1. Light of wavelength 613 nm passes through a slit 5.50 × 10 -6 m wide and fall

Question

1. Light of wavelength 613 nm passes through a slit 5.50 × 10-6 m wide and falls on a screen that is 2.78 m away. What is the distance on the screen from the center of the central bright fringe to the thrid dark fringe on either side?

2. Light waves with two different wavelengths, 632 nm and 474 nm, pass simultaneously through a single slit whose width is 7.74 × 10-5 m and strike a screen 2.00 m from the slit. Two diffraction patterns are formed on the screen. What is the distance (in cm) between the common center of the diffraction patterns and the first occurrence of the spot where a dark fringe from one pattern falls on top of a dark fringe from the other pattern?

Explanation / Answer

According to the given problem,

1.)Right, lets keep it simple...

m * = d sin (m is 3 as you are looking at the 3rd order)

so we have....

3 * (613 * 10-9) = (5.50 *10-6)*sin

= 19,53°

Now we can use trig. to find the distance.

2.78 is the straight line distance.

So, Hypotenuse* cos 19.533 = 2.78

Hypotenuse = 2.95 metres

Now use pythagoras.

2.952 = 2.782 + A2

A = 0.9873m

2.) a sin 1 = m1 minima forlonger wavelength
a sin 2 = n2 minima for shorter wavelength
m 1 = n2 since 1= 1

n / m = 1 / 2 = 632 / 474 = 1.333

n = 1.333 m
m 3, 6, 9 or some other multiple of 3 for n to bean integer
So the lowest value of m is m = 3 and n = 4

sin 1 = 3 * 6.32 * 10-7 / 7.75 *10-5 = 0.0094 = tan
tan = s / D where s = distance torequired minimum and D is distance to screen
s = 0.024 * 2 = 0.04823 m = 4.82 cm

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