A diffraction pattern is formed on a screen 150 cm away from a 0.500-mm-wide slit. Monochromatic 546.1-nm light is used. Calculate the fractional intensity I/Imax at a point on the screen 4.10 mm from the center of the principal maximum.

Solution :

The expression for the intensity of light is given by :

For a small angle, θ

sin θ = tan θ

Therefore the above equation becomes,

The given data is

λ = 546.1 nm

L = distance between the slit and the screen = 140 cm

  = 1.40 m

a = width of the slit

  =

Therefore,

Therefore the fractional intensity is  

where’s the list ? the question is incomplete

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form