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2 years ago

\textbf{Help with this question } " alt="\large \textbf{Help with this question } " align="absmiddle" class="latex-formula">~
$\textsf{For an Isosceles prism of Angle A and}$ $\textsf{Refractive index}\: \sf \mu ,\textsf{ it is found that }$ $\textsf{the angle of minimum deviation }\sf{ \delta_m = A.}$ $\textsf{Then which of the following is/are correct}$

$\textbf{[ Multiple choice ]}$

$\textsf{(a) At minimum deviation, the incident}$ $\sf{ angle \: \:i_1 }\textsf{and the retracting angle}$ $\sf{r_1}\textsf{ at the first retracting surface are}$ $\textsf{related by }\sf{ r_1 = (i_1/2) }$

$\textsf{(b) For this prism, the retracting index}$ $\sf{ \mu}\textsf{ and the angle of Prism A are related as}$
$\sf{ A = \dfrac{1}{2}cos^{-1}\bigg\{\dfrac{\mu}{2}\bigg\}}$

$\textsf{(c) For this prism, the emergent ray at the }$ $\textsf{second surface will be tangential to the }$ $\textsf{surface when the angle of incidence at the}$ $\textsf{ first surface is}\: \sf{ i_1 = sin^{-1} \bigg[sinA \sqrt{4cos²\dfrac{A}{2}-1}-cosA\bigg]} [tex] \textsf{(d) For the angle of incidence}\sf{ i_1 = A,} \textsf{the ray}$ $\textsf{ inside the prism is parallel to the base}$ $\textsf{ of the prism}$

Please provide explanation ~

Physics

1 answer:

PIT_PIT [208]2 years ago

8 0

<h3>Hi there !</h3><h2>Option A is correct </h2>

<h3> Please refer the attachment for explanation</h3><h2>Stay safe, stay healthy and blessed</h2><h2>Have a marvelous day</h2><h2>Thank you</h2>

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Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

<u />

The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

$a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2} } \\a=3.33*10^{19} [m/s^2]$

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