A В C D At which point will an image be formed? ОА Ов Ос OD

Light with a wavelength of 612 nm incident on a double slit produces a second-order maximum at an angle of 25 degree. What is th

erastova [34]

Answer:

Speration between the slits $d=2.86\ \mu m$

Explanation:

Given that,

Wavelength of light, $\lambda=612\ nm=612\times 10^{-9}\ m$

It is incident on a double slit produces a second-order maximum at an angle of 25 degree.

The condition for maxima is given as :

$dsin\theta=m\lambda$

d = seperation between the slits

m = order, m = 2

$d=\dfrac{m\lambda}{sin\theta}$

$d=\dfrac{2\times 612\times 10^{-9}\ m}{sin(25)}$

d = 0.00000289

or

$d=2.86\mu m$

So, the seperation between the silts is $2.86\ \mu m$ . Hence, this is the required solution.

3 0

3 years ago

(b) Figure 4 shows a car travelling on a motorway.

Alik [6]

Answer:

To calculate anything - speed, acceleration, all that - we need data. The more data we have, and the more accurate that data is, the more accurate our calculations will be. To collect that data, we need to measure it somehow. To measure anything, we need tools and a method. Speed is a measure of distance over time, so we'll need tools for measuring time and distance, and a method for measuring each.

Conveniently, the lamp posts in this problem are equally spaced, and we can treat that spacing as our measuring stick. To measure speed, we'll need to bring time in somehow too, and that's where the stopwatch comes in. A good method might go like this:

Press start on the stopwatch right as you pass a lamp post
Each time you pass another lamp post, press the lap button on the stopwatch
Press stop after however many lamp posts you'd like, making sure to hit stop right as you pass the last lamp post
Record your data
Calculate the time intervals for passing each lamp post using the lap data
Calculate the average of all those invervals and divide by 40 m - this will give you an approximate average speed

Of course, you'll never find an *exact* amount, but the more data points you have, the better your approximation will become.

5 0

2 years ago

10. A satellites is in a circular orbit around the earth at a height of 360 km above the earth’s surface. What is its time perio

Afina-wow [57]

Answer:

Orbital speed=8102.39m/s

Time period=2935.98seconds

Explanation:

For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)

V2R+h=g(R2(R+h)2)

V=√g(R2R+h)

V= sqrt(9.8 × (6371000)^2/(6371000+360000)

V= sqrt(9.8× (4.059×10^13/6731000)

V=sqrt(65648789.18)

V= 8102.39m/s

Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)

T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)

T=sqrt(3.40×10^21)/ (3.99×10^14)

T= sqrt(0.862×10^7)

T= 2935.98seconds

4 0