Answer:
<h2>Explanation:
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Which of the following experiments or phenomena provided evidence for the particle nature of light?
2 answers:
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Answer:
B) Within an atom, an electron can have only particular energies.
Explanation:
As we know that electrons have energy but apart from electrons we know that protons and neutrons inside the nucleus of atom will also have energy in them.
rest all the statements are true as we have
A) Electrons orbit the nucleus rather like planets orbiting the Sun.
TRUE, because electrons can move in stationary orbit around the nucleus
C) Electrons can jump between energy levels in an atom only if they receive or give up an amount of energy equal to the difference in energy between the energy levels.
Difference amount of energy is lost or absorbed by the electron in form of photons
D) An electron has a negative electrical charge.
Charge of an electron is given as
E) Electrons have very little mass compared to protons or neutrons
Mass of an electron is given as
mass of proton or neutron
Answer:
(A) therefore the image is
- 63 cm to the right of the lens
- the image size is -13.22 cm
- it is real
- it is inverted
(B) therefore the image is
- 63 cm to the right of the lens
- the image size is -13.22 cm
- it is real
- it is inverted
Explanation:
height of the insect (h) = 5.25 mm = 0.525 cm
distance of the insect (s) = 25 cm
radius of curvature of the flat left surface (R1) = ∞
radius of curvature of the right surface (R2) = -12.5 cm (because it is a planoconvex lens with the radius in the direction of the incident rays)
index of refraction (n) = 1.7
(A) we can find the location of the image by applying the formula below
where
- s' = distance of the image
- f = focal length
- but we first need to find the focal length before we can apply this formula
f =
f = 17.9 cm
now that we have the focal length we can apply
=\frac{1}{s'}
=\frac{1}{s'}
s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens
magnification = where y' is the height of the image, therefore
y' = x 0.525 = -13.22 cm
therefore the image is
- 63 cm to the right of the lens
- the image size is -13.22 cm
- it is real
- it is inverted
(B) if the lens is reversed, the radius of curvatures would be interchanged
radius of curvature of the flat left surface (R1) = ∞
radius of curvature of the right surface (R2) = 12.5 cm
we can find the location of the image by applying the formula below
where
- s' = distance of the image
- f = focal length
- but we first need to find the focal length before we can apply this formula
f =
f = 17.9 cm
now that we have the focal length we can apply
=\frac{1}{s'}
=\frac{1}{s'}
s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens
magnification = where y' is the height of the image, therefore
y' = x 0.525 = -13.22 cm
therefore the image is
- 63 cm to the right of the lens
- the image size is -13.22 cm
- it is real
- it is inverted