Question

1. A monochromatic ray of light enters a flat piece of glass (index of refraction = n) from air (index of refraction n = 1). The ray passes through the glass and re-enters the air through the bottom surface of the glass sheet. The angle it makes with the normal when it has re-entered the air is

   	a.	the same as the angle it made with the normal when it was originally in air.
   	b.	the same as the angle it made with the normal when it was in the glass.
   	c.	larger than the angle it made with the normal when it was originally in air.
   	d.	one of the preceding answers, but which one cannot be determined unless n for glass is known.

Answer 1

Using above figure and Snell's law:

Suppose incident angle of light in the air initially is 'i', then

n1*sin i = n2*sin r

n1 = refractive index of air = 1

n2 = refractive index of glass sheet = n

r = refraction angle with the normal = ?

sin r = (1/n)*sin i

Now when light exits the glass sheet, then from above figure it's incident angle will be:

i1 = r, So now using snell's law:

n2*sin i1 = n1*sin r1

i1 = r, from geometry, So (and n2 = n, & n1 = 1)

r1 = refraction angle of light exiting from glass sheet w.r.t. normal = ?

n*sin r = 1*sin r1

sin r1 = n*sin r

sin r1 = n*(1/n)*sin i

sin r1 = sin i

r1 = i

So, The angle light makes with the normal when it has re-entered the air is the same as the angle it made with the normal when it was originally in air.

Correct option is A.

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