In the usual setup, a double slit experiment has a bright spot at the center point on a screen which is equal?
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In the usual setup, a double slit experiment has a bright spot at the center point on a screen which is equal distance from each slit. But this can be changed by putting a very thin sheet of plastic in front of one of the slits. Suppose light of wavelength 540 nm is incident on a double slit experiment. When plastic with an index of refraction of (n = 1.65) is placed in front of one of the slits, the center point on the screen at a point equal distance from both slits is not a bright maximum, but a dark minimum. We want to find the (minimum) thickness of the plastic. (Part (a) and (b) should be written symbolically) a) Without the plastic in front of one of the slits, air of distance mλ gives constuctive interference. Then with the plastic of the same distance (thickness) in place, we must see destructive interference. What is the minimum thickness of the plastic in terms of the wavelength of the light in the plastic, λ'?__________ λ' b) What is λ' in terms of λ? ______λ c) Knowing that the thickness of air that causes contructive interference is the same thickness of plastic that causes destructive interference, solve for this thickness. _________nm
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Answer:
part b is 1/n i don't know about the rest
chackolates at Yahoo! Answers Visit the source
Other answers
part b is 1/n i don't know about the rest
chackola...
You said slit.
You said slit.
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