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kkurt [141]
2 years ago
10

Answer this question and show all the necessary steps to solve.

Physics
1 answer:
Liula [17]2 years ago
4 0

Answer:

√ ae/ ap = √ mp/me 1/2

Explanation:

Mass of the electron = M (Given)

Mass of the proton = M (Given)

Time = t (Given)

<h3>Electrostatic force, FE = eE (for </h3><h3>both electron and proton)</h3>

But acceleration of electron, ae =FE/me whereas

acceleration of proton, ap = FE/mp

S = 1/2at²1 = 1/2apt²2

<h3>Therefore,</h3>

t2/t1= √ ae/ ap

t2/t1= √ ae/ ap = √ mp/me 1/2

<h3>Thus, the ratio is nearly equal to</h3><h3> = √ ae/ ap = √ mp/me 1/2</h3>

<h2>so, option D is correct</h2>
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Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

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            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

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and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

              f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}

                          = 32.4 cm

Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

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