Answer:
The minimum value of width for first minima is λ
The minimum value of width for 50 minima is 50λ
The minimum value of width for 1000 minima is 1000λ
Explanation:
Given that,
Wavelength = λ
For D to be small,
We need to calculate the minimum width
Using formula of minimum width
Where, D = width of slit
= wavelength
Put the value into the formula
Here, 
should be maximum.
So. maximum value of is 1
Put the value into the formula
(b). If the minimum number is 50
Then, the width is
(c). If the minimum number is 1000
Then, the width is
Hence, The minimum value of width for first minima is λ
The minimum value of width for 50 minima is 50λ
The minimum value of width for 1000 minima is 1000λ
Answer:
± (.021 ) ohm
Explanation:
In the addition of two physical quantities , the uncertainties are simply added .
So , net uncertainty in the value of R will be
± (.007 +.014)
=± (.021 ) ohm
Answer:
configuration of string:
Node - Antinode - Node or N-A-N
This is 1/2 wavelength since a full wavelength is N-A-N-A-N
f (fundamental) = V / wavelength
F0 = 300 m/s / 1 m = 100 / sec
F1 = 300 m/s / .5 m = 600 / sec
Each increase is a multiple of the fundamental since the wavelength
increases by 1/2 wavelength to keep nodes at both ends of the string
The electromagnetic spectrum is the system of frequencies that show electromagnetic radiation, respective wavelengths, and photon energies. Some examples of frequencies found on the electromagnetic spectrum are radio waves, microwaves, infrared, optical, ultraviolet, X-rays, and gamma-rays.
