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pshichka [43]
2 years ago
14

Soap bubble interference Light of 690-nm wavelength interferes constructively when reflected from a soap bubble having refractiv

e index 1.33. Determine two possible thicknesses of the soap bubble.
Physics
1 answer:
Nikitich [7]2 years ago
5 0

Answer:

The two possible thicknesses of the soap bubble is 129 nm and 389 nm.

Explanation:

Given that,

Wavelength = 690 nm

Refractive index = 1.33

We need to calculate the two possible thicknesses of the soap bubble

Using formula of thickness

t=\dfrac{(m+\dfrac{1}{2})\lambda}{2n}

For m = 0,

Put the value into the formula

t=\dfrac{\dfrac{690\times10^{-9}}{2}}{2\times1.33}

t=129\times10^{-9}\ m

t=129\ nm

For m=1,

Put the value into the formula

t=\dfrac{\dfrac{3\times690\times10^{-9}}{2}}{2\times1.33}

t=389\times10^{-9}\ m

t=389\ nm

Hence, The two possible thicknesses of the soap bubble is 129 nm and 389 nm.

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A copper cylinder is initially at 21.1 ∘C . At what temperature will its volume be 0.163 % larger than it is at 21.1 ∘C?
fomenos

To solve this problem we will apply the concepts related to the final volume of a body after undergoing a thermal expansion. To determine the temperature, we will use the given relationship as well as the theoretical value of the volumetric coefficient of thermal expansion of copper. This is, for example to the initial volume defined as V_1, the relation with the final volume as

V_2 = V_1 +0.163\% V_1

V_2 = V_1 +0.00163V_1

V_2 = 1.00163V_1

Initial temperature = 21.1\°C

Let T be the temperature after expanding by the formula of volume expansion

we have,

V_2 = V_1 (1+\gamma \Delta t)

Where \gamma is the volume coefficient of copper 5.1*10^{-5}/C

1.00163V_1 = V_1(1+\gamma(T-21.1\°))

1.00163 = 1+5.1*10^{-5}(T-21.1\°)

0.00163 = 0.000051T-0.0010761

T = 53.0608\°C

Therefore the temperature is 53.06°C

7 0
3 years ago
A rifle fires a bullet at a velocity of 78 m/s, 40 degrees above the horizontal. Determine the height (h) above the starting pos
qwelly [4]

Answer:

H = Vy t - 1/2 g t^2  height of an object with an initial "vertical" velocity

                                 at t sec after firing

Vy = 78 m/s * sin 40 = .643 * 78 m/s = 50.1 m/s

H = 50.1 * 6 - 1/2 * 9.8 * 6^2 = 300 m - 176 m = 124 m

7 0
2 years ago
Monochromatic coherent light shines through a pair of slits. If the wavelength of the light is decreased, which of the following
Law Incorporation [45]

Answer:

he correct answers are a, b

Explanation:

In the two-slit interference phenomenon, the expression for interference is

          d sin θ= m λ                       constructive interference

          d sin θ = (m + ½) λ             destructive interference

in general this phenomenon occurs for small angles, for which we can write

           tanθ = y / L

           tan te = sin tea / cos tea = sin tea

           sin θ = y / La

un

derestimate the first two equations.

Let's do the calculation for constructive interference

         d y / L = m λ

the distance between maximum clos is and

         y = (me / d) λ

this is the position of each maximum, the distance between two consecutive maximums

         y₂-y₁ = (L   2/d) λ - (L 1 / d) λ₁          y₂ -y₁ = L / d λ

examining this equation if the wavelength decreases the value of y also decreases

the same calculation for destructive interference

         d y / L = (m + ½) κ

         y = [(m + ½) L / d] λ

again when it decreases the decrease the distance

the correct answers are a, b

7 0
2 years ago
how is the position of electrons involved in metallic bonding different from the position of electrons that form ionic and coval
Yuri [45]
While ionic bonds join metals to nonmetals, and covalent bonds join nonmetals to nonmetals, metallic bonds are responsible for the bondingbetween metal atoms. In metallic bonds, the valence electrons from the s and p orbitals of the interacting metal atoms delocalize.

I hope that this answer helps you out
7 0
3 years ago
An automobile starter motor has an equivalent resistance of 0.0500Ω and is supplied by a 12.0-V battery with a 0.0100-Ω internal
alexandr1967 [171]

Answer

given,

resistance = 0.05 Ω

internal resistance of battery = 0.01 Ω

electromotive force = 12 V

a) ohm's law

        V = IR

     and volage

   V = \epsilon - Ir

now,

   IR = \epsilon - Ir

   I(R+r) = \epsilon

   I= \dfrac{\epsilon}{R+r}

inserting the values

   I= \dfrac{12}{0.05+0.01}

      I = 200 A

b) Voltage

   V = I R

   V = 200 x 0.05

   V = 10 V

c) Power

    P = I V

    P = 200 x 10 = 2000 W

d) total resistance = 0.05 + 0.09 = 0.14 Ω

 I= \dfrac{\epsilon}{R+r}

   I= \dfrac{12}{0.14+0.01}

     I = 80 A

     V = 80 x 0.05 = 4 V

     P = 4 x 80 = 320 W

6 0
3 years ago
Read 2 more answers
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