Wave Optics and Introduction to Quantum Physics What is the minimum width of a s
ID: 1608627 • Letter: W
Question
Wave Optics and Introduction to Quantum Physics
What is the minimum width of a single slit (in microns) that will produce a first minimum for a wavelength 682 nm? Suppose you have a lens system that is to be used primarily for 685 nm red light. What is the second thinnest coating (in nanometers) of fluorite (magnesium fluoride) that would be non-reflective for this wavelength? Assume that fluorite has an index of refraction of 1.38. What is the index of refraction of a material for which the wavelength of light is 0.75 times its value in a vacuum?Explanation / Answer
Given
Question -2
wavelength Lambda = 682 nm
the condition for minimum fringe on the screen in single slit is
a sin theta = m*lambda
a is slit width , m = 1
a = lambda/sin theta
a = lambda /1
a = 682 nm
question 3
condition for minimum thickness is 2*mue*t = (m + 0.5)Lambda
t = (m+0.5)lambda /(2*mue)
where mue is refractive index and t is thickness, lambda is wavelength
t = (0+0.5)685*10^-9/(2*1.38) m
t = 1.241*10^-7 m
t = 124.1 nm
Question 4
refractive index n = c/V = L1*f/L2*f
given L2 = 0.75*L1
n = L1 /0.75L1
n = 1/0.75
n = 1.333
the refractive index of the material is n= 1.333
( the frequency of the light is same in any medium)
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