Question

1. A beam of green light is split by thin double slit with separation of 0.0500mm and incident upon a screen some distance away. The angles of the first and second maximums are .584^o and 1.17^o respectively. What is the wavelength of the green light?

2. A single, monochromatic indigo light source is shined through an etched, flat prism with a slit separation of .0250mm. The resulting interference pattern is viewed on a screen 1.25m away. The third maximum is found to be 6.6cm from the central maximum. What is the wavelength of the indigo light? Where are the first and second maxima?

Please show ALL work and explain if necessary. Thank you!

Answer 1

The equation for double slit diffraction is dsin? = ?M

Where d is the slit seperation, ? is the angle of the maximum, M is the maximum, and ? is the wavelength. We will solve for ? in each case (M = 1 and M = 2) and take the average.

M = 1

(0.05) *sin(.584) = 1 * ?

? = 5.096 * 10^-4 mm = 509.6 nano-meters

M = 2

(0.05) *sin(1.17) = 2* ?

? = 5.105 * 10^-4 mm = 510.5 nano-meters.

Your wavelenth will be the average, or (510.5 + 509.6)/2 = 510.05 nano-meters

It's also helpful to notice that this is close to what we would expect for green light, which is 520-550 nano-meters.

2) the ditance from the central maxima is n*(lambda)*distance between the screen and slit/slit seperation ... here,n=3 slit seperation is .0250mm and distnace between screena and slit is 1.25m and distacne from central maxima is 6.6cm...substituting the values lambda is calculated

maL 111人し

so, .066=3* ?*1.25/(.025*10^-3) =>?=4.4*10^-7