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3 years ago

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Light with a single wavelength falls on two slits separated by 0.510 mm. In the resulting interference pattern on a screen 2.24

m away, adjacent bright fringes are separated by 2.88 mm.What is the wavelength of the light that falls on the slits

1 answer:

scZoUnD [109]3 years ago

7 0

Answer: 655.7 nm

Explanation:

Given

The slits are separated by $d=0.510\ mm$

Distance between slits and screen is $D=2.24\ m$

Adjacent bright fringes are $\beta =2.88\ mm$ apart

Also, the distance between bright fringes is given by

$\Rightarrow \beta =\dfrac{\lambda D}{d}\quad [\lambda=\text{Wavelength of light}]\\\\\text{Insert the values}\\\\\Rightarrow 2.88\times 10^{-3}=\dfrac{\lambda \cdot 2.24}{0.510\times 10^{-3}}\\\\\Rightarrow \lambda =\dfrac{2.88\times 10^{-3}\times 0.510\times 10^{-3}}{2.24}\\\\\Rightarrow \lambda =0.6557\times 10^{-6}\ m\\\Rightarrow \lambda =655.7\ nm$

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3 years ago

An 80 kg astronaut has gone outside his space capsule to do some repair work. Unfortunately, he forgot to lock his safety tether

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Answer:

the time taken by the astronaut to reach safety = 9.8 hr

Explanation:

The equation for intensity can be written as :

$I = \frac{P}{A}$

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Replacing that into the above previous equation; we have:

$\frac{P}{Ac}=\frac{F}{A}$

$F = \frac{P}{c}$

However ; the force needed to push the astronaut is as follows:

F = ma

where ;

m = mass of the astronaut and a = its acceleration

we as well say;

$\frac{P}{c} = ma$

$a = \frac{P}{mc}$

Replacing P with 1000 W ; m with 80 kg and $3*10^{8} \ m/s$ for c

Then; a = $\frac{1000 \ W}{(80)(3.0*10^8)}$

a = $4.2*10^{-8} \ m/s$

It is also known that the battery will run for one hour and after which the battery on the laser will run out

Then to determine the change in the position after the first hour ; we have:

$\Delta x_1 = \frac{1}{2}*4.2*10^{-8} \ m/s^2 ) (1.0 \ h)^2$

$\Delta x_1 = \frac{1}{2}*4.2*10^{-8} \ m/s^2 ) (1.0 *3600 s)^2$

= 0.27 m

Furthermore, the final velocity of the astronaut is determined as:

$v_1 = at_1$

where ;

$v_1 = final \ velocity$

replacing $t_1 = 1.0 \ h$ and a = $4.2*10^{-8} \ m/s$ ; Then:

$v_1 = (4.2*10^8 \ m/s * 1.0 \ h * \frac{ 3600\ s}{1.0 \ h})$

$v_1 = 1.51 *10^{-4} \ m/s$

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To find the final distance traveled by the astronaut ;we have:

$\Delta x_2 = d - \Delta x_1$

where;

$\Delta x_2$ = the final distance

d = total distance

So;

$\Delta x_2 = 5 m - 0.27 m \\ \\ \Delta x_2 = 4.73 \ m$

The time taken to reach the final distance can be calculated as:

$t_2 = \frac{\Delta x_2 }{v_1}$

where;

$t_2$ = is the time to reach the final distance

Replacing 4.73 for ${\Delta x_2 }$ and $1.51*10^{-4}$ m/s for $v_1$

$t_2 = \frac{4.73 \ m }{1.51*10^{-4} \ m/s}$

$t_2 = 31500 \ s (\frac{1.0 \ h}{3600 \ s} )$

$t_2 = 8.8 \ h$

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Hence ; the time taken by the astronaut to reach safety = 9.8 hr

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3 years ago