1. light is observed reflecting from a thin film of oil (index = 1.25) that floa

Question

1. light is observed reflecting from a thin film of oil (index = 1.25) that floats on water (index = 1.33). What is the smallest nonzero thickness of the oil that produces constructive interferencein reflection of 600 nm light?

2. A single slit of size 0.15 mm is illuminated by light wavelength 633 nm. What is the width of the central bright fringe on a screen 2 m away from the slit?

3. A grating having 5000 lines/cm is used with light of wavelength 633 nm. How many total maxima (count central maxima plus all those on either side of the central maxima) are produced?

4. A diffraction grating has 5000 lines/cm. When light of wavelength 700 nm is incident normally on the grating, with what angle to the incident direction are first order maxima produced?

5.Unpolarized light is passes through 2 polarizers in succession. The output intensity is 12.5% of the input intensity. What is the angle between the transmission axes of the 2 polarizers?

Explanation / Answer

Number 1)

For this scenario, apply 2nt = m(wavelength)

2(1.25)(t) = 1(600 X 10-9)

t = 2.4 X 10-7 m (240 nm)

Number 2)

Apply y/L = m(wavelength)/a

y/2 = 1(633 X 10-9)/(1.5 X 10-4)

y = 8.44 X 10-3 m

Central max = 2y = 2(8.44 X 10-3) = .0169 m (1.69 cm)

Number 3)

Apply 1 = m(wavelength)/d

1 = (m)(633 X 10-9)(500000)

m = 3.16

Take the whole number and multiply by 2. Then add the central max

3 + 3 + 1 = 7 maxima

Number 4)

sin(angle) = m(wavelength)/d

sin(angle) = 1(700 X 10-9)(500000)

angle = 20.5o (In case your book does not use "small angle approximations, that angle changes to 19.3o by using tangent instead of sin)

Number 5)

Apply I/Io = .5cos2(angle)

.125 = .5cos2(angle)

angle = 60o

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