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LenaWriter [7]
3 years ago
7

A piece of curved glass has a radius of curvature of r = 10.8 m and is used to form Newton's rings, as in the drawing. Not count

ing the dark spot at the center of the pattern, there are one hundred dark fringes, the last one being at the outer edge of the curved piece of glass. The light being used has a wavelength of 652 nm in vacuum. What is the radius R of the outermost dark ring in the pattern? (Hint: Note that r is much greater than R, and you may assume that tan(θ) = θ for small angles, where θ must be expressed in radians.)
Physics
2 answers:
Sergeeva-Olga [200]3 years ago
8 0

Answer:

The Radius is 2.65cm

Explanation:

MAVERICK [17]3 years ago
7 0

Answer:

Radius of the outer most dark fringe is 2.65 cm

Solution:

As per the question:

Radius of curvature of the glass, r = 10.8 m

No. of dark fringes, n = 100

Wavelength of light, \lambda = 652\ nm = 652\times 10^{- 9}\ m

Now,

To calculate the radius R of the outermost ring:

Radius of the dark fringe of nth order is given by:

R^{2} = nr\lambda = 100\times 10.8\times 652\times 10^{- 9} = 7.042\times 10^{- 4}

R = \sqrt{7.042\times 10^{- 4}} = 0.0265\ m = 2.65\ cm

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alexdok [17]

Answer:

Answer is D. 8.04 x 10^4 J

Explanation:

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8 0
2 years ago
. A vertical electric field is set up in space to compensate for the gravitational force on a point charge. What is the required
Delicious77 [7]

(a) The required magnitude of the electric field when the point charge is an electron is 5.57 x 10⁻¹¹ N/C.

(b) The required magnitude of the electric field when the point charge is an proton is 1.02 x 10⁻⁷ N/C.

<h3>Magnitude of electric field </h3>

The magnitude of electric field is given by the following equation.

F = qE

But F = mg

mg = qE

E = mg/q

where;

  • E is the electric field
  • m is mass of the particle
  • g is acceleration due to gravity
  • q is charge of the particle
<h3>For an electron</h3>

E = (9.11 x 10⁻³¹ x 9.8)/(1.602 x 10⁻¹⁹)

E = 5.57 x 10⁻¹¹ N/C

<h3>For proton</h3>

E = (1.67 x 10⁻²⁷ x 9.8)/(1.602 x 10⁻¹⁹)

E = 1.02 x 10⁻⁷ N/C

Thus, the required vertical electric field is greater when the charge is proton.

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

4 0
2 years ago
Please need help ASAP. Would really appreciate it.
a_sh-v [17]

if i renember correctly its b

3 0
3 years ago
which fire extinguisher is most appropriate to put out a fire that involves a stack of burning newspapers
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Monoammonium phosphate effectively smothers the fire, while sodium bicarbonate induces a chemical reaction which extinguishes the fire. Fire extinguishers with a Class C rating are suitable for fires in “live” electrical equipment.

6 0
2 years ago
Read 2 more answers
Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted
galina1969 [7]

Answer:

1) When d_{o} < d_{i} (hence  d_{o} < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When d_{o} > d_{i} (and d_{o} < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}

Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}

Where:

f = Focal length of the mirror = R/2

d_{i} = Image distance from the mirror

d_{o} = Object distance from the mirror

h_{i} = Image height

h_{o} = Object height

d_{o} is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

d_{i} is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When d_{o} < d_{i} (hence  d_{o} < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When d_{o} > d_{i} (and d_{o} < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

5 0
3 years ago
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