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

the results is a decrease in air temperature with an increase of altitude.

explanation: