Light of wavelength `6000 Å` falls on a plane reflecting surface. What are the wavelength and frequency of reflected light. If angle between incident ray and reflected ray is `60^@`, what is the angle of incidence ?

Answer

`6000Å`, `5 xx 10^(14) Hz` ; `30^(@)`

Explanation

As wavelength and frequency remain unchanged on reflection, therefore, `lambda' = lambda = 6000Å`
`v' = v = ( c)/(lambda) = (3 xx 10^(8))/(6000 xx 10^(-10))`
Now, `/_i = /_r = 60^(@)`
As `/_r = /_i :. 2/_i = 60^(@)`
`I = 30^(@)`