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Fittoniya [83]
3 years ago
13

A light ray transfers from more to less dense medium at certain condition , if we repeat this experiment by increasing the angle

of incidence in more dense medium to double its value in the 1st condition so the value of the relative refractive index between the 2 media is
increased to double
decreased to half
still constant
there is no relative refractive index between more and less dense medium
Physics
1 answer:
evablogger [386]3 years ago
4 0

Answer:

The correct option is;

Still constant

Explanation:

The relative refractive index ₁n₂ between the two medium can be as follows;

_1n_2 = \dfrac{The \ speed \ of \ light \ in \ medium \ 1}{The \ speed \ of \ light \ in \ medium \ 2}  = \dfrac{c_1}{c_2}

Therefore, given that the speed of light in medium 1 is constant and the speed of light on medium 2 is also constant, the relative refractive index ₁n₂ = c₁/c₂ is always constant.

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Answer:

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(b) 5.003 m

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Explanation:

(a) The minimum value of x will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.

(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

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h^{2} = 5^{2} + 0.17^{2}   \\h = \sqrt{} 25.03\\h= 5.002 m

<em>Hence, the maximum distance is 5.002 m</em>

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F=\frac{9 x 10^{9} x3.2x10^{-19}x 6.4x10^{-19}  }{0.17^{2} }

F= 6.38×10^{-26} N

(d) For maximum magnitude, we use the maximum distance calculated in (b)

Maximum Distance = 5.002 m

Using the formula for electrostatic force again:

F =  \frac{9x10^{9}x3.2x10^{-19}x6.4x10^{-19}   }{5.002^{2} } }

F= 7.37×10^{-29 N

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