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Sergeeva-Olga [200]
2 years ago
5

A diffraction-limited laser of length L and aperture diameter 0.50 cm generates light of wavelength 700nm . If the beam is direc

ted at the surface and the radius of the illuminated area on the surface is approximately 45cm, how far away is the surface
Physics
1 answer:
Hoochie [10]2 years ago
4 0

The distance of the surface illuminated by the light of the given wavelength is 5,269.32 m.

<h3>Distance of the surface</h3>

The distance of the surface illuminated by the light of the given wavelength is calculated as follows;

D = (rd)/(0.61λ)

where;

  • r is the radius of the illuminated area = 45 cm = 0.45 m
  • d is the diameter of the diameter = 0.5 cm = 0.005 m
  • λ is wavelength = 700 nm = 700 x 10⁻⁹ m

D = (0.45 x 0.005) / (0.61 x 700 x 10⁻⁹)

D = 5,269.32 m

Thus, the distance of the surface illuminated by the light of the given wavelength is 5,269.32 m.

Learn more about wavelength here: brainly.com/question/10728818

#SPJ1

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28

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3 years ago
Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at
dedylja [7]

Answer:

50.91 \mu C

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for q_{1} and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted byq_{1} on q. Then we can know the magnitude of the force exerted by q_{2} about q, finally this will allow us to know the magnitude of q_{2}

q_{1} exerts a force on q in +y direction, and q_{2} exerts a force on q in -y direction.

F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\

The net force on q is:

F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}

Rewriting for q_{2}:

q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C

8 0
3 years ago
If it requires 8.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re
jenyasd209 [6]

Answer:

The amount of work done required to stretch spring by additional 4 cm is 64 J.

Explanation:

The energy used for stretching spring is given by the relation :

E = \frac{1}{2}kx^{2}           .......(1)

Here k is spring constant and x is the displacement of spring from its equilibrium position.

For stretch spring by 2.0 cm or 0.02 m, we need 8.0 J of energy. Hence, substitute the suitable values in equation (1).

8 = \frac{1}{2}\timesk\times k \times(0.02)^{2}

k = 4 x 10⁴ N/m

Energy needed to stretch a spring by 6.0 cm can be determine by the equation (1).

Substitute 0.06 m for x and 4 x 10⁴ N/m for k in equation (1).

E = \frac{1}{2}\times4\times10^{4}\times (0.06)^{2}

E = 72 J

But we already have 8.0 J. So, the extra energy needed to stretch spring by additional 4 cm is :

E = ( 72 - 8 ) J = 64 J

7 0
3 years ago
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