A 5.00 Ω, a 10.0 Ω, and a 15.0 Ω resistor are connected in
1 answer:
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Answer:
spacing between the slits is 405.32043 × m
Explanation:
Given data
wavelength = 610 nm
angle = 2.95°
central bright fringe = 85%
to find out
spacing between the slits
solution
we know that spacing between slit is
I = 4 × cos²∅/2
so
I/4 = cos²∅/2
here I/4 is 85 % = 0.85
so
0.85 = cos²∅/2
cos∅/2 = √0.85
∅ = 2 × 0.921954
∅ = 45.56°
∅ = 45.56° ×π/180 = 0.7949 rad
and we know that here
∅ = 2π d sinθ / wavelength
so
d = ∅× wavelength / ( 2π sinθ )
put all value
d = 0.795 × 610× / ( 2π sin2.95 )
d = 405.32043 × m
spacing between the slits is 405.32043 × m
Ans: Beat frequency = 
= 4Hz
Explanation:
The beat frequency is equal to the absolute value of the difference in frequency of the two waves. In other words, the number of beats per second is equal to the difference in frequency. It is due to the destructive and constructive interference. <span>According to this interference, sound will be soft or loud.
Hence. the formula is:
</span>Beat frequency =
<span>
Since,
</span>
Therefore,
Beat frequency =
=> Beat frequency =
-i
Explanation:
The beat frequency is equal to the absolute value of the difference in frequency of the two waves. In other words, the number of beats per second is equal to the difference in frequency. It is due to the destructive and constructive interference. <span>According to this interference, sound will be soft or loud.
Hence. the formula is:
</span>Beat frequency =
<span>
Since,
</span>
Therefore,
Beat frequency =
=> Beat frequency =
-i