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3 years ago

6

The index of refraction for a certain type of glass is 1.641 for blue light and 1.603 for red light. When a beam of white light

(one that contains all colors) enters a plate of glass from the air at an incidence angle of 40.05∘, what is the absolute value of ????, the angle in the glass between blue and red parts of the refracted beams?

1 answer:

exis [7]3 years ago

6 0

Answer:

The angle between the emergent blue and red light is $0.566^{o}$

Explanation:

We have according to Snell's law

$n_{1}sin(\theta _{i})=n_{2}sin(\theta _{r})$

Since medium from which light enter's is air thus $n_{1}=1$

Thus for blue incident light we have

$1\times sin(40.05)=1.641\times sin(\theta _{rb})\\\\\therefore \theta _{rb}=sin^{-1}(\frac{sin(40.05)}{1.64})\\\\\theta _{rb}=23.10$

Similarly using the same procedure for red light we have

$1\times sin(40.05)=1.603\times sin(\theta _{rr})\\\\\therefore \theta _{rr}=sin^{-1}(\frac{sin(40.05)}{1.603})\\\\\theta _{rr}=23.66^{o}$

Thus the absolute value of angle between the refracted blue and red light is

$|23.66-23.10|=0.566^{o}$

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Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

Explanation:

A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

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Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

If the value is less than zero; then we conclude that the assumption is wrong.

Then, the flow is said to be in the opposite direction

Formula for the change in specific entropy can be calculated as:

$s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \ -------> \ \ \ 2$

where;

$s_1, s_2 , s^0(T_2), s^0(T1)$ are specific entropies

R = universal gas constant

$P_1$ = pressure at location 1

$P_2$ = pressure at location 2

We obtain the specific properties of air at temperature at $T_1$ = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)

$s^0(T1)$ = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature $T_2$ = 22°C + 273) K = 295 K

From the table A- 22

$s^0(T_2)$ = 1.68515 kJ/kg . K

R = $\frac{8.314 kJ}{28.97 kg.K}$

$P_1 =$ 0.95 bar

$P_2 =$ 0.8 bar

Now replacing our values into equation (2) from above; we have;

$s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )$

$s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97} \ In (\frac{0.8}{0.95} )$

$s_2-s_1 = 1.68515 -1.8279+ 0.0493$

$s_2-s_1 =-0.0934 \ kJ/kg.K$

Equating our result to equation (1)

$s_2-s_1 \geq 0\\-0.0934 \leq 0$

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

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Susan gently pushes the tip of her finger against the eraser on her pencil and the pencil does not move. Which of the

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2 years ago

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Hall emf, $V'_{Hall} = 69\ mV = 0.069\ V$

Now,

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$v_{d} = \frac{0.02}{0.10} = 0.2\ m/s$

Now, the expression for the electric field is given by:

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Thus eqn (1) becomes

$V_{Hall}d = dBv_{d}sin\theta$

where

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ch4aika [34]

Answer:

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F = ma

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3 years ago