∙∙(a) Light passes through three parallel slabs of different...

$\bullet$$\bullet$ (a) Light passes through three parallel slabs of different thicknesses and refractive indexes. The light is incident in the first slab and finally refracts into the third slab. Show that the middle slab has no effect on the final direction of the light. That is, show that the direction of the light in the third slab is the same as if the light had passed directly from the first slab into the third slab. (b) Generalize this result to a stack of $N$ slabs. What determines the final direction of the light in the last slab?

Key Concepts

Parallel Interfaces in Layered Media

When multiple slabs are arranged with parallel interfaces, the intermediate layers do not alter the angular direction of light. This is because the deviations induced by entering and leaving each slab exactly counterbalance one another. Thus, the final trajectory of the light ray depends solely on the first and last media, regardless of any intervening layers.

Generalization to Multiple Layers

The principle observed for three slabs can be generalized to a stack of N slabs. When the interfaces are parallel, the refraction effects at all intermediate boundaries cancel out due to Snell’s Law, making the final direction of light determined exclusively by the refractive indices at the first and the final boundaries. The cumulative effect of the intermediate slabs becomes irrelevant, as only the initial and final refractive conditions impact the ultimate trajectory of the ray.

Snell’s Law

Snell’s Law governs the behavior of light as it passes from one medium to another, relating the sine of the angle of incidence to the sine of the angle of refraction via the ratio of the refractive indices. In the context of layered media, it ensures that when light enters and then exits a medium with parallel interfaces, the changes at the entry and exit points cancel out, leaving the final direction of the light determined only by the incident and final refractive indices.

Transcript

00:01 In the first part of this problem, we have to show that the direction of emerging light is independent of the middle slab.

00:08 In order to show this relation, we apply the snail's light different interfaces.

00:13 So at first interface we can write it as n sine theta equals to n.

00:22 Sine theta a.

00:24 We call this equation as equation number one.

00:26 Here n is the refractive index of the top medium.

00:29 Now, where theta is the incidence angle, n .a is the refractive index of the first slide, and theta a is the refructural angle of refraction in first slide.

00:42 So for second interface, we can avoid it as, we can write the signal size, n .a, sine theta a equals to n b, sine theta b.

00:59 We call this equation as equation number two.

01:02 As this nb is the refracto index of second slab and a theta b is the angle of affraction for this refraction similarly we can write the relation for third interfaces and b sine theta b equals to n c sine theta c we call this equation as equation number three now using equation number one and 2 we can write n sine theta equals to n b sine theta b we call this equation as equation number 4.

01:48 Now using equation number 3 and 4 we can write n sine theta equals to nc sine theta c.

02:01 Our solving this question for theta c we can write it is theta c equals to arc sine of n divided by nc and then sine theta.

02:19 This theta c describes the direction of the emerging white and we see that this theta c is independent of the refractive index and the angles of the middle slabs...

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