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2 years ago

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A nearsighted person cannot see objects further away than 10 m clearly. This person wants to be able to see objects 300 m away w

ithout trouble. What is the focal length of the lens you would prescribe? (Ignore the distance between the person’s eye and the lens.)
a. 9.7 m
b. – 9.7 m
c. No such lens can be prescribed.
d. 10.3 m
e. – 30 m
f. 30 m
g. – 10.3 m

2 answers:

k0ka [10]2 years ago

5 0

Answer:

d. 10.3 m

Explanation:

For a nearsighted person,

1/f = (1/v)+(1/u).................... Equation 1

f = focal, v = image distance, u = object distance.

f = vu/(v+u)................. Equation 2

Given: v = -10 m, u = 300 m

Substitute into equation 2

f = (-10×300)/(-10+300)

f = -3000/290

f = -10.34 m

therefore a concave lens of focal length 10.34 m is required.

The right option is d. 10.3 m

icang [17]2 years ago

4 0

Answer:

d.-10.3m

Explanation:

Note for short sightedness the focal length is negative

Let do be object distance=10m

And di= image distance=-300m

Using lens formula

F=do*di/do-di= 10*300/10-300=-10.3m

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An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the ea

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Answer:

Q1) Time taking by particle to travel the 40 km wrt. earth = $1.34\times10^{-6}$ sec.

Q2) The distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

Q3) The time taking by particle to travel from where it is created to the surface of the earth = $1.285\times10^{-5}$ sec.

Explanation:

Given :

Speed of particle wrt. earth $v=0.99537c$

Distance between where particle is created and earth surface = 40 km

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Where $x = 40\times10^{3} m$ , $v = 0.99537c$ , we know speed of light $c = 3 \times10^{8}$

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$= \frac{40 \times10^{3} }{0.99537\times3\times10^{8} }$

$t = 1.34\times10^{-6} sec$

∴ Thus, time taking by particle to travel the 40 km wrt. earth $t = 1.34\times10^{-6} sec$

According to the lorentz transformation,

⇒ $l = l_{o} \sqrt{1-\frac{v^2}{c^2} }$

Where $l =$ improper length, $l_{o} =$ proper length (distance measured wrt. rest frame) = 40 km

$l = 40 \sqrt{1-\frac{v^2}{c^2 }$

$l = 40 \times 0.096$

$l = 3.84$ km

∴ Thus, the distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

According to the time dilation,

$\Delta t = \frac{\Delta t_{o} }{\sqrt{1-\frac{v^2}{c^2} } }$

Where $\Delta t =$ improper time (wrt. earth frame time) $=1.34\times10^{-6} sec$ , $\Delta t _{o} =$ proper time (wrt. particle frame).

$1.34\times10^{-6} = \frac{ \Delta t_{o}}{0.096}$

$\Delta t_{o} = 1.285 \times10^{-5} sec$

Thus, the time taking by particle to travel from where it is created to the surface of the earth $= 1.285 \times10^{-5} sec$ .

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