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Helga [31]
3 years ago
7

Maurice directs a beam of light on a sheet of glass at an angle of 51°. What is the angle of refraction in the glass? The refrac

tive index of glass is 1.46.
A: 22 degrees
B: 29 degrees
C: 32 degrees
D: 36 degrees
Physics
2 answers:
alexgriva [62]3 years ago
5 0

Answer:

Angle of refraction in the glass is 32 degrees

Explanation:

It is given that,

Maurice directs a beam of light on a sheet of glass at an angle, θ₁ = 51°

The refractive index of the glass is, n₂ = 1.46

We know that the refractive index of the air is, n₁ = 1

We have to find angle of refraction in glass i.e. θ₂

According to Snell's law :

\dfrac{n_1}{n_2}=\dfrac{sin\ \theta_2}{sin\ \theta_1}

Putting all the values we get :

\dfrac{1}{1.46}=\dfrac{sin\ \theta_2}{sin51}  

\theta_2=sin^{-1}(\dfrac{sin(51)}{1.46})

\theta_2=32\ ^0

Hence, the correct option is (c) " 32 degrees ".  

iVinArrow [24]3 years ago
4 0
Maurice directs a beam of light on a sheet of glass at an angle of 51°. The refractive index of glass is <span>1.46. </span>The angle of refraction in the glass is 29 degrees. The answer is letter B.
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A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and
Mariulka [41]

Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

q_{out}=q_{2}-q_{1}

q_{out}=-6.43-(-4.00)

q_{out}=-2.43\ \mu C

The charge on the inner surface is q.

q+(-2.43)=-6.43

q=-6.43+2.43= 4.00\ \mu C

We need to calculate the electric field outside the shell

Using formula of electric field

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}

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(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

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