A diffraction-limited lens can focus light to a 10-μ m-diameter spot on a screen. Do the following actions make the spot diameter larger, smaller, or leave it unchanged? a. Decreasing the wavelength of light: b. Decreasing the lens diameter c. Decreasing the focal length: d. Decreasing the lens-to-screen distance:

a) the size of the sport DECREASES  , b)the spot size is increased

c)  it has no effect  and d) spot size DECREASES

Explanation:

The expression for constructive interference in diffraction is

         a sin θ = m λ

the chaff of the spot is given by the minimum interference m = 1

           a sin θ = λ

let's use trigonometry to find the angle

         tan λ= y / L

         tan θ = sin θ / cos t = sin θ

         sin θ = y / L

          a y / L = lam

in the case of materials with circular dome, the problem must result in polar coordinates, with which a numerical constant is introduced

          y = 1.22 L / a lam

let's analyze each electrode action in this equation

a) Decrease lam

         y ’= 1.22 L / a lm2

          lam2 <lamo

          y '<y

therefore the size of the sport DECREASES

b) decreases the lens diameter

       a '<a

         y ’= 1.22 L / a’ lam

         y '> y

the spot size is increased

c) Decrease the focal length

The expression does not depend on the focal length, so it has no effect

d) decrease in L

         y ’= 1.22 L’ / a lam

spot size DECREASES

he . if i may ask what i your question exactly?

explanation:

real, smaller and upside-down.   so the answer should be b.