Please help answer the following: 1. In a Young\'s double-slit experiment, a scr
ID: 1304482 • Letter: P
Question
Please help answer the following:
1. In a Young's double-slit experiment, a screen is placed 5m behind a double slit of .250mm seperation. If light of 589nm wavelength shines on the slit, find the value of theta corresponding to the first three darkk fringes.
2. In a double-slit experiment the location of the second bright fringe on the screen is found to be at 12.5cm. If the distance to the screen is 7m and the speration between the slits is .176mm, find the wavelength of the light used.
3. A light wave of 589nm wavelength in air passes from air into water and then into glass. Find the wavelength of water and the glass.
4. We want to prevent any light of 700nm wavelength from being reflected from a lens. What thickness of magnesium fluoride should be deposited on this lens? For this thickness is there any wavelength of light that gives maximum reflection?
5. A source of light of lambda=633nm shines on a single slit and a diffraction pattern is found on a wall 1m away. The first dark fringe is found at x = 5cm from the central maximum. Find the width of the slit.
6. A first dark fringe is observed at angular displacement theta = 15degrees due to light of frequency 6x10^14Hz passing through a diffraction grating. Find the separation of the slits in the grating.
Explanation / Answer
1. d sin theta = mL
sin theta = 3* 589nm/0.25mm
sin theta = 0.007068
theta = 0.4049 deg
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2. apply Y = mLR/d
L = Yd/mR = 0.125 * 0.176mm/2*7)
L = 157.14 nm
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3. L = Lo/n
L = 589nm/1.33 = 442.85 nm
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4.thicklness t = L/2n
t = 700nm/(2*1.33)
t = 263.15 nm
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5. apply Y = mLR/d
so d = mLR/Y
d = 1* 633nm*1/0.005
d = 0.1266 mm
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6. apply d sin theta = mL
V = Lf
L = 3e8/6*10^14
L = 500 nm
d = 1* 500nm/sin 15
d = 1.931 um
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