e) Suppose you are doing a double slit diffraction experiment with monochromatic

Question

e) Suppose you are doing a double slit diffraction experiment with monochromatic (single frequency) light. Suddenly the atmosphere changes from air to a gas with a higher index of refraction. What will happen to the pattern?

the distance between the maxima will decrease, the color of the light will change to a higher frequency   the distance between the maxima will decrease, the color of the light will remain the same   the distance between the maxima will increase, the color of the light will change to a higher frequency the distance between the maxima will remain the same, the color of the light will remain the same the distance between the maxima will decrease, the color of the light will change to a lower frequency the distance between the maxima will increase, the color of the light will remain the same the distance between the maxima will increase, the color of the light will change to a lower frequency the distance between the maxima will remain the same, the color of the light will change to a higher frequency the distance between the maxima will remain the same, the color of the light will change to a lower frequency

Be careful of units!

In procedure 2: suppose a beam of monochromatic (single frequency) light, wavelenth 540 nm falls on a double slit whose slits are 0.0451 mm apart. The screen is 78.7 cm from the slits.

a) Calculate the distance from the central maximum to the first maximum.
d0 to 2 =  cm

b) Calculate the distance from the central maximum to the tenth minimum.
d0 to 10 =  cm

Suppose procedure 2 is done with a diffraction grating with 642,000 lines per centimeter.

a) Find d, the distance between the lines.
d =  cm

b) Light hits this diffraction grating, and goes onto a screen distance 91.1 cm away, creating a diffraction pattern where the maxima are 41.7 mm apart. Find the wavelength of the light being used.
NOTE: Depending on the numbers, the light may not be visible!
? =  nm

Explanation / Answer

so e) y = m lambda L/d

and lambda = lambda0/n

so if n increases, lambda goes down so y will get smallers

but frequency doesnt change

the distance between the maxima will decrease, the color of the light will remain the same

2)

y = 1*540.0E-9*.787/(0.0451E-3)= 9.4E-3 m = 0.94 cm

b) now y = 10*540.0E-9*.787/(0.0451E-3) = 94.2E-3 m = 9.42 cm

3) a) d = 1cm/lines = 1/642000= 1.56E-6 cm

y = m lambda L/d

so
41.7E-3 = lambda*0.911/1.56E-8

lambda = 7.14E-10 m = 0.714 nm

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