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0.23 mm far apart are the second-order fringes for these two wavelengths on a screen 1.5 m away.
<h3>Given wavelengths 710nm and 660nm,0.65mm apart two slits, and a screen 1.5m away.</h3>
Position of n the order fringe = n λ D / d
for n = 2
position = 2 λ D / d
λ = 710 nm , D = 1.5m
d = .65 x 10⁻³
position 1 = 2 x 710 x 10⁻⁹ x 1.5 / .65 x 10⁻³
= 3276.92 x 10⁻⁶ m
= 3.276x 10⁻³ m
= 3.276mm .
For λ = 660 nm
position = 2 λ D / d
λ = 660 nm , D = 1.5 m
d = .65 x 10⁻³
position 2 = 2 x 660 x 10⁻⁹ x 1.5 / .65 x 10⁻³
= 3046.15 x 10⁻⁶ m
= 3.046 x 10⁻³ m
= 3.046 mm .
Difference between their position
= 3.276mm ₋ 3.046 mm
= 0.23 mm .
To know more about Fringes refer to: brainly.com/question/15649748
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Answer:
9.875
Explanation:
w=f×s
395=40×s
make s the subject of the formula
s=395/40
=9.875
Answer: 4nmeter
Explanation: The two observer a and b will measure the same wavelength since the speed of the space craft is very small compared with the speed of light c. That is
V which is the speed of space craft 15000km/s = 15000000m/s
Comparing this with the speed of light c 3*EXP(8)m/s we have
15000000/300000000
= 0.05=0.1
Therefore the speed of the space craft V in terms of the speed of light c is 0.1c special relativity does not apply to object moving at such speed. So the wavelength would not be contracted it will remain same for both observers.
The answer is C because all the other choices would cause disaster like pollute the water if you just throw it in the sink and it would smell terrible if you put it in the garbage etc...
