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marusya05 [52]
3 years ago
11

When light is incident normally on the interface between two transparent optical media, the intensity of the reflected light is

given by the expression(equation 1)In this equation S1 represents the average magnitude of the Poynting vector in the incident light (the incident intensity), S’1 is the reflected intensity, and n1 and n2 are the refractive indices of the two media. The fraction of incident intensity is 0.0246 or 2.46 precent for 589-nm light.For light normally incident on an interface between vacuum and a transparent medium of index n is given by S2/S1 = 4n/(n + 1)2 (equation 2). (a) Light travels perpendicularly through a diamond slab, surrounded by air, with parallel surfaces of entry and exit. Apply the tramsmission fraction in equation 2 to find the approximate overall transmission through the slab of diamond, as a percentage. Ignore light relected back and forth within the slab.(b) Light travels perpendicularly through a diamond slab, surrounded by air, with parallel surfaces of entry and exit. The intensity of the transmitted light is what fraction of the incident intensity? Include the effects of light reflected back and forth inside the slab.If someone could answer part (a) and (a) I would gladly appreciate it.
Physics
1 answer:
Alex777 [14]3 years ago
8 0

i need to ask sum so oycixgixigdtdidit

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5 0
3 years ago
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Which change would cause the largest increase in a mountain climber’s gravitational potential energy?
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<span>carrying twice the weight and climbing twice as high</span>
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3 years ago
A circular cross section, d = 25 mm, experiences a torque load, T = 25 N·m, and a shear force, V = 85 kN. Calculate the shear st
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using the equation of torsion we Have

\frac{T}{I_{p}}=\frac{\tau }{r}\\\\\therefore \tau =\frac{T}{I_{p}}\times r

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\tau= shear stress at a distance 'r' from the center

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I_{p} = polar moment of inertia of the section

r = radial distance from the center

Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.

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\tau _{max}=\frac{4}{3}\times \frac{85\times 10^{3}}{0.25\times \pi \times (25\times 10^{-3})^{2}}\\\\\therefore \tau _{max}=230.88Mpa\approx 231Mpa

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