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Marizza181 [45]
3 years ago
14

The diagram below shows a ray of light moving from left to right, through the interface between two mediums. Snell's law is used

to determine the direction of the refracted ray. If the index of refraction in the first medium is 1.5 and the index of refraction in the second medium is 1.0, what is the angle of refraction when the angle of incidence is 28°?
A. 25°

B. 18°

C. 42°

D. 45°

Physics
1 answer:
makkiz [27]3 years ago
7 0
The angle of refraction is <span>45°.</span>
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Answer:

10.78 s

Explanation:

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F^{\to}= qE^{\to} +q (v^{\to} + B^{\to}) \\ \\  F^{\to} = (9.12 \times 10^{-6}) *278 (-\hat k) +9.12 *10^{-6} *2.1 *0.18 (\hat i * \hat k) \\ \\  F^{\to} = -2.535 *10^{-3} \hat k -3.447*10^{-6} \hat j

F = ma

∴

a ^{\to}= \dfrac{F^{\to}}{m}

a ^{\to}= \dfrac{-1}{3.57\times 10^{-3}}(2.535*10^{-3}\hat k + 3.447*10^{-6} \hat j)

a ^{\to}=-0.710 \hat k -9.656*10^{-4} \hat j

At time t(sec; the partiCle velocity becomes v(t) = 3.78 v_o

The velocity of the charge after the time t(sec) is expressed by using the formula:

v^{\to}= v_{o \ \hat i} + a^{\to }t \\ \\  \implies (2.1)\hat i -0.710 t \hat k -9.656 \times 10^{-4} t \hat j = 3.78 v_o \\ \\ \implies (2.1)^2 +(0.710\ t)^2+ (9.656 *10^{-4}t )^2 = (3.78 *2.1^2 \\ \\ \implies 4.41 +0.5041 t^2 +9.324*10^{-7} t^2 = 63.012 \\ \\ \implies 4.41 +0.5041 t^2 = 63.012\\  \\  0.5041t^2 = 63.012-4.41 \\ \\ t^2 = \dfrac{58.602}{0.5041} \\ \\ t^2 = 116.25 \\ \\ t = \sqrt{116.25} \\ \\  \mathbf{t = 10.78 \ s}

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