Hz

T = \frac{1}{f} = \frac{1}{3 Hz} = 0.33 s

Hence, the period is 0.33 s.

2) The amplitude is the distance between the equilibrium position and the maximum position traveled by the spring. Since the spring is moving up and down over a distance of 20 cm, then the amplitude is:          

A = \frac{20 cm}{2} = 10 cm  

Therefore, the amplitude is 10 cm.          

I hope it helps you!                    

5 0
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