The distance between slit and the screen is 1.214m.
To find the answer, we have to know about the width of the central maximum.
<h3>How to find the distance between slit and the screen?</h3>
- It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.
- We have the expression for slit width w as,
where, d is the distance between slit and the screen, and a is the slit width.
- Thus, distance between slit and the screen is,
Thus, we can conclude that, the distance between slit and the screen is 1.214m.
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Answer:
d = 68.5 x 10⁻⁶ m = 68.5 μm
Explanation:
The complete question is as follows:
An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is 1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?
The answer can be given by using the formula derived from Young's Double Slit Experiment:
where,
d = slit separation = ?
λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m
L = distance from screen (detector) = 1.7 m
y = distance between bright fringes = 15.7 mm = 0.0157 m
Therefore,
<u>d = 68.5 x 10⁻⁶ m = 68.5 μm</u>
Explanation:
since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.
We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.
range can be calculated by the formula :-
u is the velocity during its take off and is the angle at which its thrown
Given that
- u = 8m/ s
- = 40°
calculating range using the above formula
value of sin 80 = 0. 985
Hence,
It depends, You have to have the length and the width of the crest wave.
Answer:
A. 1.64 J
Explanation:
First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:
where
n is the number of moles
m = 1.4 mg = 0.0014 g is the mass of mercury
Mm = 200.6 g/mol is the molar mass of mercury
Substituting, we find
Now we have to find the number of atoms contained in this sample of mercury, which is given by:
where
n is the number of moles
is the Avogadro number
Substituting,
atoms
The energy emitted by each atom (the energy of one photon) is
where
h is the Planck constant
c is the speed of light
is the wavelength
Substituting,
And so, the total energy emitted by the sample is