This is a Fraunhofer single slit experiment, where the light passing through the slit produces an interference pattern on the screen, and where the dark bands (minima of diffraction) are located at a distance of

from the center of the pattern. In the formula, m is the order of the minimum,

the wavelenght,

the distance of the screen from the slit and

the width of the slit.
In our problem, the distance of the first-order band (m=1) is

. The distance of the screen is D=86 cm while the wavelength is

. Using these data and re-arranging the formula, we can find a, the width of the slit:
Answer:
Tension on tendon = 1669800N
Explanation:
Detailed explanation and calculation is shown in the image below
Answer:
0.000003782 m
0.000001891 m
0.000001197125 m
Explanation:
= Wavelength = 248 nm
D = Diameter of beam = 1 cm
f = Focal length = 0.625 cm
The angle is given by

The width is given by

The required width is 0.000003782 m
Minimum resolvable line separation is given by

The minimum resolvable line separation between adjacent lines is 0.000001891 m
when 

The new minimum resolvable line separation between adjacent lines is

It's hard to tell exactly what's happening in that 110 cm that you marked over the wave. What is under the ends of the long arrow ? How many complete waves ? I counted 4.5 complete waves ... maybe ?
If there are 4.5 complete waves in 110cm, then the length of 1 wave is (110/4.5)=24.44cm.
Frequency = speed/wavelength
Frequency = 2m/s /0.2444m
Frequency = 8.18 Hz
c] Electrons
Electron gets transferred