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1 year ago

how wide must a narrow slit be if the first diffraction minimum occurs at ±12° with laser light of 633 nm?

Physics

1 answer:

Charra [1.4K]1 year ago

8 0

The width of narrow slit will be 3 μm if the diffraction minima occurs with 633 nm wavelength.

The angular width of the central maximum is the angle between the two first order minima (on either side of the center).
The diffraction formula's central maximum width is inversely related to the slit width.
The central maximum widens as the slit width gets smaller, and gets narrower as it gets larger. From this behavior, it can be concluded that light bends more as the aperture's size decreases.
Monochromatic light is passed through a single slit with a finite width in a single-slit experiment, and a similar pattern is seen on the screen.
As we move farther from the central maximum in the single-slit diffraction pattern, compared to the double-slit diffraction pattern, the width and intensity decrease.

Δ=2L/α

Δ=3 x 10⁻⁶m

=3micro meter.

To study more on Monochromatic light -

<u>brainly.com/question/14896284</u>

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A merry-go-round with a a radius of R = 1.99 m and moment of inertia I = 194 kg-m2 is spinning with an initial angular speed of

Aleksandr [31]

Answer:

Part 1)

$L_1 = 185.2 kg m^2/s^2$

Part 2)

$L_2 = 663.07 kg m^2/s^2$

Part 3)

$L = 663.07 kg m^2/s^2$

Part 4)

$\omega = 1.83 rad/s$

Part 5)

$F_c = 453.6 N$

Explanation:

Part a)

Initial angular momentum of the merry go round is given as

$L_1 = I \omega$

here we know that

$I = 194 kg m^2$

$\omega = 1.47 rad/s^2$

now we have

$L_1 = 194 \times 1.47$

$L_1 = 185.2 kg m^2/s^2$

Part b)

Angular momentum of the person is given as

$L = mvR$

so we have

$m = 68 kg$

$v = 4.9 m/s$

R = 1.99 m

so we have

$L_2 = (68)(4.9)(1.99)$

$L_2 = 663.07 kg m^2/s^2$

Part 3)

Angular momentum of the person is always constant with respect to the axis of disc

so it is given as

$L = 663.07 kg m^2/s^2$

Part 4)

By angular momentum conservation of the system we will have

$L_1 + L_2 = (I_1 + I_2)\omega$

$185.2 + 663.07 = (194 + 68(1.99^2))\omega$

$848.27 = 463.28 \omega$

$\omega = 1.83 rad/s$

Part 5)

Force required to hold the person is centripetal force which act towards the center

so we will have

$F_c = m\omega^2 R$

$F_c = 68(1.83^2)(1.99)$

$F_c = 453.6 N$

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3 years ago

Which of the following occurs on March 21?

iren [92.7K]

Vernal equinox, for spring.

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3 years ago

Describe another model in science that you have learned about that you think may have been revised over time as scientists have

Degger [83]

Scientists gather a new evidence like let me say for example jj Thompson said that an atoms is the least particles of an element but according to our definition an atom is the smallest particles of an element so you see

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3 years ago

Two 2.3 kg bodies, A and B, collide. The velocities before the collision are and . After the collision, . What are (a) the x-com

myrzilka [38]

Missing details in the question:

The velocities before collision are

$u_A=(40i+49j) m/s$

$u_B=(35i+11j) m/s$

After the collision:

$v_A=(14i+21j) m/s$

(a) 61 m/s

We can solve the problem by simply treating separately the x- and the y-components of the motion.

Here we want to analzye the motion along x. We have:

$u_A = 40 m/s$ is the initial velocity of A along the x-direction

$u_B = 35 m/s$ is the initial velocity of B along the x-direction

$v_A = 14 m/s$ is the final velocity of A along the x-direction

$v_B = ?$ is the final velocity of B along the x-direction

Since the total momentum along the x-direction must be conserved, we can write

$mu_A + mu_B = mv_A + mv_B$

where

m = 2.3 kg is the mass of the two bodies. Since the mass is the same, we can eliminate it from the equation,

$u_A + u_B = v_A + v_B$

And so, we find the final velocity of B along the x-direction:

$v_B = u_A + u_B - v_A=40+35-14=61 m/s$

(b) 39 m/s

Similarly to what we did in part a), here we analyze the conservation of momentum along the y-direction.

We have:

$u_A = 49 m/s$ is the initial velocity of A along the y-direction

$u_B = 11 m/s$ is the initial velocity of B along the y-direction

$v_A = 21 m/s$ is the final velocity of A along the y-direction

$v_B = ?$ is the final velocity of B along the y-direction

Since the total momentum along the y-direction must be conserved, we can write

$mu_A + mu_B = mv_A + mv_B$

Since the mass is the same, we can eliminate it from the equation,

$u_A + u_B = v_A + v_B$

And so, we find the final velocity of B along the y-direction:

$v_B = u_A + u_B - v_A=49+11-21=39 m/s$

c) +615 J

Here we have to find the total kinetic energy before and after the collision first.

First, we have to find the speed of each object before and after the collision. We have:

$u_A = \sqrt{40^2+49^2}=63.2 m/s\\u_B = \sqrt{35^2+11^2}=36.7 m/s\\v_A = \sqrt{14^2+21^2}=25.2 m/s\\v_B = \sqrt{61^2+39^2}=72.4 m/s$

So, the total kinetic energy before the collision was

$K_i = \frac{1}{2}mu_A^2+\frac{1}{2}mu_B^2 = \frac{1}{2}(2.3)(63.2)^2+\frac{1}{2}(2.3)(36.7)^2=6143 J$

While after the collision

$K_f = \frac{1}{2}mv_A^2+\frac{1}{2}mv_B^2 = \frac{1}{2}(2.3)(25.2)^2+\frac{1}{2}(2.3)(72.4)^2=6758 J$

So, the change in kinetic energy is

$\Delta K = K_f - K_i = 6758-6143 = +615 J$

(note that the system cannot gain kinetic energy in the collision, unless there is an external force acting on it)

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katrin [286]

Its Thales of Miletus

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