A 0.45-mm-wide slit is illuminated by light of wavelength 590 nm. What is the width (in mm) of the central maximum on a screen 1.5 m behind the slit?

And if we convert this into mm we got:

Step-by-step explanation:

For this case we have the following info givenL

represent the wavelength

represent the width of the single slit

represent the longitude

And we want to find the maximum width of the central maximum in mm, so we can use the following formula:

And if we solve for w the width we got:

And replacing we got:

And if we convert this into mm we got:

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.

answer: c

(2, -4)

goes 7 to the

(9, 4)

easy