Ask question

Ask question

All categories

2 years ago

7

Diffraction spreading for a flashlight is insignificant compared with other limitations in its optics, such as spherical aberrat

ions in its mirror. To show this, calculate the minimum angular spreading in rad of a flashlight beam that is originally 5.85 cm in diameter with an average wavelength of 580 nm

1 answer:

lyudmila [28]2 years ago

5 0

Answer:

$\theta_{min} = 1.21 \times 10^{-5}\ rad$

Explanation:

given,

diameter of the beam (d)= 5.85 cm

= 0.0585 m

average wavelength of the(λ) = 580 n m

angle of of spreading = ?

according to the Rayleigh Criterion the minimum angular spreading, for a circular aperture, is

$\theta_{min} = 1.22\ \dfrac{\lambda}{d}$

$\theta_{min} = 1.22\ \dfrac{580 \times 10^{-9}}{0.0585}$

$\theta_{min} = 1.22\times 9.145 \times 10^{-6}$

$\theta_{min} = 1.21 \times 10^{-5}\ rad$

the minimum angle of spreading is $\theta_{min} = 1.21 \times 10^{-5}\ rad$

You might be interested in

Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of

Charra [1.4K]

To solve the problem it is necessary to apply the Torque equations and their respective definitions.

The Torque is defined as,

$\tau = I \alpha$

Where,

I=Inertial Moment

$\alpha =$ Angular acceleration

Also Torque with linear equation is defined as,

$\tau = F*d$

Where,

F = Force

d= distance

Our dates are given as,

R = 30 cm = 0.3m

m = 1.5 kg

F = 20 N

r = 4.0 cm = 0.04 m

t = 4.0s

Therefore matching two equation we have that,

$d*F = I\alpha$

For a wheel the moment inertia is defined as,

I= mR2, replacing we have

$d*F= \frac{mR^2a}{R}$

$d*F= mRa$

$a = \frac{rF}{ mR}$

$a = \frac{0.04*20}{1.5*0.3}$

$a=1.77 m/s^2$

Then the velocity of the wheel is

$V = a *t \\V=1.77*4 \\V=7.11 m/s$

Therefore the correct answer is D.

4 0

2 years ago

The banking angle in a turn on the Olympic bobsled track is not constant, but increases upward from the horizontal. Coming aroun

zlopas [31]

Answer:

the curve inclination is increased so that a weight component helps keep the car on track

Explanation:

In the sledging competition these devices go at quite high speeds over 100 km/h, so when reaching the curves the friction force is not enough to keep the car on the track. For this reason, the curve inclination is increased so that a weight component helps keep the car on track.

In general we can solve Newton's second law for this case, with the condition of no friction, it is found that

V² = r g tan θ

Where V is the maximum velocity, r is the radius of the curve a, θ is the angle of the inclination

8 0

2 years ago

Consider N non-interacting diatomic molecules stuck on a metal surface. Each molecule can either lie flat on the surface, in whi

Art [367]

Answer:

Detailed solution given in diagram

7 0

2 years ago

A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p

kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

$y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2$

$V(t) = V_0 + a * t$ ,

Where $y_0$ its our initial height, $V_0$ our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

$y(t) = 38 \frac{m}{s} * t - \frac{1}{2} g t^2$

$V(t) = 38 \frac{m}{s} - g * t$

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

$0 m = 38 \frac{m}{s} - g * t$

and obtain:

$t = 38 \frac{m}{s} / g$

$t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}$

$t = 3.9 s$

Plugin this time on our first equation we find:

$y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2$

$y=73.67 m$

6 0

2 years ago

The electric potential in a region that is within 2.00 mm of the origin of a rectangular coordinate system is given by V=Axl+Bym

Fynjy0 [20]

Answer:

Given the potential, $V = Ax^l+By^m+Cz^n+D$

The components of the electric field are:

$E_x = \frac{-dV}{dx} = -Alx^l^-^1$

$E_y = \frac{-dV}{dy} = - Bmy^m^-^1$

$E_z = \frac{-dV}{dz} = - nCzn^n^-^1$

Let's calculate the potential difference for all given points.

$V(0, 0, 0) = 10V => Ax^l+By^m+Cz^n+D = 10$

$=> D = 10$

$V(1, 0, 0) = 4V => A + 10 = 4$

Solving for A, we have:

$A = 4 - 10$

$A = -6$

$V(0, 1, 0) = 6V => B + 10 = 6$

Solving for B, we have:

$B = 6 - 10$

$B = -4$

$V(0, 0, 1) = 8V => C + 10 = 4$

Solving for C, we have:

$C = 8 - 10$

$C = -2$

For all given points, let's calculate the magnitude of electric field as follow:

$E_x (1, 0, 0) = 16 => - Alx^l^-^1 = 16$

$Al = -16$

Solving for l, we have:

$l = \frac{-16}{A}$

From above, A = -6

$l = \frac{-16}{-6}$

$l = \frac{8}{3}$

$E_y (0, 1, 0) = 16=> Bmy^m^-^1 = 16$

$Bm = -16$

Solving for m, we have:

$m = \frac{-16}{A}$

From above, B = -4

$m = \frac{-16}{-4}$

$m = 4$

$E_y (0, 0, 1) = 16=> nCz^n^-^1 = 16$

$nC = - 16$

Solving for n, we have:

$n = \frac{-16}{C}$

From above, C = -2

$n = \frac{-16}{-2}$

$n = 8$

4 0

2 years ago