Question

fill in the table. answer the questions.

Part 2. Single Slit Diffraction 1 T 1 DT SINGLE SLIT L If the viewing screen is far away, the rays heading for any point on the screen are essentially parallel. Consider the waves emanating from the upper half and lower half of the slit. Destructive interference, a dark fringe, occurs if the path difference from any point in the upper half of slit and the corresponding points in the lower half of the slit is an integer pultiple of 1/2 so that the total electric field is zero. The angle at which this occurs can be seen from the diagram to be D sin 01 [first minimum) The intensity is a maximum at 0= 0 and decreases to zero at the angle given by equation (3). At a larger angle there will be a bright line, but not nearly as bright as the central spot at 00. As the path difference becomes an integer multiple of 1/2, there will again be a minimum of zero intensity when D sin mami

IDK the measured values. You have to figure it out thru a simulation.

Answer 1

For Red light -

Slit width = D = 1600nm = 1.6um

Distance from slit to screen = L = 50cm

Distance from central to first dark spot = x = 20.3cm

Wavelength calculated = lamda(c) = D*x/L = 1.6*10^-6 * 20.3*10^-2 / 50*10^-2 = 649.6nm

since,        sin@ = x/L

wavelength measured = lamda(m) = 650 nm

%error = | lamda(c) - lamda(m) | / lamda(m) * 100 = |649.6 - 650|/650 *100= 0.06%

For violet light -

Slit width = D = 1.6um

Distance from slit to screen = L = 50cm

Distance from central to first dark spot = x = 13.5cm

Wavelength calculated = D*x / L = 1.6*10^-6 * 13.5*10^-2 / 50*10^-2 = 432nm

wavelength measured = 430nm

%error = |lamda(c) - lamda(m)| / lamda(m) * 100 = |432 - 430|/430 * 100 = 0.46%

1. The pattern got stretched as wavelength was increased. Thus longer wavelength is diffracted more than shorter wavelength.

2. This happened as distance of minima from the central maxima is directly proportional to the wavelength. Hence when wavelength is increased the pattern stretches out and all maxima and minima move further away from central maxima.

3. When d i.e. width of the slit is increased the diffraction pattern shrinks. The maximas and minimas come closer to the central maxima. When d is decreased the pattern expands/stretches. Thus narrow slit(d decreased) diffracts more than wider slits(d increased)

4. d = lamda/sin@

sin @ = x/L

d is inversely proportional to x.

The increase in d will result in decrease of distance between 1st minima(as per the given formula but all other maximas and minimas will experience identical result) and central maxima.

Whereas decrease of d will result in increase of the x and pattern stretches out.

For d = 40 um

For d = 20 um

5. A ring is observed around the sun due to diffraction.

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