# Light through the corner of a glass block

1. Apr 6, 2005

### T7

Hi,

I am trying to prove algebraically that light cannot pass across the corner of a right-angled block of glass of refractive index n=1.5. Would someone be willing to let me know whether the method I have used below is valid or not?

'i1' denotes the angle of incidence that the light ray makes with the corner of the glass on entry, 'r1' its angle of refraction. 'i2' denotes the angle of incidence the light within the glass makes against the side perpendicular to the side of its entry, and 'r2' denotes the angle of refraction.

If the light ray is to escape, r2 < 90.

Now n = (sin i1)/(sin r1) ... [1]

And (1/n) = (sin i2)/(sin r2) ... [2]

Since i2 is simply made by the refracted ray with the normal of the perpendicular side,

i2 = 90 - r1.

Therefore sin i2 = cos r1.

From [2], sin r1 = sin i1 / n
From [1], cos r1 = sin r2 / n

Given that sin^2 r1 + cos^2 r1 = 1

(sin r2)^2/n^2 + (sin i1)^2/n^2 = 1

Therefore

sin r2 = sqr(n^2 - (sin i1)^2)

But there are no values of i1 (the angle of incidence) such that 0 <= sin r2 < 1 therefore no value of i1 will produce an angle such that 0 <= r < 90.

Not an elegant proof, perhaps, but is it valid? Anyone care to suggest a better one (just using algebra)?

Cheers.

2. Feb 6, 2012

### jdjw

Hi,

I like your answer, but it's a bit hard to follow the layout. From what I can see, it sees mathematically valid, but I think the 'easier' answer is something like this:

Critical angle for glass is:
$$C=\sin^{-1}(\frac{1}{1.5})$$
$$C=41.8^{\circ}$$
Therefore the refracted ray inside the block must meet the edge of the glass to air interface at less than or equal to 41.8 degrees. What is the angle of incidence required for this?

$$1 \sin \theta = 1.5 \sin 41.8$$
$$\theta = \sin^{-1}(1.5 \sin 41.8)$$

This is not possible, since Sine is always between 1 and 0. If we were to decrease the angle of incidence, we will increase the angle at which the light meets the glass to air boundary, so there is no solution that gives a possible answer - the ray must totally internally refract instead.