Anarrow single slit is illuminated with angle infrared laser with λ = 1152 nm. the center of the tenth dark band (minimum) lies at the angle of 6° off the central axis.
(a)- what is the width of the slit?
(b)-how does this angle change if we immerse the whole experiment in water? (n = 1.33).

a) 110 *10^-6 m (110 μm); b) 82.86*10^-6 m (82.86 μm).

Explanation: In order to explain this problem we have to consider the expresion for the dark fringes in a difraction pattern for a single narrow slit. It is given by:

a*sin (θ)= m*λ where a is the slit width. θ is the angle corresponding the m dark fringe from the central axis. λ is the wavelength of the incident light.

Then we have m=10 and θ=6° so;

a=10*1152*10^-9/Sin(6°)=110 *10^-6 m

Finally if the whole system is inmmersed in water (n=1.33), we have to add the refractive index in the path difference then: a*n*sin(θ)

a*n*sin (θ)= m*λ  then

a= m*λ/(n*sin (θ))=10*1152*10^-9/1.33*Sin(6°)= 82.86* 10^-6 m

answer: d. a cme increases the amount of energy in the solar wind

explanation:

coronal mass ejections, also called cmes, are large clouds of plasma and magnetic field that erupt from the sun. only when the cloud is aimed at earth will the cme hit earth and therefore cause impacts. high-speed solar wind streams come from areas on the sun known as coronal holes.