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ASHA 777 [7]
3 years ago
10

Estimate how far apart the rays of deepest red and deepest violet light are as they exit the bottom surface. assume nred = 1.57

and nviolet = 1.60.

Physics
1 answer:
Harlamova29_29 [7]3 years ago
6 0
We begin by noting that the angle of incidence is the one that's taken with respect to the normal to the surface in question. In this case the angle of incidence is 30. The material is Flint Glass according to the original question. The refractive indez of air n1=1, the refractive index of red in flint glass is nred=1.57, finally for violet in the glass medium is nviolet=1.60. Snell's Law dictates:
n_1sin(\theta_1)=n_2sin(\theta_2)
Where \theta_2 differs for each wavelenght, that means violet and red will have different refractive indices in the glass.
In the second figure provided details are given on which are the angles in question, \Delta x is the distance between both rays.
\theta_{2red}=Asin(\frac{sin(30)}{1.57})\approx 18.5705
\theta_{2violet}=Asin(\frac{sin(30)}{1.60})\approx 18.21
At what distance d from the incidence normal will the beams land at the bottom?
For violet we have:
d_{violet}=h.tan(\theta_{2violet})\approx 0.0132m
For red we have:
d_{red}=h.tan(\theta_{2red})\approx 0.0134m
We finally have:
\Delta x=d_{red}-d_{violet}\approx2.8\times10^{-4}m


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3 years ago
A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
EleoNora [17]

Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

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Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

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