Upon reflection, light undergoes a 180degree phase change: always. if the incide
ID: 1515769 • Letter: U
Question
Upon reflection, light undergoes a 180degree phase change: always. if the incident medium has the higher index of refraction. if the Incident medium has the lower index of refraction. whenever the incident angle is less than the critical angle. A beam of polarized light of intensity I_0 passes through a sheet of ideal polarizing material. The polarization axis of the beam and the transmiss on axi the sheet differ by 30degree. What is the intensity of the emerging light? 0.87 I_0 0.75 I_0 0.02 I_0 0.96 I_0 0.15 I_0 What is the surface temperature of a distant star (which emits light as if it were a blakbody) where the peak wavelength is 595 nm? 18.200 K 19.500 K 4870 K 9,700 K 1620 K Violet light (gamma = 402 nm) is incident on a piece of potassium (small letter phi = 2.20 eV). What is the maximum kinetic energy of the ejected photoelectrons? (h = 6.63 times 10^-34 J midot s, c = 3.00 times 10^8 m/s, 1 eV = 1.60 times 10^-19 J, and 1 nm = 10^-9 m) 2.75 eV 0.89 eV 3.09 eV 5.29 eV 1.03 eVExplanation / Answer
24. b
upon reflection light undergoes 180 degrees phase change , when the incident medium has the highere incex of refraction ( total internal reflection i > c)
25. b
from Malus law I= I_0 cos^2 theta
given thet is 30 degrees, I = I_0 cos^230 = 0.75(I_0 )
26. c
from wein's displacement law Lambda max = b/T, b- wein's displacement constant2.8977729*10^-3 mk
given peak wavelength is 595 nm ==> T = b/ lambda = 2.8977729*10^-3/(595*10^-9) =4870.20 k
27. b
photo electric equation E = W +k.e
maximum kinetic energy is , E = hc / lambda , k.e= (hc / lambda) - W
= (6.626*10^-34*3*10^8) /(402*10^-9) -(2.20)= 0.89 eV
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