Question

a) How wide in m is a single slit that produces its first minimum for 635-nm light at an angle of 19.0°?

b) At what angle in degrees will the second minimum be?

Answer 1

given
wavelength, lamda = 635 nm = 635*10^-9 m
for first minimum, theta1 = 19.0 degrees

a) let a is the slit size.

we know, for first minimum, a*sin(theta1) = lamda

==> a = lamda/sin(theta1)

= 635*10^-9/sin(19)

= 1.95*10^-6 m <<<<<<<------------Answer

b) for second minimum, a*sin(theta2) = 2*lamda

sin(theta2) = 2*lamda/a

= 2*635*10^-9/(1.95*10^-6)

= 0.65128

theta2 = sin^-1(0.65128)

= 40.6 degrees <<<<<<<<<<<<<--------------------Answer