When a man stands near the edge of an empty drainage ditch of depth 2.80 m, he c

Question

When a man stands near the edge of an empty drainage ditch of depth 2.80 m, he can barely see the boundary between the opposite wall and bottom of the ditch as in Figure (a) shown below. The distance from his eyes to the ground is h = 1.96 m. (Assume theta= 31.2 degree.) What is the horizontal distance d from the man to the edge of the drainage ditch? d = m After the drainage ditch is filled with water as in Figure (b) shown above, what is the maximum distance x the man can stand from the edge and still see the same boundary? x = m A laser beam is incident at an angle of 32.0degrees to the vertical onto a solution of com syrup in water. If the beam is refracted to 18.72 degree to the vertical, what is the index of refraction of the syrup solution? Suppose the light is red, with wavelength 632.8 nm in a vacuum. Find its wavelength in the solution, nm Find its frequency in the solution. Hz Find its speed in the solution.

Explanation / Answer

1)

a) from figure

tan(theta) = d/h

d = h*tan(theta)

= 1.96*tan(31.2)

= 1.187 m

b) let theta_r is the angle of refraction in the air.

Apply Snell's law

sin(theta_i)/sin(theta_r) = n2/n1

sin(theta_r) = sin(theta_i)*n1/n2

= sin(31.2)*1.33/1

= 0.689

theta_r = sin^-1(0.689)

= 43.55 degrees

now use, tan(43.55) = x/h

x = h*tan(43.55)

= 1.96*tan(43.55)

= 1.86 m

9)

a) Let n is the refractive index of solution.

Apply Snell's law

sin(theta_i)/sin(theta_r) = n2/n1

==> sin(32)/sin(18.72) = n/1

==> n = sin(32)/sin(18.72)

= 1.65

b) lamda = lamdao/n

= 632.8/1.65

= 383.2 nm

c) frequency reamins same.

f = c/lamdao

= 3*10^8/(632.8*10^-9)

= 4.74*10^14 hz

d) v = c/n

= 3*10^8/(1.65)

= 1.818*10^8 m/s

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