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blagie [28]
2 years ago
9

A thin film of soap with n = 1.34 hanging in the air reflects dominantly red light with λ = 642 nm. What is the minimum thicknes

s of the film? Tries 0/20 Now this film is on a sheet of glass, with n = 1.55. What is the wavelength of the light in air that will now be predominantly reflected?
Physics
1 answer:
wariber [46]2 years ago
6 0

Answer:

The minimum thickness of the film and the wavelength of the light in air are 1.197\times10^{-7}\ m and 371 nm.

Explanation:

Given that,

Refractive index of soap= 1.34

Refractive index  of glass= 1.55

Wavelength = 642 nm

(I). We need to calculate the minimum thickness

Using formula of thickness

t=\dfrac{(2m+1)\lambda}{4n}

Where, m = 0 for constrictive

Put the value into the formula

t=\dfrac{(642\times10^{-9})}{4\times1.34}

t=1.197\times10^{-7}\ m

(II). We need to calculate the wavelength

Using formula of wavelength

\lambda=\dfrac{2nt}{m}

Where, m = 1

Put the value into the formula

\lambda=\dfrac{2\times1.55\times1.197\times10^{-7}}{1}

\lambda=371\ nm

Hence, The minimum thickness of the film and the wavelength of the light in air are 1.197\times10^{-7}\ m and 371 nm.

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1 year ago
What energy comes from a rining bell
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7 0
3 years ago
The a992 steel rod bc has a diameter of 50 mm and is used as a strut to support the beam. determine the maximum intensity w of t
EastWind [94]

Answer:

w = 11.211 KN/m

Explanation:

Given:

diameter, d = 50 mm

F.S = 2

L = 3

Due to symmetry, we have:

Ay = By = \frac{w * 6}{2} = 3w

P_c_r = 3w * F.S = 3w * 2.0 = 6w

I = \frac{\pi}{64} (0.05)^4 = 3.067*10^-^7

To find the maximum intensity, w, let's take the Pcr formula, we have:

P_c_r = \frac{\pi^2 E I}{(KL)^2}

Let's take k = 1

E = 200*10^9

Substituting figures, we have:

6w = \frac{\pi^2 * 200*10^9 * 3.067*10^-^7}{(1 * 3)^2}

Solving for w, we have:

w = \frac{67266.84}{6}

w = 11211.14 N/m = 11.211 KN/m

Since Area, A= pi * (0.05)²

\sigma _c_r = \frac{w}{A}

\sigma _c_r = \frac{11.211}{\pi (0.05)^2} = 1.4 MPA < \sigma y. This means it is safe

The maximum intensity w = 11.211KN/m

3 0
3 years ago
The electric field in the region between the plates of a parallel plate capacitor has a magnitude of 6.3 105 V/m. If the plate s
xz_007 [3.2K]

Answer:

3.528×10² V.

Explanation:

potential difference: This is the work done when one coulomb of charge moves from one point to another in an electric field. The S.I unit of potential difference is volt. The formula of potential difference is given as,

V = kq/r..................... Equation 1

And

E = kq/r² .................. Equation 2

Comparing equation 1 and equation 2,

V = E×r............................. Equation 3

Where V = potential difference, E = Electric field between the plate of the capacitor, r = distance between the plate.

Given: E = 6.3×10⁵ V/m, r = 0.56 mm = 0.00056 m.

Substitute into equation 3,

V = 6.3×10⁵×0.00056

V = 3.528×10² V.

Hence the potential difference of the plate = 3.528×10² V.

4 0
2 years ago
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