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Mademuasel [1]
3 years ago
14

A beam of unpolarized light in air strikes a flat piece of glass at an angle of incidence of 54.2 degrees. If the reflected beam

is completely polarized, what is the index of refraction of the glass?
A. 1.60
B. 1.39
C. 1.52
D. 2.48
Physics
2 answers:
Stels [109]3 years ago
6 0

Answer:

correct option is B. 1.39

Explanation:

given data

angle of incidence (θ) = 54.2 degrees

to find out

index of refraction of the glass

solution

we know that here reflected beam is completely polarized

so angle of incidence = angle of polarized    ....................1

for reflective index we apply here Brewster law that is

μ = tan(θ)     ...............2

put here θ value we get

μ = tan(54.2)

μ = 1.386

so correct option is B. 1.39

marishachu [46]3 years ago
5 0

Answer:

1.386

Explanation:

Angle of incidence, i = 54.2°

Let n be the refractive index of glass.

According to the Brewster's law

n = tan ip

where, ip is the polarising angle.

n = tan 54.2°

n = 1.386

Thus, the refractive index of glass is 1.386.

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A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it
Degger [83]

Answer:

a)\omega_1=8.168\,rad.s^{-1}

b)n_1=7.735 \,rev

c)\alpha_1 =0.6864\,rad.s^{-2}

d)\alpha_2=4.1454\,rad.s^{-2}

e)t_2=1.061\,s

Explanation:

Given that:

  • initial speed of turntable, N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}
  • full speed of rotation, N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}
  • time taken to reach full speed from rest, t_1=11.9\,s
  • final speed after the change,  N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}
  • no. of revolutions made to reach the new final speed,  n_2=11\,rev

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

\omega=\frac{2\pi.N}{60}\, rad.s^{-1}

where N = angular speed in rpm.

putting the respective values from case 1 we've

\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}

\omega_1=8.168\,rad.s^{-1}

(c)

using the equation of motion:

\omega_1=\omega_0+\alpha . t_1

here α is the angular acceleration

78=0+\alpha_1\times 11.9

\alpha_1 = \frac{8.168 }{11.9}

\alpha_1 =0.6864\,rad.s^{-2}

(b)

using the equation of motion:

\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1

8.168^2=0^2+2\times 0.6864\times n_1

n_1=48.6003\,rad

n_1=\frac{48.6003}{2\pi}

n_1=7.735\, rev

(d)

using equation of motion:

\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2

12.5664^2=8.168^2+2\alpha_2\times 11

\alpha_2=4.1454\,rad.s^{-2}

(e)

using the equation of motion:

\omega_2=\omega_1+\alpha_2 . t_2

12.5664=8.168+4.1454\times t_2

t_2=1.061\,s

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An aircraft maintenance technician walks past a tall hangar door that acts like a single slit for sound entering the hangar. Out
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Answer:

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Explanation:

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λ = ?

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λ = wavelength incident on the single slit

θ = angular position of the mth minima

But, λ = v / f

λ = 340 / 610 = 0.557m

θ = sin⁻(mλ/d)

θ = sin⁻ [(1 * 0.557) / 0.840]

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Explanation:

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