You are under water (a = 1.33) shining a light towards the surface at an angle (

Question

You are under water (a = 1.33) shining a light towards the surface at an angle (to the normal) of 49.5 degree. Can your friend, somewhere above the surface, see it? Yes, if he is looking in the right direction No Not enough information is given to determine this Oil (n = 1.48) is floating on water (n = 1.33). A beam of light is incident (from air) on the oil at an angle of 25 degree. What is the refracted angle in the water? A beam of light is incident upon a flat piece of glass (n = 1.5) at an angle of incidence of 45 degrees. Part of the beam is transmitted and part is reflected. What is the angle between the reflected and transmitted rays? 45 degrees 73 degrees 28 degrees None of these. A monochromatic light source emits a wavelength of 600 nm in air. What is its wavelength in glass (n = 1.5)?

Explanation / Answer

3.

using snell's law:

n1*sin i = n2*sin r

r = arcsin (1*(sin 45 deg)/1.5) = 28.12 deg

we know that

angle of incidence with normal = angle of reflected ray with normal

i = f = 45 deg

Now

i + f + A = 180 deg

A = 180 - (45 + 28.12) = 106.88 deg

A = angle between reflected and transmitted ray.

Correct option is none of these.

4.

lambda = c/f

c = speed of light

v = speed of light in glass

v = c/n

So,

wavelength in glass will be

lambda1 = v/f = c/nf = lambda/n

lambda1 = 600/1.5 = 400 nm

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